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Unformatted text preview: 5E04(pp 222231) 1/17/06 2:40 PM Page 222 CHAPTER 4
Scientists have tried to
explain how rainbows are
formed since the time of
Aristotle. In the project on
page 232, you will be able
to use the principles of
differential calculus to
explain the formation,
location, and colors of
the rainbow. A pplications of Differentiation 5E04(pp 222231) 1/17/06 2:40 PM Page 223 We have already investigated some of the applications of
derivatives, but now that we know the differentiation rules
we are in a better position to pursue the applications of differentiation in greater depth. Here we learn how derivatives
affect the shape of a graph of a function and, in particular, how they help us locate maximum and minimum values of functions. Many practical problems require us to minimize a cost or maximize an area or somehow ﬁnd
the best possible outcome of a situation. In particular, we will be able to investigate
the optimal shape of a can and to explain the location of rainbows in the sky.  4.1 Maximum and Minimum Values
Some of the most important applications of differential calculus are optimization problems, in which we are required to ﬁnd the optimal (best) way of doing something. Here are
examples of such problems that we will solve in this chapter:
■ ■ ■ ■ What is the shape of a can that minimizes manufacturing costs?
What is the maximum acceleration of a space shuttle? (This is an important
question to the astronauts who have to withstand the effects of acceleration.)
What is the radius of a contracted windpipe that expels air most rapidly during
a cough?
At what angle should blood vessels branch so as to minimize the energy expended
by the heart in pumping blood? These problems can be reduced to ﬁnding the maximum or minimum values of a function.
Let’s ﬁrst explain exactly what we mean by maximum and minimum values.
1 Definition A function f has an absolute maximum (or global maximum) at c
f x for all x in D, where D is the domain of f . The number f c is called
if f c
the maximum value of f on D. Similarly, f has an absolute minimum at c if
fc
f x for all x in D and the number f c is called the minimum value of f
on D. The maximum and minimum values of f are called the extreme values of f . Figure 1 shows the graph of a function f with absolute maximum at d and absolute
minimum at a. Note that d, f d is the highest point on the graph and a, f a is the lowest point.
y f(d)
FIGURE 1 Minimum value f(a),
maximum value f(d) f(a)
a 0 b c d e x 223 5E04(pp 222231) 224 ❙❙❙❙ 1/17/06 2:40 PM Page 224 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION In Figure 1, if we consider only values of x near b [for instance, if we restrict our attention to the interval a, c ], then f b is the largest of those values of f x and is called a
local maximum value of f . Likewise, f c is called a local minimum value of f because
fc
f x for x near c [in the interval b, d , for instance]. The function f also has a local
minimum at e. In general, we have the following deﬁnition.
2 Definition A function f has a local maximum (or relative maximum) at c if
fc
f x when x is near c. [This means that f c
f x for all x in some open
interval containing c.] Similarly, f has a local minimum at c if f c
f x when
x is near c. EXAMPLE 1 The function f x
cos x takes on its (local and absolute) maximum value
of 1 inﬁnitely many times, since cos 2 n
1 for any integer n and 1 cos x 1 for
all x. Likewise, cos 2 n 1
1 is its minimum value, where n is any integer. y y=≈ 0 x 2, then f x
f 0 because x 2 0 for all x. Therefore, f 0
0
is the absolute (and local) minimum value of f . This corresponds to the fact that the
origin is the lowest point on the parabola y x 2. (See Figure 2.) However, there is no
highest point on the parabola and so this function has no maximum value.
EXAMPLE 2 If f x x FIGURE 2 Minimum value 0, no maximum x 3, shown in Figure 3, we see that this
function has neither an absolute maximum value nor an absolute minimum value. In fact,
it has no local extreme values either.
E XAMPLE 3 From the graph of the function f x y y=˛ E XAMPLE 4 The graph of the function
0 x fx 3x 4 16 x 3 18 x 2 1 x 4 is shown in Figure 4. You can see that f 1
5 is a local maximum, whereas the
absolute maximum is f 1
37. (This absolute maximum is not a local maximum
because it occurs at an endpoint.) Also, f 0
0 is a local minimum and f 3
27
is both a local and an absolute minimum. Note that f has neither a local nor an absolute
maximum at x 4. FIGURE 3 No minimum, no maximum y
(_1, 37) y=3x$16˛+18≈ (1, 5)
_1 1 2 3 4 5 x (3, _27) FIGURE 4 We have seen that some functions have extreme values, whereas others do not. The
following theorem gives conditions under which a function is guaranteed to possess
extreme values. 5E04(pp 222231) 1/17/06 2:40 PM Page 225 S ECTION 4.1 MAXIMUM AND MINIMUM VALUES ❙❙❙❙ 225 3 The Extreme Value Theorem If f is continuous on a closed interval a, b , then f
attains an absolute maximum value f c and an absolute minimum value f d at
some numbers c and d in a, b . The Extreme Value Theorem is illustrated in Figure 5. Note that an extreme value can
be taken on more than once. Although the Extreme Value Theorem is intuitively very plausible, it is difﬁcult to prove and so we omit the proof.
y FIGURE 5 0 y a c db 0 x y a c d=b 0 x a c¡ d c™ b x Figures 6 and 7 show that a function need not possess extreme values if either hypothesis (continuity or closed interval) is omitted from the Extreme Value Theorem.
y y 3 1 0 1 2 x 0 2 x FIGURE 6 y
{c, f (c)} {d, f (d)}
0 FIGURE 8 c d x FIGURE 7 This function has minimum value
f(2)=0, but no maximum value. This continuous function g has
no maximum or minimum. The function f whose graph is shown in Figure 6 is deﬁned on the closed interval [0, 2]
but has no maximum value. (Notice that the range of f is [0, 3). The function takes on values arbitrarily close to 3, but never actually attains the value 3.) This does not contradict
the Extreme Value Theorem because f is not continuous. [Nonetheless, a discontinuous
function could have maximum and minimum values. See Exercise 13(b).]
The function t shown in Figure 7 is continuous on the open interval (0, 2) but has neither a maximum nor a minimum value. [The range of t is 1, . The function takes on
arbitrarily large values.] This does not contradict the Extreme Value Theorem because the
interval (0, 2) is not closed.
The Extreme Value Theorem says that a continuous function on a closed interval has a
maximum value and a minimum value, but it does not tell us how to ﬁnd these extreme
values. We start by looking for local extreme values.
Figure 8 shows the graph of a function f with a local maximum at c and a local minimum
at d. It appears that at the maximum and minimum points the tangent lines are horizontal
and therefore each has slope 0. We know that the derivative is the slope of the tangent line,
so it appears that f c
0 and f d
0. The following theorem says that this is always
true for differentiable functions. 5E04(pp 222231) 226 ❙❙❙❙ 1/17/06 2:40 PM Page 226 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION  Fermat’s Theorem is named after Pierre
Fermat (1601–1665), a French lawyer who took
up mathematics as a hobby. Despite his amateur
status, Fermat was one of the two inventors of
analytic geometry (Descartes was the other). His
methods for ﬁnding tangents to curves and maximum and minimum values (before the invention
of limits and derivatives) made him a forerunner
of Newton in the creation of differential calculus. 4 Fermat’s Theorem If f has a local maximum or minimum at c, and if f c exists, then f c 0. Proof Suppose, for the sake of deﬁniteness, that f has a local maximum at c. Then,
according to Deﬁnition 2, f c
f x if x is sufﬁciently close to c. This implies that if
h is sufﬁciently close to 0, with h being positive or negative, then fc fc h and therefore
fc 5 h fc 0 We can divide both sides of an inequality by a positive number. Thus, if h
sufﬁciently small, we have
fc h
h fc 0 and h is 0 Taking the righthand limit of both sides of this inequality (using Theorem 2.3.2), we get
fc lim h
h hl0 fc lim 0 0 hl0 But since f c exists, we have
fc lim fc h
h hl0 fc lim fc h
h hl0 fc and so we have shown that f c
0.
If h 0, then the direction of the inequality (5) is reversed when we divide by h :
fc h
h fc 0 h 0 fc h
h So, taking the lefthand limit, we have
fc hl0 fc h
h fc lim hl0 fc 0 We have shown that f c
0 and also that f c
0. Since both of these inequalities
must be true, the only possibility is that f c
0.
We have proved Fermat’s Theorem for the case of a local maximum. The case of a
local minimum can be proved in a similar manner, or we could use Exercise 70 to
deduce it from the case we have just proved (see Exercise 71). y y=˛ 0 lim x The following examples caution us against reading too much into Fermat’s Theorem.
0 and solving for x.
We can’t expect to locate extreme values simply by setting f x
FIGURE 9 If ƒ=˛, then f ª(0)=0 but ƒ
has no maximum or minimum. x 3, then f x
3x 2, so f 0
0. But f has no maximum or minimum at 0, as you can see from its graph in Figure 9. (Or observe that x 3 0 for x 0
but x 3 0 for x 0.) The fact that f 0
0 simply means that the curve y x 3 has a
EXAMPLE 5 If f x 5E04(pp 222231) 1/17/06 2:40 PM Page 227 S ECTION 4.1 MAXIMUM AND MINIMUM VALUES ❙❙❙❙ 227 horizontal tangent at 0, 0 . Instead of having a maximum or minimum at 0, 0 , the
curve crosses its horizontal tangent there.
y EXAMPLE 6 The function f x
x has its (local and absolute) minimum value at 0, but
that value can’t be found by setting f x
0 because, as was shown in Example 6 in
Section 3.2, f 0 does not exist. (See Figure 10.) y= x
0  x F IGURE 10 If ƒ= x , then f(0)=0 is a
minimum value, but f ª(0) does not exist. WARNING Examples 5 and 6 show that we must be careful when using Fermat’s
Theorem. Example 5 demonstrates that even when f c
0 there need not be a maximum
or minimum at c. (In other words, the converse of Fermat’s Theorem is false in general.)
Furthermore, there may be an extreme value even when f c does not exist (as in
Example 6).
Fermat’s Theorem does suggest that we should at least start looking for extreme values
of f at the numbers c where f c
0 or where f c does not exist. Such numbers are
given a special name.
■ 6 Definition A critical number of a function f is a number c in the domain of f
such that either f c
0 or f c does not exist.  Figure 11 shows a graph of the function f
in Example 7. It supports our answer because
there is a horizontal tangent when x 1.5 and
a vertical tangent when x 0. x3 5 4 EXAMPLE 7 Find the critical numbers of f x
SOLUTION The Product Rule gives fx 3
5 x 25 4 x3 5 x 1 3.5 34 _0.5 5 _2 FIGURE 11 x. x
5x 2 5x
5 34 x
5x 2 5 x3 5 12 8 x
5x 2 5 [The same result could be obtained by ﬁrst writing f x
4 x 3 5 x 8 5.] Therefore,
3
fx
0 if 12 8 x 0, that is, x 2 , and f x does not exist when x 0. Thus, the
critical numbers are 3 and 0.
2
In terms of critical numbers, Fermat’s Theorem can be rephrased as follows (compare
Deﬁnition 6 with Theorem 4):
7 If f has a local maximum or minimum at c, then c is a critical number of f. To ﬁnd an absolute maximum or minimum of a continuous function on a closed interval,
we note that either it is local [in which case it occurs at a critical number by (7)] or it occurs
at an endpoint of the interval. Thus, the following threestep procedure always works.
The Closed Interval Method To ﬁnd the absolute maximum and minimum values of a
continuous function f on a closed interval a, b :
1. Find the values of f at the critical numbers of f in a, b .
2. Find the values of f at the endpoints of the interval.
3. The largest of the values from Steps 1 and 2 is the absolute maximum value;
the smallest of these values is the absolute minimum value. 5E04(pp 222231) 228 ❙❙❙❙ 1/17/06 2:40 PM Page 228 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION EXAMPLE 8 Find the absolute maximum and minimum values of the function x3 fx SOLUTION Since f is continuous on 3x 2 [ 1
2 y=˛3≈+1
(4, 17) x3 fx 20 x 4 , 4], we can use the Closed Interval Method: fx y 1
2 1 3x 2 3x 2 1 6x 3x x 2 Since f x exists for all x, the only critical numbers of f occur when f x
0, that is,
x 0 or x 2. Notice that each of these critical numbers lies in the interval ( 1 , 4).
2
The values of f at these critical numbers are 15 f0 1 f2 3 10 The values of f at the endpoints of the interval are 5
1
_1 0
_5 f( 2
3 1
2 ) 1
8 f4 17 x 4 Comparing these four numbers, we see that the absolute maximum value is f 4
17
and the absolute minimum value is f 2
3.
Note that in this example the absolute maximum occurs at an endpoint, whereas the
absolute minimum occurs at a critical number. The graph of f is sketched in Figure 12. (2, _3) FIGURE 12 If you have a graphing calculator or a computer with graphing software, it is possible
to estimate maximum and minimum values very easily. But, as the next example shows,
calculus is needed to ﬁnd the exact values.
EXAMPLE 9 (a) Use a graphing device to estimate the absolute minimum and maximum values of
the function f x
x 2 sin x, 0 x 2 .
(b) Use calculus to ﬁnd the exact minimum and maximum values.
SOLUTION
8 0
_1 FIGURE 13 2π (a) Figure 13 shows a graph of f in the viewing rectangle 0, 2 by 1, 8 . By moving the cursor close to the maximum point, we see that the ycoordinates don’t change
very much in the vicinity of the maximum. The absolute maximum value is about 6.97
and it occurs when x 5.2. Similarly, by moving the cursor close to the minimum point,
we see that the absolute minimum value is about 0.68 and it occurs when x 1.0. It is
possible to get more accurate estimates by zooming in toward the maximum and minimum points, but instead let’s use calculus.
(b) The function f x
x 2 sin x is continuous on 0, 2 . Since f x
1 2 cos x ,
we have f x
0 when cos x 1 and this occurs when x
3 or 5 3. The values
2
of f at these critical points are
f and 3 f5 3 3
5
3 2 sin 3 2 sin 5
3 0.684853 s3 3
5
3 s3 6.968039 5E04(pp 222231) 1/17/06 2:40 PM Page 229 S ECTION 4.1 MAXIMUM AND MINIMUM VALUES ❙❙❙❙ 229 The values of f at the endpoints are
f0 0 and f2 2 6.28 Comparing these four numbers and using the Closed Interval Method, we see that the
absolute minimum value is f
3
3 s3 and the absolute maximum value is
f5 3
5 3 s3. The values from part (a) serve as a check on our work.
EXAMPLE 10 The Hubble Space Telescope was deployed on April 24, 1990, by the space
shuttle Discovery. A model for the velocity of the shuttle during this mission, from liftoff
at t 0 until the solid rocket boosters were jettisoned at t 126 s, is given by vt 0.001302 t 3 0.09029 t 2 23.61t 3.083 (in feet per second). Using this model, estimate the absolute maximum and minimum
values of the acceleration of the shuttle between liftoff and the jettisoning of the boosters.
SOLUTION We are asked for the extreme values not of the given velocity function, but
rather of the acceleration function. So we ﬁrst need to differentiate to ﬁnd the acceleration: at vt d
0.001302 t 3
dt 0.003906 t 2 0.09029 t 2 0.18058 t 23.61t 3.083 23.61 We now apply the Closed Interval Method to the continuous function a on the interval
0 t 126. Its derivative is
at 0.007812 t The only critical number occurs when a t
t1 0.18058 0: 0.18058
0.007812 23.12 Evaluating a t at the critical number and at the endpoints, we have
a0 23.61 a t1 21.52 a 126 62.87 So the maximum acceleration is about 62.87 ft s2 and the minimum acceleration is
about 21.52 ft s2.  4.1 Exercises 1. Explain the difference between an absolute minimum and a local minimum.
2. Suppose f is a continuous function deﬁned on a closed interval a, b . (a) What theorem guarantees the existence of an absolute maximum value and an absolute minimum value for f ?
(b) What steps would you take to ﬁnd those maximum and
minimum values? 5E04(pp 222231) ❙❙❙❙ 230 1/17/06 2:41 PM Page 230 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 3–4  For each of the numbers a, b, c, d, e, r, s, and t, state
whether the function whose graph is shown has an absolute maximum or minimum, a local maximum or minimum, or neither a
maximum nor a minimum. 3. 9. Absolute maximum at 5, absolute minimum at 2, local maximum at 3, local minima at 2 and 4
10. f has no local maximum or minimum, but 2 and 4 are critical numbers y ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 11. (a) Sketch the graph of a function that has a local maximum at 2 and is differentiable at 2.
(b) Sketch the graph of a function that has a local maximum
at 2 and is continuous but not differentiable at 2.
(c) Sketch the graph of a function that has a local maximum
at 2 and is not continuous at 2.
0 4. a b c d e r s x t 12. (a) Sketch the graph of a function on [ 1, 2] that has an absolute maximum but no local maximum.
(b) Sketch the graph of a function on [ 1, 2] that has a local
maximum but no absolute maximum. y 13. (a) Sketch the graph of a function on [ 1, 2] that has an absolute maximum but no absolute minimum.
(b) Sketch the graph of a function on [ 1, 2] that is discontinuous but has both an absolute maximum and an absolute
minimum. t
a 0 ■ ■ ■ b cd ■ ■ e ■ ■ r ■ s ■ x ■ ■ one local minimum, and no absolute minimum.
(b) Sketch the graph of a function that has three local minima,
two local maxima, and seven critical numbers. ■ 5–6  Use the graph to state the absolute and local maximum and
minimum values of the function. 5. 14. (a) Sketch the graph of a function that has two local maxima, y 15–30  Sketch the graph of f by hand and use your sketch to
ﬁnd the absolute and local maximum and minimum values of f .
(Use the graphs and transformations of Sections 1.2 and 1.3.) 15. f x 8 16. f x y=ƒ 3 17. f x
1
0 18. f x
19. f x x 1 20. f x
6. y 21. f x y=ƒ 3x, x 1 2 x, x 5 2 x 2 2 x 2 2 x 2 2 x 2 x, 0
x, 0
x, 0
x, 0
2 x, 3 x 2 2 22. f x 0 ■ ■ 7–10 ■ ■ ■ ■ ■ ■ ■ ■ Sketch the graph of a function f that is continuous on
[1, 5] and has the given properties.
 7. Absolute minimum at 2, absolute maximum at 3, ■ 0 t 1 1 t, 0 t 1 sin , 2 26. f
■ 1 t, 25. f x 1, 24. f t
1 x 23. f t
1 1 tan , 27. f x 1 sx 28. f x 1 x3 local maximum at 2, local minimum at 4 x 5 2
4 2 29. f x 1x
2x 4 if 0
if 2 30. f x x2
2 if 1 x 0
if 0 x 1 local minimum at 4
8. Absolute minimum at 1, absolute maximum at 5, 2 ■ ■ ■ x2
■ ■ x
x ■ 2
3 ■ ■ ■ ■ ■ ■ 5E04(pp 222231) 1/17/06 2:41 PM Page 231 S ECTION 4.1 MAXIMUM AND MINIMUM VALUES 31– 44  31. f x 5x 33. f x x3 3x2 4 3 35. s t 3t 37. t x 2x 39. t t 23 5t 41. F x
43. f
■ 6t 45–56 2 x4 5 x 4 3
sx 2 44. t sin
■ st 1 42. G x
2 ■ x ■ ■ 4 ■ ■ V 1
z 13 40. t t
2 x z
z2 1 kg of water at a temperature T is given approximately by the
formula x x 38. t x
53 2 x2 x3 36. f z 3
t x 34. f x 24 x 2 cos
■ 32. f x 4x 4t 3 1 x 46. f x x 3 47. f x 2x 3 48. f x x3 6x 2 9x 49. f x x4 2x 2 3, 50. f x
51. f x x 12 x
3x ■ ■ 2 1 52. f x x2
x2 4
,
4 53. f t t s4 54. f t 3
st 8 t, 55. f x sin x 56. f x x ■ xa 1 on mission STS49, the purpose of which was to install a new
perigee kick motor in an Intelsat communications satellite. The
table gives the velocity data for the shuttle between liftoff and
the jettisoning of the solid rocket boosters. 1, 2
0, 8
0, 3
Event ,
■ ■ x b, 0 ■ ■ ■ ■ ■ ■ 1. x ; 58. Use a graph to estimate the critical numbers of
x3 fx ; 59–62 3x 2 2 correct to one decimal place. Time (s) Velocity (ft s) Launch
Begin roll maneuver
End roll maneuver
Throttle to 89%
Throttle to 67%
Throttle to 104%
Maximum dynamic pressure
Solid rocket booster separation 0
10
15
20
32
59
62
125 0
185
319
447
742
1325
1445
4151  (a) Use a graph to estimate the absolute maximum and minimum
values of the function to two decimal places.
(b) Use calculus to ﬁnd the exact maximum and minimum values.
59. f x x3 60. f x x 4 61. f x x sx 62. f x
■ 99.33 where t is measured in years since midyear 1984, so
0 t 10, and I t is measured in 1987 dollars and scaled
such that I 3
100. Estimate the times when food was
cheapest and most expensive during the period 1984–1994. 57. If a and b are positive numbers, ﬁnd the maximum value of fx 0.6270 t 0.06561t 3 ; 66. On May 7, 1992, the space shuttle Endeavour was launched 2 cos x, ■ 0.001438t 4 0.4598t 2 It 4, 4 t2, 0.00009045t 5 It 2, 3 cos x, ■ 1, 4 2, 0, 2 , x2 ■ “basket” of foods) between 1984 and 1994 is given by the
function 2, 3 1, 1, 2 1,
x 65. A model for the foodprice index (the price of a representative 12 x 3 cos where is a positive constant called the coefﬁcient of friction
and where 0
2. Show that F is minimized when
.
tan 0, 3 3x 2 sin 0, 3 5,
1, W F Find the absolute maximum and absolute minimum
values of f on the given interval.
3x 2 0.0000679T 3 by a force acting along a rope attached to the object. If the rope
makes an angle with the plane, then the magnitude of the
force is t ■ 0.0085043T 2 64. An object with weight W is dragged along a horizontal plane  45. f x 0.06426 T Find the temperature at which water has its maximum density. 23 x 999.87 tan
■ 231 63. Between 0 C and 30 C, the volume V (in cubic centimeters) of Find the critical numbers of the function.
2 ❙❙❙❙ ■ 8x
3x 3 3x 2 x
x, 3 0 x 2 x2 cos x
■ 1,
3 (a) Use a graphing calculator or computer to ﬁnd the cubic
polynomial that best models the velocity of the shuttle for
the time interval t
0, 125 . Then graph this polynomial.
(b) Find a model for the acceleration of the shuttle and use it to
estimate the maximum and minimum values of the acceleration during the ﬁrst 125 seconds. 2
■ sin x ,
■ ■ 0 x
■ 67. When a foreign object lodged in the trachea (windpipe) forces 2
■ ■ ■ ■ ■ a person to cough, the diaphragm thrusts upward causing an
increase in pressure in the lungs. This is accompanied by a 5E04(pp 232241) 232 ❙❙❙❙ 1/17/06 2:37 PM Page 232 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION contraction of the trachea, making a narrower channel for the
expelled air to ﬂow through. For a given amount of air to
escape in a ﬁxed time, it must move faster through the narrower channel than the wider one. The greater the velocity
of the airstream, the greater the force on the foreign object.
X rays show that the radius of the circular tracheal tube
contracts to about twothirds of its normal radius during a
cough. According to a mathematical model of coughing, the
velocity v of the airstream is related to the radius r of the
trachea by the equation
vr k r0 rr 1
20 2 r r 68. Show that 5 is a critical number of the function tx 2 x 5 3 but t does not have a local extreme value at 5.
69. Prove that the function fx x 101 x 51 x 1 has neither a local maximum nor a local minimum.
70. If f has a minimum value at c, show that the function tx f x has a maximum value at c. 71. Prove Fermat’s Theorem for the case in which f has a local r0 minimum at c. where k is a constant and r0 is the normal radius of the trachea.
The restriction on r is due to the fact that the tracheal wall
stiffens under pressure and a contraction greater than 1 r0 is
2
prevented (otherwise the person would suffocate).
(a) Determine the value of r in the interval 1 r0 , r0 at which v
2
has an absolute maximum. How does this compare with
experimental evidence?
(b) What is the absolute maximum value of v on the interval?
(c) Sketch the graph of v on the interval 0, r0 . [ 72. A cubic function is a polynomial of degree 3; that is, it has the form
fx ax 3 bx 2 cx d where a 0.
(a) Show that a cubic function can have two, one, or no critical
number(s). Give examples and sketches to illustrate the
three possibilities.
(b) How many local extreme values can a cubic function have? APPLIED PROJECT
The Calculus of Rainbows
Rainbows are created when raindrops scatter sunlight. They have fascinated mankind since
ancient times and have inspired attempts at scientiﬁc explanation since the time of Aristotle. In
this project we use the ideas of Descartes and Newton to explain the shape, location, and colors
of rainbows.
åA
from
Sun 1. The ﬁgure shows a ray of sunlight entering a spherical raindrop at A. Some of the light is
∫ B ∫ O D(å ) ∫ ∫ å
to
observer C reﬂected, but the line AB shows the path of the part that enters the drop. Notice that the light
is refracted toward the normal line AO and in fact Snell’s Law says that sin
k sin ,
where is the angle of incidence, is the angle of refraction, and k 4 is the index of
3
refraction for water. At B some of the light passes through the drop and is refracted into the
air, but the line BC shows the part that is reﬂected. (The angle of incidence equals the angle
of reﬂection.) When the ray reaches C, part of it is reﬂected, but for the time being we are
more interested in the part that leaves the raindrop at C. (Notice that it is refracted away
from the normal line.) The angle of deviation D
is the amount of clockwise rotation that
the ray has undergone during this threestage process. Thus Formation of the primary rainbow D 2 2 4 Show that the minimum value of the deviation is D
138 and occurs when
59.4 .
The signiﬁcance of the minimum deviation is that when
59.4 we have D
0, so
D
0. This means that many rays with
59.4 become deviated by approximately
the same amount. It is the concentration of rays coming from near the direction of minimum 5E04(pp 232241) 1/17/06 2:37 PM Page 233 A PPLIED PROJECT THE CALCULUS OF RAINBOWS ❙❙❙❙ 233 deviation that creates the brightness of the primary rainbow. The following ﬁgure shows
that the angle of elevation from the observer up to the highest point on the rainbow is
180
138
42 . (This angle is called the rainbow angle.)
rays from Sun 138°
rays from Sun 42° observer
2. Problem 1 explains the location of the primary rainbow, but how do we explain the colors? Sunlight comprises a range of wavelengths, from the red range through orange, yellow,
green, blue, indigo, and violet. As Newton discovered in his prism experiments of 1666, the
index of refraction is different for each color. (The effect is called dispersion.) For red light
the refractive index is k 1.3318 whereas for violet light it is k 1.3435. By repeating the
calculation of Problem 1 for these values of k, show that the rainbow angle is about 42.3 for
the red bow and 40.6 for the violet bow. So the rainbow really consists of seven individual
bows corresponding to the seven colors.
3. Perhaps you have seen a fainter secondary rainbow above the primary bow. That results from C
∫ D the part of a ray that enters a raindrop and is refracted at A, reﬂected twice (at B and C ), and
refracted as it leaves the drop at D (see the ﬁgure). This time the deviation angle D
is the
total amount of counterclockwise rotation that the ray undergoes in this fourstage process.
Show that
D
2
6
2 ∫ ∫ å
to
observer ∫
from
Sun ∫
å ∫ and D has a minimum value when B cos A Formation of the secondary rainbow k2 1
8 Taking k 4 , show that the minimum deviation is about 129 and so the rainbow angle for
3
the secondary rainbow is about 51 , as shown in the ﬁgure. 42° 51° 4. Show that the colors in the secondary rainbow appear in the opposite order from those in the primary rainbow. 5E04(pp 232241) 234 ❙❙❙❙ 1/17/06 2:37 PM Page 234 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION  4.2 The Mean Value Theorem
We will see that many of the results of this chapter depend on one central fact, which is
called the Mean Value Theorem. But to arrive at the Mean Value Theorem we ﬁrst need
the following result.
Rolle’s Theorem Let f be a function that satisﬁes the following three hypotheses:  Rolle’s Theorem was ﬁrst published in
1691 by the French mathematician Michel Rolle
(1652–1719) in a book entitled Méthode pour
résoudre les égalitéz. Later, however, he became
a vocal critic of the methods of his day and
attacked calculus as being a “collection of
ingenious fallacies.” 1. f is continuous on the closed interval a, b .
2. f is differentiable on the open interval a, b .
3. f a fb
Then there is a number c in a, b such that f c 0. Before giving the proof let’s take a look at the graphs of some typical functions that satisfy the three hypotheses. Figure 1 shows the graphs of four such functions. In each case
it appears that there is at least one point c, f c on the graph where the tangent is horizontal and therefore f c
0. Thus, Rolle’s Theorem is plausible.
y 0 y a c¡ c™ b x 0 y y a (a) c b x (b) 0 a c¡ c™ b x 0 a (c) c b x (d) FIGURE 1
 Take cases Proof There are three cases:
CASE I f ( x) ■ Then f x k, a constant
0, so the number c can be taken to be any number in a, b . CASE II
f ( x) > f ( a) for some x in ( a, b) [as in Figure 1(b) or (c)]
By the Extreme Value Theorem (which we can apply by hypothesis 1), f has a maximum value somewhere in a, b . Since f a
f b , it must attain this maximum value at
a number c in the open interval a, b . Then f has a local maximum at c and, by hypothesis 2, f is differentiable at c. Therefore, f c
0 by Fermat’s Theorem.
■ CASE III f ( x) < f ( a) for some x in ( a, b) [as in Figure 1(c) or (d)]
By the Extreme Value Theorem, f has a minimum value in a, b and, since f a
f b , it
attains this minimum value at a number c in a, b . Again f c
0 by Fermat’s Theorem.
■ EXAMPLE 1 Let’s apply Rolle’s Theorem to the position function s
f t of a moving
object. If the object is in the same place at two different instants t a and t b, then
fa
f b . Rolle’s Theorem says that there is some instant of time t c between a
and b when f c
0; that is, the velocity is 0. (In particular, you can see that this is
true when a ball is thrown directly upward.) 5E04(pp 232241) 1/17/06 2:37 PM Page 235 S ECTION 4.2 THE MEAN VALUE THEOREM  Figure 2 shows a graph of the function
fx
x 3 x 1 discussed in Example 2.
Rolle’s Theorem shows that, no matter how much
we enlarge the viewing rectangle, we can never
ﬁnd a second xintercept.
3 _2 2 EXAMPLE 2 Prove that the equation x 3 FIGURE 2 235 0 has exactly one real root. 1 SOLUTION First we use the Intermediate Value Theorem (2.5.10) to show that a root exists.
Let f x
x 3 x 1. Then f 0
1 0 and f 1
1 0. Since f is a polynomial, it is continuous, so the Intermediate Value Theorem states that there is a number c
between 0 and 1 such that f c
0. Thus, the given equation has a root.
To show that the equation has no other real root, we use Rolle’s Theorem and argue
by contradiction. Suppose that it had two roots a and b. Then f a
0 f b and, since
f is a polynomial, it is differentiable on a, b and continuous on a, b . Thus, by Rolle’s
Theorem, there is a number c between a and b such that f c
0. But 3x 2 fx
_3 x ❙❙❙❙ 1 1 for all x (since x 2 0 ) so f x can never be 0. This gives a contradiction. Therefore, the equation can’t have two real roots.
Our main use of Rolle’s Theorem is in proving the following important theorem, which
was ﬁrst stated by another French mathematician, JosephLouis Lagrange.
The Mean Value Theorem Let f be a function that satisﬁes the following hypotheses:
1. f is continuous on the closed interval a, b .
2. f is differentiable on the open interval a, b .  The Mean Value Theorem is an example of
what is called an existence theorem. Like the
Intermediate Value Theorem, the Extreme Value
Theorem, and Rolle’s Theorem, it guarantees that
there exists a number with a certain property,
but it doesn’t tell us how to ﬁnd the number. Then there is a number c in a, b such that
fb
b fc 1 fa
a or, equivalently,
fb 2 fa fcb a Before proving this theorem, we can see that it is reasonable by interpreting it geometrically. Figures 3 and 4 show the points A a, f a and B b, f b on the graphs of two differentiable functions. The slope of the secant line AB is
mAB 3 y fb
b fa
a y P¡ P { c, f(c)} B P™ A A{ a, f(a)}
B { b, f(b)}
0 a FIGURE 3 c b x 0 a FIGURE 4 c¡ c™ b x 5E04(pp 232241) 236 ❙❙❙❙ 1/17/06 2:37 PM Page 236 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION which is the same expression as on the right side of Equation 1. Since f c is the slope of
the tangent line at the point c, f c , the Mean Value Theorem, in the form given by
Equation 1, says that there is at least one point P c, f c on the graph where the slope of
the tangent line is the same as the slope of the secant line AB. In other words, there is a
point P where the tangent line is parallel to the secant line AB.
y Proof We apply Rolle’s Theorem to a new function h deﬁned as the difference between
h(x) A y=ƒ f and the function whose graph is the secant line AB. Using Equation 3, we see that the
equation of the line AB can be written as ƒ y 0 x
f(a)+ fa fb
b fa
a x a y B fa fb
b fa
a x a fa
a x x f(b)f(a)
(xa)
ba F IGURE 5 or as
So, as shown in Figure 5,
hx 4 fx fb
b fa a First we must verify that h satisﬁes the three hypotheses of Rolle’s Theorem.
1. The function h is continuous on a, b because it is the sum of f and a ﬁrstdegree polynomial, both of which are continuous.
2. The function h is differentiable on a, b because both f and the ﬁrstdegree polynomial are differentiable. In fact, we can compute h directly from Equation 4:
 The Mean Value Theorem was ﬁrst formulated by JosephLouis Lagrange (1736–1813),
born in Italy of a French father and an Italian
mother. He was a child prodigy and became a
professor in Turin at the tender age of 19.
Lagrange made great contributions to number
theory, theory of functions, theory of equations,
and analytical and celestial mechanics. In particular, he applied calculus to the analysis of the
stability of the solar system. At the invitation of
Frederick the Great, he succeeded Euler at the
Berlin Academy and, when Frederick died,
Lagrange accepted King Louis XVI’s invitation to
Paris, where he was given apartments in the
Louvre. Despite all the trappings of luxury and
fame, he was a kind and quiet man, living only
for science. hx
(Note that f a and f b fb
b fx
fa fa
a a are constants.) b ha fa fa fb
b fa
a a a hb fb fa fb
b fa
a b a fb 3. fa fb fa Therefore, h a 0 0 hb. Since h satisﬁes the hypotheses of Rolle’s Theorem, that theorem says there is a number c in a, b such that h c
0. Therefore
0
and so hc
fc fc
fb
b fb
b fa
a fa
a EXAMPLE 3 To illustrate the Mean Value Theorem with a speciﬁc function, let’s consider fx x3 x, a 0, b 2. Since f is a polynomial, it is continuous and differentiable 5E04(pp 232241) 1/17/06 2:37 PM Page 237 S ECTION 4.2 THE MEAN VALUE THEOREM ❙❙❙❙ 237 for all x, so it is certainly continuous on 0, 2 and differentiable on 0, 2 . Therefore, by
the Mean Value Theorem, there is a number c in 0, 2 such that
y y=˛ x
B f2
Now f 2 6, f 0 f0 fc2 0, and f x 3x 2 1, so this equation becomes 6
O
c FIGURE 6 2 x 3c 2 6c 2 12 0 2 which gives c 2 4 , that is, c
2 s3. But c must lie in 0, 2 , so c 2 s3.
3
Figure 6 illustrates this calculation: The tangent line at this value of c is parallel to the
secant line OB.
EXAMPLE 4 If an object moves in a straight line with position function s
average velocity between t a and t b is fb
b f t , then the fa
a and the velocity at t c is f c . Thus, the Mean Value Theorem (in the form of Equation 1) tells us that at some time t c between a and b the instantaneous velocity f c
is equal to that average velocity. For instance, if a car traveled 180 km in 2 hours, then
the speedometer must have read 90 km h at least once.
In general, the Mean Value Theorem can be interpreted as saying that there is a number at which the instantaneous rate of change is equal to the average rate of change over
an interval.
The main signiﬁcance of the Mean Value Theorem is that it enables us to obtain information about a function from information about its derivative. The next example provides
an instance of this principle.
EXAMPLE 5 Suppose that f 0 3 and f x 5 for all values of x. How large can f 2 possibly be?
SOLUTION We are given that f is differentiable (and therefore continuous) everywhere.
In particular, we can apply the Mean Value Theorem on the interval 0, 2 . There exists a
number c such that
f2
f0
fc2 0 so f2 f0 2f c 3 2f c We are given that f x
5 for all x, so in particular we know that f c
ing both sides of this inequality by 2, we have 2 f c
10, so
f2 3 2f c 3 10 5. Multiply 7 The largest possible value for f 2 is 7.
The Mean Value Theorem can be used to establish some of the basic facts of differential calculus. One of these basic facts is the following theorem. Others will be found in the
following sections.
5 Theorem If f x 0 for all x in an interval a, b , then f is constant on a, b . Proof Let x 1 and x 2 be any two numbers in a, b with x 1 x 2 . Since f is differentiable on a, b , it must be differentiable on x 1, x 2 and continuous on x 1, x 2 . By 5E04(pp 232241) 238 ❙❙❙❙ 1/17/06 2:37 PM Page 238 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION applying the Mean Value Theorem to f on the interval x 1, x 2 , we get a number c such
that x 1 c x 2 and
f x2 6 Since f x f x1 0 for all x, we have f c
f x2 f x1 f c x2 x1 0, and so Equation 6 becomes
0 or f x2 f x1 Therefore, f has the same value at any two numbers x 1 and x 2 in a, b . This means that
f is constant on a, b . 7 Corollary If f x
on a, b ; that is, f x Proof Let F x fx t x for all x in an interval a, b , then f
tx
c where c is a constant.
t x . Then
Fx fx tx 0 for all x in a, b . Thus, by Theorem 5, F is constant; that is, f
NOTE ■ t is constant t is constant. Care must be taken in applying Theorem 5. Let
fx x
x 1
1 if x
if x 0
0 The domain of f is D
x x 0 and f x
0 for all x in D. But f is obviously not a
constant function. This does not contradict Theorem 5 because D is not an interval. Notice
that f is constant on the interval 0,
and also on the interval
,0 .  4.2 Exercises 1–4  Verify that the function satisﬁes the three hypotheses of
Rolle’s Theorem on the given interval. Then ﬁnd all numbers c that
satisfy the conclusion of Rolle’s Theorem. 1. f x 4x 2. f x x3 3x 2 3. f x sin 2 x, 4. f x
■ x2 x sx ■ ■ 2x 0, 2 1, 1 6,
■ 5, y =ƒ 6, 0
■ ■ ■ ■ ■ ■ ■ 1 x 2 3. Show that f 1
f 1 but there is no
0. Why does this not
number c in 1, 1 such that f c
contradict Rolle’s Theorem? 5. Let f x x 1 2. Show that f 0
f 2 but there is no
number c in 0, 2 such that f c
0. Why does this not contradict Rolle’s Theorem? 6. Let f x conclusion of the Mean Value Theorem for the interval 0, 8 .
y 0, 4 1, 7. Use the graph of f to estimate the values of c that satisfy the ■ 1
0 1 x 8. Use the graph of f given in Exercise 7 to estimate the values of c that satisfy the conclusion of the Mean Value Theorem for
the interval 1, 7 . 5E04(pp 232241) 1/17/06 2:37 PM Page 239 S ECTION 4.2 THE MEAN VALUE THEOREM ; 9. (a) Graph the function f x x 4 x in the viewing
rectangle 0, 10 by 0, 10 .
(b) Graph the secant line that passes through the points 1, 5
and 8, 8.5 on the same screen with f .
(c) Find the number c that satisﬁes the conclusion of the Mean
Value Theorem for this function f and the interval 1, 8 .
Then graph the tangent line at the point c, f c and notice
that it is parallel to the secant line. ; 10. (a) In the viewing rectangle 3, 3 by 5, 5 , graph the
function f x
x 3 2 x and its secant line through the
points 2, 4 and 2, 4 . Use the graph to estimate
the xcoordinates of the points where the tangent line is
parallel to the secant line.
(b) Find the exact values of the numbers c that satisfy the conclusion of the Mean Value Theorem for the interval 2, 2
and compare with your answers to part (a). 11–14  Verify that the function satisﬁes the hypotheses of the
Mean Value Theorem on the given interval. Then ﬁnd all numbers
c that satisfy the conclusion of the Mean Value Theorem. 11. f x 3x 2 12. f x x3 13. f x 3
sx, 14. f x
■ ■ x 1, 239 22. (a) Suppose that f is differentiable on and has two roots.
Show that f has at least one root.
(b) Suppose f is twice differentiable on and has three roots.
Show that f has at least one real root.
(c) Can you generalize parts (a) and (b)? 23. If f 1 10 and f x
possibly be? 24. Suppose that 3 18 f8 f2 2 for 1 x 4, how small can f 4 fx
5 for all values of x. Show that
30. 25. Does there exist a function f such that f 0 and f x 1, f 2 4, 2 for all x ? 26. Suppose that f and t are continuous on a, b and differentiable t x for
on a, b . Suppose also that f a
t a and f x
a x b. Prove that f b
t b . [Hint: Apply the Mean
Value Theorem to the function h f t.]
27. Show that s1 x 1 1
2 x if x 0. 28. Suppose f is an odd function and is differentiable everywhere. Prove that for every positive number b, there exists a number c
in b, b such that f c
f b b. 1, 1
0, 2 29. Use the Mean Value Theorem to prove the inequality 0, 1
sin a x
x 5, 2x ❙❙❙❙ , 2 ■ 30. If f x
■ ■ ■ ■ ■ ■ ■ ■ ■ 15. Let f x x 1 . Show that there is no value of c such that
f3
f0
f c 3 0 . Why does this not contradict the
Mean Value Theorem? 16. Let f x x 1 x 1 . Show that there is no value of c
such that f 2
f0
f c 2 0 . Why does this not contradict the Mean Value Theorem? 17. Show that the equation 1 2x x3 4x 5 0 has exactly one real root.
18. Show that the equation 2 x 1 sin x 0 has exactly one real root.
19. Show that the equation x 3 in the interval sin b a for all a and b b 1, 4 15 x c 0 has at most one root 2, 2 . 20. Show that the equation x 4 that f x
31. Let f x c (c a constant) for all x, use Corollary 7 to show
c x d for some constant d.
1 x and tx 1
x
1 if x
1
x 0 if x 0 Show that f x
t x for all x in their domains. Can we conclude from Corollary 7 that f t is constant?
32. At 2:00 P.M. a car’s speedometer reads 30 mi h. At 2:10 P.M. it reads 50 mi h. Show that at some time between 2:00 and 2:10
the acceleration is exactly 120 mi h2.
33. Two runners start a race at the same time and ﬁnish in a tie. 4x c 0 has at most two real roots.
21. (a) Show that a polynomial of degree 3 has at most three real roots.
(b) Show that a polynomial of degree n has at most n real
roots. Prove that at some time during the race they have the same
speed. [Hint: Consider f t
tt
h t , where t and h are
the position functions of the two runners.]
34. A number a is called a ﬁxed point of a function f if f a Prove that if f x
1 for all real numbers x, then f has at
most one ﬁxed point. a. 5E04(pp 232241) 240 ❙❙❙❙ 1/17/06 2:37 PM Page 240 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION  4.3 How Derivatives Affect the Shape of a Graph y Many of the applications of calculus depend on our ability to deduce facts about a function f from information concerning its derivatives. Because f x represents the slope of
the curve y f x at the point x, f x , it tells us the direction in which the curve proceeds
at each point. So it is reasonable to expect that information about f x will provide us with
information about f x . D
B What Does f Say about f ? A C
x 0 FIGURE 1 To see how the derivative of f can tell us where a function is increasing or decreasing, look
at Figure 1. (Increasing functions and decreasing functions were deﬁned in Section 1.1.)
Between A and B and between C and D, the tangent lines have positive slope and so
fx
0. Between B and C, the tangent lines have negative slope and so f x
0. Thus,
it appears that f increases when f x is positive and decreases when f x is negative. To
prove that this is always the case, we use the Mean Value Theorem.
Increasing/Decreasing Test  Let’s abbreviate the name of this test to
the I/D Test. (a) If f x
(b) If f x 0 on an interval, then f is increasing on that interval.
0 on an interval, then f is decreasing on that interval. Proof
Resources / Module 3
/ Increasing and Decreasing Functions
/ IncreasingDecreasing Detector (a) Let x 1 and x 2 be any two numbers in the interval with x1 x2 . According to the deﬁnition of an increasing function (page 21) we have to show that f x1
f x2 .
Because we are given that f x
0, we know that f is differentiable on x1, x2 . So,
by the Mean Value Theorem there is a number c between x1 and x2 such that
f x2 1 f x1 Now f c
0 by assumption and x 2
Equation 1 is positive, and so
f x2 f x1 x1 0 f c x2 x1 0 because x 1 x 2 . Thus, the right side of or f x2 f x1 This shows that f is increasing.
Part (b) is proved similarly.
EXAMPLE 1 Find where the function f x 3x 4 4x 3 24 x 12 x x 12 x 2 5 is increasing and where it is decreasing.
SOLUTION Module 4.3A guides you in determining
properties of the derivative f by examining the graphs of a variety of functions f . fx 12 x 3 12 x 2 2x 1 To use the I D Test we have to know where f x
0 and where f x
0. This
depends on the signs of the three factors of f x , namely, 12 x, x 2, and x 1. We
divide the real line into intervals whose endpoints are the critical numbers 1, 0, and
2 and arrange our work in a chart. A plus sign indicates that the given expression is
positive, and a minus sign indicates that it is negative. The last column of the chart gives
the conclusion based on the I D Test. For instance, f x
0 for 0 x 2, so f is
decreasing on (0, 2). (It would also be true to say that f is decreasing on the closed interval
0, 2 .) 5E04(pp 232241) 1/17/06 2:37 PM Page 241 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH 20 _2 Interval
3 _30 1
0 x
x
x
x 12 x x 2 x 1 fx ❙❙❙❙ 241 f 1 decreasing on ( , 1)
increasing on ( 1, 0)
decreasing on (0, 2)
increasing on (2, ) 0
2
2 The graph of f shown in Figure 2 conﬁrms the information in the chart. FIGURE 2 Recall from Section 4.1 that if f has a local maximum or minimum at c, then c must be
a critical number of f (by Fermat’s Theorem), but not every critical number gives rise to
a maximum or a minimum. We therefore need a test that will tell us whether or not f has
a local maximum or minimum at a critical number.
You can see from Figure 2 that f 0
5 is a local maximum value of f because f
increases on
1, 0 and decreases on 0, 2 . Or, in terms of derivatives, f x
0 for
1 x 0 and f x
0 for 0 x 2. In other words, the sign of f x changes from
positive to negative at 0. This observation is the basis of the following test.
The First Derivative Test Suppose that c is a critical number of a continuous function f .
(a) If f changes from positive to negative at c, then f has a local maximum at c.
(b) If f changes from negative to positive at c, then f has a local minimum at c.
(c) If f does not change sign at c (for example, if f is positive on both sides of c
or negative on both sides), then f has no local maximum or minimum at c.
The First Derivative Test is a consequence of the I D Test. In part (a), for instance, since
the sign of f x changes from positive to negative at c, f is increasing to the left of c and
decreasing to the right of c. It follows that f has a local maximum at c.
It is easy to remember the First Derivative Test by visualizing diagrams such as those
in Figure 3.
y y fª(x)>0 fª(x)<0
fª(x)<0 0 x c fª(x)>0 0 (a) Local maximum x c (b) Local minimum y y fª(x)<0
fª(x)>0 fª(x)<0
fª(x)>0 0 FIGURE 3 c x (c) No maximum or minimum 0 c x (d) No maximum or minimum 5E04(pp 242251) 242 ❙❙❙❙ 1/17/06 2:43 PM Page 242 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION EXAMPLE 2 Find the local minimum and maximum values of the function f in Example 1.
SOLUTION From the chart in the solution to Example 1 we see that f x changes from
negative to positive at 1, so f 1
0 is a local minimum value by the First Derivative Test. Similarly, f changes from negative to positive at 2, so f 2
27 is also a
local minimum value. As previously noted, f 0
5 is a local maximum value because
f x changes from positive to negative at 0. EXAMPLE 3 Find the local maximum and minimum values of the function tx x 2 sin x 0 x 2 SOLUTION To ﬁnd the critical numbers of t, we differentiate: tx 1 2 cos x 1
So t x
0 when cos x
3 and 4 3.
2 . The solutions of this equation are 2
Because t is differentiable everywhere, the only critical numbers are 2 3 and 4 3 and
so we analyze t in the following table.  The + signs in the table come from the fact
1
that t x
0 when cos x
2 . From the
graph of y cos x, this is true in the indicated
intervals. Interval 2
4 0
3
3 x
x
x tx
2
4
2 1 2 cos x t 3
3 increasing on (0, 2 3)
decreasing on (2 3, 4 3)
increasing on (4 3, 2 ) Because t x changes from positive to negative at 2 3, the First Derivative Test tells us
that there is a local maximum at 2 3 and the local maximum value is
6 t2 3 2
3 2 sin 2
3 2
3 2 s3
2 Likewise, t x changes from negative to positive at 4 0 2π t4 3 4
3 2 sin 4
3 4
3 2 2
3 s3 3.83 3 and so s3
2 4
3 s3 2.46 FIGURE 4 y=x+2 sin x is a local minimum value. The graph of t in Figure 4 supports our conclusion. What Does f Say about f ?
Explore concavity on a roller coaster.
Resources / Module 3
/ Concavity
/ Introduction Figure 5 shows the graphs of two increasing functions on a, b . Both graphs join point A
to point B but they look different because they bend in different directions. How can we
distinguish between these two types of behavior? In Figure 6 tangents to these curves have
been drawn at several points. In (a) the curve lies above the tangents and f is called concave upward on a, b . In (b) the curve lies below the tangents and t is called concave
downward on a, b . 5E04(pp 242251) 1/17/06 2:43 PM Page 243 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH y y B g A A
a 0 x b a x b (a) FIGURE 5 (b) y y B B
g f
A A 0 FIGURE 6 243 B f 0 ❙❙❙❙ 0 x (a) Concave upward x (b) Concave downward Definition If the graph of f lies above all of its tangents on an interval I , then it is called concave upward on I . If the graph of f lies below all of its tangents on I, it
is called concave downward on I .
Figure 7 shows the graph of a function that is concave upward (abbreviated CU) on the
intervals b, c , d, e , and e, p and concave downward (CD) on the intervals a, b , c, d ,
and p, q .
y D
B 0a F IGURE 7 b CD P C c CU d CD e CU p CU q x CD Let’s see how the second derivative helps determine the intervals of concavity. Looking
at Figure 6(a), you can see that, going from left to right, the slope of the tangent increases.
This means that the derivative f is an increasing function and therefore its derivative f
is positive. Likewise, in Figure 6(b) the slope of the tangent decreases from left to right,
so f decreases and therefore f is negative. This reasoning can be reversed and suggests
that the following theorem is true. A proof is given in Appendix F with the help of the
Mean Value Theorem.
Concavity Test (a) If f x
(b) If f x 0 for all x in I , then the graph of f is concave upward on I .
0 for all x in I , then the graph of f is concave downward on I . 5E04(pp 242251) 244 ❙❙❙❙ 1/17/06 2:43 PM Page 244 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION EXAMPLE 4 Figure 8 shows a population graph for Cyprian honeybees raised in an apiary.
How does the rate of population increase change over time? When is this rate highest?
Over what intervals is P concave upward or concave downward?
P
80
Number of bees
(in thousands) 60
40
20
0 3 6 FIGURE 8 9 12 15 t 18 Time (in weeks) SOLUTION By looking at the slope of the curve as t increases, we see that the rate of
increase of the population is initially very small, then gets larger until it reaches a maximum at about t 12 weeks, and decreases as the population begins to level off. As the
population approaches its maximum value of about 75,000 (called the carrying capacity), the rate of increase, P t , approaches 0. The curve appears to be concave upward on
(0, 12) and concave downward on (12, 18). In Example 4, the population curve changed from concave upward to concave downward at approximately the point (12, 38,000). This point is called an inﬂection point of the
curve. The signiﬁcance of this point is that the rate of population increase has its maximum
value there. In general, an inﬂection point is a point where a curve changes its direction of
concavity.
Definition A point P on a curve y f x is called an inﬂection point if f is continuous there and the curve changes from concave upward to concave downward or
from concave downward to concave upward at P. For instance, in Figure 7, B, C, D, and P are the points of inﬂection. Notice that if a
curve has a tangent at a point of inﬂection, then the curve crosses its tangent there.
In view of the Concavity Test, there is a point of inﬂection at any point where the second derivative changes sign.
EXAMPLE 5 Sketch a possible graph of a function f that satisﬁes the following conditions:
(i f0 0, f2 3, f4 ii fx 0 for 0 x 4, iii fx 0 for x 2, fx 6,
fx f0 f4 0 for x
0 for x 0 0 and for x 4 2 SOLUTION Condition (i) tells us that the graph has horizontal tangents at the points 0, 0
and 4, 6 . Condition (ii) says that f is increasing on the interval 0, 4 and decreasing on 5E04(pp 242251) 1/17/06 2:43 PM Page 245 S ECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH y
6 (2, 3) 0 dec 2 x 4 inc
CD The Second Derivative Test Suppose f is continuous near c. (a) If f c
(b) If f c FIGURE 9 y f 0, then f has a local minimum at c.
0, then f has a local maximum at c. For instance, part (a) is true because f x
0 near c and so f is concave upward near
c. This means that the graph of f lies above its horizontal tangent at c and so f has a local
minimum at c. (See Figure 10.)
x 4 4 x 3 with respect to concavity, points of inﬂection,
and local maxima and minima. Use this information to sketch the curve. ƒ f(c)
c 0 and f c
0 and f c EXAMPLE 6 Discuss the curve y P 0 Another application of the second derivative is the following test for maximum and
minimum values. It is a consequence of the Concavity Test. dec CU f ª(c)=0 245 the intervals
, 0 and 4, . It follows from the I/D Test that f 0
0 is a local
minimum and f 4
6 is a local maximum.
Condition (iii) says that the graph is concave upward on the interval
, 2 and
concave downward on 2, . Because the curve changes from concave upward to concave downward when x 2, the point 2, 3 is an inﬂection point.
We use this information to sketch the graph of f in Figure 9. Notice that we made the
curve bend upward when x 2 and bend downward when x 2. (4, 6) 3 ❙❙❙❙ SOLUTION If f x
x FIGURE 10
f ·(c)>0, f is concave upward x x4 4 x 3, then
fx 4x 3 fx 12 x 2 12 x 2 4x 2 x 3 24 x 12 x x 2 To ﬁnd the critical numbers we set f x
0 and obtain x 0 and x
Second Derivative Test we evaluate f at these critical numbers:
f0 0 f3 36 3. To use the 0 0 and f 3
0, f 3
27 is a local minimum. Since f 0
0, the
Since f 3
Second Derivative Test gives no information about the critical number 0. But since
fx
0 for x 0 and also for 0 x 3, the First Derivative Test tells us that f does
not have a local maximum or minimum at 0. [In fact, the expression for f x shows that
f decreases to the left of 3 and increases to the right of 3.]
Since f x
0 when x 0 or 2, we divide the real line into intervals with these
numbers as endpoints and complete the following chart.
Interval
( , 0)
(0, 2)
(2, ) fx 12 x x 2 Concavity
upward
downward
upward The point 0, 0 is an inﬂection point since the curve changes from concave upward to
concave downward there. Also 2, 16 is an inﬂection point since the curve changes
from concave downward to concave upward there. 5E04(pp 242251) 246 ❙❙❙❙ 1/17/06 2:43 PM Page 246 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION y Using the local minimum, the intervals of concavity, and the inﬂection points, we
sketch the curve in Figure 11. y=x$4˛
(0, 0)
2 inﬂection
points x 3 NOTE The Second Derivative Test is inconclusive when f c
0. In other words, at
such a point there might be a maximum, there might be a minimum, or there might be neither (as in Example 6). This test also fails when f c does not exist. In such cases the First
Derivative Test must be used. In fact, even when both tests apply, the First Derivative Test
is often the easier one to use.
■ (2, _16) x2 3 6 EXAMPLE 7 Sketch the graph of the function f x
(3, _27) fx x Since f x
0 when x
numbers are 0, 4, and 6.
Interval
0
4 In Module 4.3B you can practice using
graphical information about f to determine the shape of the graph of f .  Try reproducing the graph in Figure 12
with a graphing calculator or computer. Some
machines produce the complete graph, some
produce only the portion to the right of the
yaxis, and some produce only the portion
between x 0 and x 6. For an explanation
and cure, see Example 7 in Section 1.4. An
equivalent expression that gives the correct
graph is
x2 13 6
6 x
x 6 x 13 . SOLUTION You can use the differentiation rules to check that the ﬁrst two derivatives are FIGURE 11 y x x
x
x
x 4 x 13 4
6 x
x 8 fx 23 x 43 4 and f x does not exist when x x1 3 6 x 23 fx 0
4
6
6 x 53 0 or x 6, the critical f
decreasing on (
, 0)
increasing on (0, 4)
decreasing on (4, 6)
decreasing on (6, ) To ﬁnd the local extreme values we use the First Derivative Test. Since f changes
from negative to positive at 0, f 0
0 is a local minimum. Since f changes from
positive to negative at 4, f 4
2 5 3 is a local maximum. The sign of f does not change
at 6, so there is no minimum or maximum there. (The Second Derivative Test could be
used at 4, but not at 0 or 6 since f does not exist at either of these numbers.)
Looking at the expression for f x and noting that x 4 3 0 for all x, we have
fx
0 for x 0 and for 0 x 6 and f x
0 for x 6. So f is concave downward on
, 0 and 0, 6 and concave upward on 6, , and the only inﬂection point
is 6, 0 . The graph is sketched in Figure 12. Note that the curve has vertical tangents at
0, 0 and 6, 0 because f x l as x l 0 and as x l 6.
y
4 (4, 2%?# ) 3
2 13 0 FIGURE 12 6 1 2 3 4 5 y=x @ ?#(6x)! ?# 7x 5E04(pp 242251) 1/17/06 2:43 PM Page 247 S ECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH  4.3
1–2 (a)
(b)
(c)
(d)
(e)
1. ❙❙❙❙ 247 Exercises  Use the given graph of f to ﬁnd the following.
The largest open intervals on which f is increasing.
The largest open intervals on which f is decreasing.
The largest open intervals on which f is concave upward.
The largest open intervals on which f is concave downward.
The coordinates of the points of inﬂection. 7. The graph of the second derivative f of a function f is shown. State the xcoordinates of the inﬂection points of f . Give reasons for your answers.
y y=f·(x) y 0 12 4 6 x 8 4 8. The graph of the ﬁrst derivative f of a function f is shown. (a) On what intervals is f increasing? Explain.
(b) At what values of x does f have a local maximum or minimum? Explain.
(c) On what intervals is f concave upward or concave downward? Explain.
(d) What are the xcoordinates of the inﬂection points of f ?
Why? 2 0 2. 4 2 6 x 8 y y y=fª(x)
2
0
0 2 4 1 8x 6 3 5 7 9 x _2 9. Sketch the graph of a function whose ﬁrst and second deriva tives are always negative.
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 10. A graph of a population of yeast cells in a new laboratory cul3. Suppose you are given a formula for a function f . (a) How do you determine where f is increasing or decreasing?
(b) How do you determine where the graph of f is concave
upward or concave downward?
(c) How do you locate inﬂection points? ture as a function of time is shown.
(a) Describe how the rate of population increase varies.
(b) When is this rate highest?
(c) On what intervals is the population function concave
upward or downward?
(d) Estimate the coordinates of the inﬂection point. 4. (a) State the First Derivative Test. (b) State the Second Derivative Test. Under what circumstances is it inconclusive? What do you do if it fails? 700
600
Number 500
400
of
yeast cells 300
200
100 5–6  The graph of the derivative f of a function f is shown.
(a) On what intervals is f increasing or decreasing?
(b) At what values of x does f have a local maximum or minimum? 5. 6. y y=fª(x) 0 y y=fª(x) ■ ■ 2 ■ 4 ■ 6 ■ x ■ 0 ■ 2 ■ ■ 4 ■ 6 ■ 4 6 8 10 12 14 16 18
Time (in hours) 11–16
0 2  (a) Find the intervals on which f is increasing or decreasing.
(b) Find the local maximum and minimum values of f .
(c) Find the intervals of concavity and the inﬂection points. x ■ 11. f x x3 12 x 1 12. f x 5 3x 2 x3 5E04(pp 242251) ❙❙❙❙ 248 1/17/06 Page 248 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 13. f x x4 15. f x x 16. f x cos2 x ■ 2:44 PM ■ 2x2 0 x 2 sin x, 0 x x 2 27. y 3 y=fª(x) 3 2 sin x, ■ x2 14. f x 3 ■ ■ ■ 2 2
■ ■ ■ ■ ■ ■ 0 17–19  Find the local maximum and minimum values of f using
both the First and Second Derivative Tests. Which method do you
prefer? 17. f x x5 5x 3 19. f x x s1 28. ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 0 0 and f 2
0 and f 6 0 for all x 1,
0 if x 1 or x
f2
0 if x
0 if 0
0 if 1 fx
fx
fx
24. f 1 .
5, what can you say about f ?
0, what can you say about f ? f 29–40 (a)
(b)
(c)
(d) f
f
f
f
■ 27–28 x
x
x
x
■ 0, 2, lim f x f0
f2
0 if 0 x
0 if 2 x
0 if 0 x
0 if 1 x
■ f4
2 or 4
4 or x
1 or 3
3 or x
■ 0 if x
fx , xl2 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 1 or x 29. f x ■ x4
3x 5 35. A x x sx 37. C x x 13 39. f 3 2x 3 33. h x 2 cos 40. f t 2, ■ 0 if x 2 t ■ ; 41–42 3x 2 30. f x 3 3
x cos 2 ,
■ x 38. B x 4 cos t,
■ x2 36. G x 5x 3 200 34. h x 6x 2 2 32. t x 12 x 0 2 t ■ ■ x3 3x
8x 3
1 x4 3 4 sx 3x 23 x 2
2
■ ■ ■ ■ ■ ■  (a) Use a graph of f to estimate the maximum and minimum
values. Then ﬁnd the exact values.
(b) Estimate the value of x at which f increases most rapidly. Then
ﬁnd the exact value. 0,
f6
x 6,
6,
x 5,
5, f x
■  Find the intervals of increase or decrease.
Find the local maximum and minimum values.
Find the intervals of concavity and the inﬂection points.
Use the information from parts (a)–(c) to sketch the graph.
Check your work with a graphing device if you have one. 31. f x fx ■  f4
0,
0 or 2 x 4,
x 2 or x 4,
x 3, f x
0 if x 1,
x3 fx
0 if x
1,
1 if x
2,
2, f x
0, inﬂection point 0, 1 2 26. f 0 vertical asymptote x
3, f x
0 if 1 0 if x 25. f x ■ , f
1
0,
x
0 if 1
0 if 2 x fx
fx 2 _2 22–26  Sketch the graph of a function that satisﬁes all of the
given conditions. 23. f 0 8x ■ ■ 21. Suppose f is continuous on fx 6 2 x 4 x 1 3.
(b) What does the Second Derivative Test tell you about the
behavior of f at these critical numbers?
(c) What does the First Derivative Test tell you? 22. f x 4 y=fª(x) 4 20. (a) Find the critical numbers of f x (a) If f 2
(b) If f 6 8x y x x2 6 _2 x 18. f x 4 2 fx
■ ■ ■ ■ The graph of the derivative f of a continuous function f is shown.
(a) On what intervals is f increasing or decreasing?
(b) At what values of x does f have a local maximum or
minimum?
(c) On what intervals is f concave upward or downward?
(d) State the xcoordinate(s) of the point(s) of inﬂection.
(e) Assuming that f 0
0, sketch a graph of f. 41. f x x1
sx 2 1 42. f x x ■ ; 43–44 ■ 2 cos x, ■ ■ 0 ■ x
■ 2
■ ■ ■ ■ ■  (a) Use a graph of f to give a rough estimate of the intervals of
concavity and the coordinates of the points of inﬂection.
(b) Use a graph of f to give better estimates.
43. f x cos x 1
2 cos 2 x, 0 x 2 ■ 5E04(pp 242251) 1/17/06 2:44 PM Page 249 S ECTION 4.4 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES 44. f x
■ C AS ■ x3 x 2 ■ 4 ■ ■ x3 ■ 10x
sx ■ 2 249 53. Suppose f is differentiable on an interval I and f x ■ ■ ■ ■ ■ ■ ■ 0 for
all numbers x in I except for a single number c. Prove that f is
increasing on the entire interval I . ■ 45–46  Estimate the intervals of concavity to one decimal place
by using a computer algebra system to compute and graph f . 45. f x ❙❙❙❙ 5
4 ■ x
x3 46. f x ■ ■ ■ ■ ■ 1 3 x2
1 x2
■ 54–56  Assume that all of the functions are twice differentiable
and the second derivatives are never 0. 5
4
■ 54. (a) If f and t are concave upward on I , show that f t is concave upward on I .
(b) If f is positive and concave upward on I , show that the
function t x
f x 2 is concave upward on I . ■ 47. Let K t be a measure of the knowledge you gain by studying 55. (a) If f and t are positive, increasing, concave upward func K7
for a test for t hours. Which do you think is larger, K 8
or K 3
K 2 ? Is the graph of K concave upward or concave
downward? Why? tions on I , show that the product function f t is concave
upward on I .
(b) Show that part (a) remains true if f and t are both
decreasing.
(c) Suppose f is increasing and t is decreasing. Show, by
giving three examples, that f t may be concave upward,
concave downward, or linear. Why doesn’t the argument in
parts (a) and (b) work in this case? 48. Coffee is being poured into the mug shown in the ﬁgure at a constant rate (measured in volume per unit time). Sketch a
rough graph of the depth of the coffee in the mug as a function
of time. Account for the shape of the graph in terms of concavity. What is the signiﬁcance of the inﬂection point? 56. Suppose f and t are both concave upward on , . Under
what condition on f will the composite function h x
f tx
be concave upward? ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 57. Show that a cubic function (a thirddegree polynomial) always has exactly one point of inﬂection. If its graph has three
xintercepts x 1, x 2, and x 3, show that the xcoordinate of the
inﬂection point is x 1 x 2 x 3 3. ; 58. For what values of c does the polynomial
49. Show that tan x fx tan x x for 0 x
2. [Hint: Show that
x is increasing on 0, 2 .] 50. Prove that, for all x 1,
2 sx 59. Prove that if c, f c is a point of inﬂection of the graph of f 3 1
x ax 3 b x 2 c x d that has a
local maximum value of 3 at 2 and a local minimum value
of 0 at 1. 51. Find a cubic function f x 52. For what values of a and b does the function fx x3 ax 2 have a local maximum when x
when x
1?  4.4 Px
x 4 c x 3 x 2 have two inﬂection points? One inﬂection point? None? Illustrate by graphing P for several values
of c. How does the graph change as c decreases? bx 2 3 and a local minimum 0.
and f exists in an open interval that contains c, then f c
[Hint: Apply the First Derivative Test and Fermat’s Theorem to
the function t f .]
x 4, then f 0
inﬂection point of the graph of f . 60. Show that if f x 61. Show that the function t x 0, but 0, 0 is not an x x has an inﬂection point at 0, 0 but t 0 does not exist.
62. Suppose that f is continuous and f c fc
0, but
fc
0. Does f have a local maximum or minimum at c ?
Does f have a point of inﬂection at c ? Limits at Infinity; Horizontal Asymptotes
In Sections 2.2 and 2.4 we investigated inﬁnite limits and vertical asymptotes. There we
let x approach a number and the result was that the values of y became arbitrarily large
(positive or negative). In this section we let x become arbitrarily large (positive or negative) and see what happens to y. We will ﬁnd it very useful to consider this socalled end
behavior when sketching graphs. 5E04(pp 242251) 250 ❙❙❙❙ 1/17/06 2:44 PM Page 250 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION x
0
1
2
3
4
5
10
50
100
1000 Let’s begin by investigating the behavior of the function f deﬁned by fx
1
0
0.600000
0.800000
0.882353
0.923077
0.980198
0.999200
0.999800
0.999998 x2
x2 fx 1
1 as x becomes large. The table at the left gives values of this function correct to six decimal
places, and the graph of f has been drawn by a computer in Figure 1.
y y=1 0 1 y= x ≈1
≈+1 FIGURE 1 As x grows larger and larger you can see that the values of f x get closer and closer
to 1. In fact, it seems that we can make the values of f x as close as we like to 1 by taking
x sufﬁciently large. This situation is expressed symbolically by writing
lim xl x2
x2 1
1 1 In general, we use the notation
lim f x xl L to indicate that the values of f x become closer and closer to L as x becomes larger and
larger.
1 Definition Let f be a function deﬁned on some interval a, lim f x xl . Then L means that the values of f x can be made arbitrarily close to L by taking x sufﬁciently large.
Another notation for lim x l f x L is f x lL as xl The symbol does not represent a number. Nonetheless, the expression lim f x
xl
often read as L is “the limit of f x , as x approaches inﬁnity, is L”
or “the limit of f x , as x becomes inﬁnite, is L” or “the limit of f x , as x increases without bound, is L” The meaning of such phrases is given by Deﬁnition 1. A more precise deﬁnition, similar
to the , deﬁnition of Section 2.4, is given at the end of this section. 5E04(pp 242251) 1/17/06 2:44 PM Page 251 S ECTION 4.4 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES ❙❙❙❙ 251 Geometric illustrations of Deﬁnition 1 are shown in Figure 2. Notice that there are
many ways for the graph of f to approach the line y L (which is called a horizontal
asymptote) as we look to the far right of each graph.
y y y=L y y=ƒ y=L y=ƒ y=ƒ y=L
0 0 x 0 x x FIGURE 2 Examples illustrating lim ƒ=L
x` Referring back to Figure 1, we see that for numerically large negative values of x, the
values of f x are close to 1. By letting x decrease through negative values without bound,
we can make f x as close as we like to 1. This is expressed by writing
lim xl x2
x2 1
1 1 The general deﬁnition is as follows.
2 Definition Let f be a function deﬁned on some interval lim f x , a . Then L xl means that the values of f x can be made arbitrarily close to L by taking x sufﬁciently large negative.
Again, the symbol
is often read as y y=ƒ does not represent a number, but the expression lim f x
xl L “the limit of f x , as x approaches negative inﬁnity, is L”
Deﬁnition 2 is illustrated in Figure 3. Notice that the graph approaches the line y
we look to the far left of each graph. y=L
0 L as x 3 y y Definition The line y L is called a horizontal asymptote of the curve f x if either
lim f x y=ƒ xl y=L 0 x L x _` lim f x xl For instance, the curve illustrated in Figure 1 has the line y
tote because FIGURE 3 Examples illustrating lim ƒ=L or lim xl x2
x2 1
1 1 L 1 as a horizontal asymp 5E04(pp 252261) 252 ❙❙❙❙ 1/17/06 3:26 PM Page 252 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION The curve y f x sketched in Figure 4 has both y
1 and y 2 as horizontal asymptotes because
lim f x
1
and
lim f x
2
xl xl y
2 y=2 0 y=_1 y=ƒ
x _1 F IGURE 4
y EXAMPLE 1 Find the inﬁnite limits, limits at inﬁnity, and asymptotes for the function f
whose graph is shown in Figure 5.
SOLUTION We see that the values of f x become large as x l 1 from both sides, so 2 lim f x xl 1 0 x 2 Notice that f x becomes large negative as x approaches 2 from the left, but large positive as x approaches 2 from the right. So
lim f x and x l2 F IGURE 5 lim f x x l2 Thus, both of the lines x
1 and x 2 are vertical asymptotes.
As x becomes large, it appears that f x approaches 4. But as x decreases through
negative values, f x approaches 2. So
lim f x xl This means that both y
EXAMPLE 2 Find lim
xl 4 4 and y and lim f x xl 2 2 are horizontal asymptotes. 1
1
and lim .
xl
x
x SOLUTION Observe that when x is large, 1 x is small. For instance, 1
100 y y=Δ 0 FIGURE 6 1
1
lim =0, lim =0
x `x
x _` x x 0.01 1
10,000 0.0001 1
1,000,000 0.000001 In fact, by taking x large enough, we can make 1 x as close to 0 as we please. Therefore,
according to Deﬁnition 1, we have
1
lim
0
xl
x
Similar reasoning shows that when x is large negative, 1 x is small negative, so we also
have
1
lim
0
xl
x
It follows that the line y 0 (the xaxis) is a horizontal asymptote of the curve y
(This is an equilateral hyperbola; see Figure 6.) 1 x. 5E04(pp 252261) 1/17/06 3:26 PM Page 253 SECTION 4.4 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES ❙❙❙❙ 253 Most of the Limit Laws that were given in Section 2.3 also hold for limits at inﬁnity. It
can be proved that the Limit Laws listed in Section 2.3 (with the exception of Laws 9 and
10) are also valid if “x l a” is replaced by “x l ” or “ x l
.” In particular, if we
combine Laws 6 and 11 with the results of Example 2, we obtain the following important
rule for calculating limits.
4 Theorem If r 0 is a rational number, then
1
xr lim xl If r 0 0 is a rational number such that x r is deﬁned for all x, then
1
xr lim xl 0 EXAMPLE 3 Evaluate lim xl 3x 2
5x 2 x
4x 2
1 and indicate which properties of limits are used at each stage.
SOLUTION As x becomes large, both numerator and denominator become large, so it isn’t
obvious what happens to their ratio. We need to do some preliminary algebra.
To evaluate the limit at inﬁnity of any rational function, we ﬁrst divide both the
numerator and denominator by the highest power of x that occurs in the denominator.
(We may assume that x 0, since we are interested only in large values of x.) In this
case the highest power of x in the denominator is x 2, so we have 3x 2
x
4x 2
1 2 1
x 2
x2 lim 5 3x
lim
x l 5x 2 x
x2
4x
x2 lim 3 2 4
x 1
x2 lim xl 5x 2 xl y xl y=0.6
0 x lim 3 lim xl 4 lim 3
5 0
0 xl 0
0 xl lim 1 1
x xl lim 5 1 3
xl 5 1
x
4
x 2
x2
1
x2 (by Limit Law 5) 1
x2
1
lim
xl x2 2 lim
xl 1
x (by 1, 2, and 3) (by 7 and Theorem 4) 3
5
FIGURE 7 y= 3≈x2
5≈+4x+1 A similar calculation shows that the limit as x l
is also 3 . Figure 7 illustrates the
5
results of these calculations by showing how the graph of the given rational function
approaches the horizontal asymptote y 3 .
5 5E04(pp 252261) 254 ❙❙❙❙ 1/17/06 3:27 PM Page 254 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION E XAMPLE 4 Find the horizontal and vertical asymptotes of the graph of the function s2x 2 1
3x 5 fx SOLUTION Dividing both numerator and denominator by x and using the properties of
limits, we have lim xl s2x
3x 2 1
x2 2 1 lim 5 xl 3 lim 1
x2 2 xl (since sx 2 5
x lim 2 xl 5
x lim 3 xl s2 0
3 50 lim 3 xl x for x 0) 1
x2
1
5 lim
xl x
lim xl s2
3 Therefore, the line y s 2 3 is a horizontal asymptote of the graph of f .
In computing the limit as x l
, we must remember that for x 0, we have
x
x. So when we divide the numerator by x, for x 0 we get
sx 2
1
s2x 2
x Therefore s2x
3x lim xl 1
s2x 2
sx 2 1 2 1 2
1
x2 2 1 lim 5 xl 3 y 5
x 3
2 1
x2 1
x2 lim xl 5 lim
xl 1
x s2
3 Thus, the line y
s 2 3 is also a horizontal asymptote.
A vertical asymptote is likely to occur when the denominator, 3x 5, is 0, that is,
when x 5 . If x is close to 5 and x 5 , then the denominator is close to 0 and 3x 5
3
3
3
is positive. The numerator s 2 x 2 1 is always positive, so f x is positive. Therefore œ„
2
y= 3 x
œ„
2 y=_ 3 lim xl 5 3 x= 5
3 If x is close to 5 but x
3 5
3 , then 3x
lim FIG URE 8 œ„„„„„„
2≈+1
y=
3x5 xl 5 3 The vertical asymptote is x 5
3 5 s2x 2 1
3x 5
0 and so f x is large negative. Thus
s2x 2 1
3x 5 . All three asymptotes are shown in Figure 8. 5E04(pp 252261) 1/17/06 3:28 PM Page 255 S ECTION 4.4 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES ( EXAMPLE 5 Compute lim s x 2
xl ❙❙❙❙ 255 x). 1 SOLUTION Because both sx 2  We can think of the given function as having
a denominator of 1. 1 and x are large when x is large, it’s difﬁcult to see what
happens to their difference, so we use algebra to rewrite the function. We ﬁrst multiply
numerator and denominator by the conjugate radical:
lim (s x 2 x) 1 xl lim (s x 2 xl x2 1
sx 2 1 lim xl x) 1 sx 2
sx 2 x2
x lim xl 1
1 x
x
1 sx 2 1 x The Squeeze Theorem could be used to show that this limit is 0. But an easier method is
to divide numerator and denominator by x. Doing this and using the Limit Laws, we obtain
lim (s x 2 1 xl x) lim xl 1
sx 1 2 y lim y=œ„„„„„x
≈+1 xl 1
x
sx 2 1 1
0 x x lim xl 1 x
1 0
0 x FIGURE 9 s1 1
x2 1 0 Figure 9 illustrates this result.
EXAMPLE 6 Evaluate lim sin
xl  The problemsolving strategy for Example 6
is introducing something extra (see page 58).
Here, the something extra, the auxiliary aid, is
the new variable t. 1 1
x SOLUTION If we let t 1
.
x 1 x, then t l 0 as x l . Therefore
lim sin xl 1
x lim sin t tl0 0 (See Exercise 69.)
EXAMPLE 7 Evaluate lim sin x.
xl SOLUTION As x increases, the values of sin x oscillate between 1 and 1 inﬁnitely often
and so they don’t approach any deﬁnite number. Thus, lim x l sin x does not exist. Infinite Limits at Infinity
The notation
lim f x xl is used to indicate that the values of f x become large as x becomes large. Similar meanings are attached to the following symbols:
lim f x xl lim f x xl lim f x xl 5E04(pp 252261) 256 ❙❙❙❙ 1/17/06 3:28 PM Page 256 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION E XAMPLE 8 Find lim x 3 and lim x 3.
xl y xl SOLUTION When x becomes large, x 3 also becomes large. For instance,
y=˛ 0 10 3 x 100 3 1000 1000 3 1,000,000 1,000,000,000 In fact, we can make x 3 as big as we like by taking x large enough. Therefore, we can
write
lim x 3
xl Similarly, when x is large negative, so is x 3. Thus
lim x 3 xl FIGURE 10 lim x #=`, lim x #=_`
x` x _` x 3 in Figure 10. These limit statements can also be seen from the graph of y
EXAMPLE 9 Find lim x 2 x. xl  SOLUTION Note that we cannot write
lim x 2 lim x 2 x xl lim x xl xl The Limit Laws can’t be applied to inﬁnite limits because
can’t be deﬁned). However, we can write
lim x 2 xl because both x and x
EXAMPLE 10 Find lim
xl x lim x x is not a number ( 1 xl 1 become arbitrarily large and so their product does too.
x2
3 x
.
x SOLUTION As in Example 3, we divide the numerator and denominator by the highest
power of x in the denominator, which is just x : lim xl because x 1l and 3 x x2
3
1l x
x lim xl x
3
x 1
1 1 as x l . The next example shows that by using inﬁnite limits at inﬁnity, together with intercepts,
we can get a rough idea of the graph of a polynomial even without computing derivatives.
EXAMPLE 11 Sketch the graph of y and its limits as x l and as x l SOLUTION The yintercept is f 0 found by setting y 0: x 2, x 2 4 x 1 3 x 1 by ﬁnding its intercepts . 2413 1
16 and the xintercepts are
1, 1. Notice that since x 2 4 is positive, the function 5E04(pp 252261) 1/17/06 3:28 PM Page 257 S ECTION 4.4 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES ❙❙❙❙ 257 doesn’t change sign at 2; thus, the graph doesn’t cross the xaxis at 2. The graph crosses
the axis at 1 and 1.
When x is large positive, all three factors are large, so
lim x 2 xl 4 x 1 3 x 1 When x is large negative, the ﬁrst factor is large positive and the second and third factors
are both large negative, so
lim x xl 2 4 x 1 3 x 1 Combining this information, we give a rough sketch of the graph in Figure 11. The use
of derivatives would enable us to sketch a more accurate graph by giving the precise
location of maximum and minimum points and inﬂection points, but for this particular
function the computations would be extremely complex.
y 0 _1 1 2 x y=(x2)$ (x +1)#(x1)
_16 FIGURE 11 Precise Definitions
Deﬁnition 1 can be stated precisely as follows. 5 Definition Let f be a function deﬁned on some interval a, lim f x xl means that for every . Then L 0 there is a corresponding number N such that
fx L whenever x N In words, this says that the values of f x can be made arbitrarily close to L (within a
distance , where is any positive number) by taking x sufﬁciently large (larger than N ,
where N depends on ). Graphically it says that by choosing x large enough (larger than
some number N ) we can make the graph of f lie between the given horizontal lines 5E04(pp 252261) 258 ❙❙❙❙ 1/17/06 3:29 PM Page 258 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION yL
and y L
as in Figure 12. This must be true no matter how small we
choose . Figure 13 shows that if a smaller value of is chosen, then a larger value of N
may be required.
y y=ƒ y=L +∑
∑
L∑
y=L ∑ ƒ is
in here 0 x N FIGURE 12 lim ƒ=L when x is in here x` y y=ƒ
y=L+∑ L y=L∑
0 FIGURE 13 x N lim ƒ=L
x` Similarly, a precise version of Deﬁnition 2 is given by Deﬁnition 6, which is illustrated
in Figure 14.
6 Definition Let f be a function deﬁned on some interval lim f x xl means that for every , a . Then L 0 there is a corresponding number N such that
fx L whenever x N y y=ƒ
y=L+∑
L
y=L∑
FIGURE 14 x 0 N lim ƒ=L x _` In Example 3 we calculated that
lim xl 3x 2
5x 2 x
4x 2
1 3
5 In the next example we use a graphing device to relate this statement to Deﬁnition 5 with
L 3 and
0.1.
5 5E04(pp 252261) 1/17/06 3:29 PM Page 259 SECTION 4.4 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES ❙❙❙❙ 259 E XAMPLE 12 Use a graph to ﬁnd a number N such that 3x 2
5x 2 x
4x 2
1 0.6 0.1 whenever x N S OLUTION We rewrite the given inequality as 3x 2
5x 2 0.5 2
1 0.7 We need to determine the values of x for which the given curve lies between the horizontal lines y 0.5 and y 0.7. So we graph the curve and these lines in Figure 15. Then
we use the cursor to estimate that the curve crosses the line y 0.5 when x 6.7. To
the right of this number the curve stays between the lines y 0.5 and y 0.7. Rounding to be safe, we can say that 1
y=0.7
y=0.5
y= x
4x 3≈x2
5≈+4x+1 0 FIGURE 15 3x 2
5x 2 15 In other words, for x
4x 2
1 0.6 0.1 whenever 0.1 we can choose N E XAMPLE 13 Use Deﬁnition 5 to prove that lim
xl x 7 7 (or any larger number) in Deﬁnition 5.
1
x 0. SOLUTION
1. Preliminary analysis of the problem ( guessing a value for N ). Given 0, we want to ﬁnd N such that
1
x 0 whenever In computing the limit we may assume x
1
x x N 0, in which case
1
x 0 1
x Therefore, we want
1
x
that is, whenever N whenever 1 x x x N This suggests that we should take N 1 .
2. Proof (showing that this N works). Given
Then
1
x Thus 1
x 0 0 1
x 0, we choose N
1
x whenever 1
N x N 1 . Let x N. 5E04(pp 252261) 260 ❙❙❙❙ 1/17/06 3:30 PM Page 260 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Therefore, by Deﬁnition 5,
lim xl 1
x 0 Figure 16 illustrates the proof by showing some values of
of N .
y y and the corresponding values y ∑=1
∑=0.2
0 x N=1 0 ∑=0.1
N=5 x 0 N=10 x FIGURE 16 Finally we note that an inﬁnite limit at inﬁnity can be deﬁned as follows. The geometric illustration is given in Figure 17.
y 7 y=M M Definition Let f be a function deﬁned on some interval a, . Then lim f x xl 0 N means that for every positive number M there is a corresponding positive number
N such that
fx
M
whenever
xN x FIGURE 17 lim ƒ=` Similar deﬁnitions apply when the symbol x`  4.4 is replaced by . (See Exercise 70.) Exercises 1. Explain in your own words the meaning of each of the following.
(a) lim f x
xl 5 (b) lim f x
xl (f) The equations of the asymptotes
y 3 2. (a) Can the graph of y f x intersect a vertical asymptote?
Can it intersect a horizontal asymptote? Illustrate by
sketching graphs.
(b) How many horizontal asymptotes can the graph of y f x
have? Sketch graphs to illustrate the possibilities. 1
x 1 3. For the function f whose graph is given, state the following. (a) lim f x (b) lim f x (c) lim f x (d) lim f x x l2 xl 1 (e) lim f x
xl xl 1 xl 4. For the function t whose graph is given, state the following. (a) lim t x (b) lim t x (c) lim t x (d) lim t x xl x l3 xl x l0 5E04(pp 252261) 1/17/06 3:31 PM Page 261 S ECTION 4.4 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES 23. lim (sx 2 (e) lim t x
xl 2 (f) The equations of the asymptotes sx 2 ax xl bx ) xl 3
26. lim sx xl xl y 27. lim ( x sx ) xl 29. lim x 4
x 2 31. lim
xl ■ 2
x 1 7–8  Evaluate the limit and justify each step by indicating the
appropriate properties of limits. ■ ■ 9–32  9. lim
xl 11. lim ■ xl 15. lim
ul xl ■ ■ ■ xl x2
7 x
2x 2 x3
2x 2 yl 5x 3 12. lim u2 x 14. lim 4 4u 4 5
2 2u 2 tl 1 s9x 6 x
xl
x3 1
x
19. lim
2
xl
4
sx
17. lim 21. lim (s9x 2
xl ■ 10. lim 3 1 x ■ ■ ■ ■ ■ 1
x ■ ■ ■ ■ ■ ■ ■ x) 1 ; 35–40  Find the horizontal and vertical asymptotes of each
curve. Check your work by graphing the curve and estimating the
asymptotes. ■ 36. y x
x 4
x3
3x 2 10 x
4
sx 4 ■ ■ 4
1 x3
x3 1
x
x9
s4x 2 3x 40. F x 1
■ x2
x2 38. y x 35. y 39. h x
■ x ■ ■ ■ ■ ■ ■ 2
■ ■ 41. Find a formula for a function f that satisﬁes the following 1
2x xl s3x 2 8 x 6 s3x 2 3x 1
to estimate the value of lim x l f x to one decimal place.
(b) Use a table of values of f x to estimate the limit to four
decimal places.
(c) Find the exact value of the limit. 37. y 12 x 3 5x
2
1
4x 2 3x 3 8. lim Find the limit. xl 13. lim 4
8
■ 32. lim x sin ; 34. (a) Use a graph of f x x to estimate the value of lim x l f x correct to two decimal
places.
(b) Use a table of values of f x to estimate the limit to four
decimal places. x
5x x5
x4 by graphing the function f x
sx 2 x 1 x.
(b) Use a table of values of f x to guess the value of the limit.
(c) Prove that your guess is correct. ; 6. (a) Use a graph of xl ■ x4 xl xl by evaluating the function f x
x 2 2 x for x 0, 1, 2, 3, 4, 5,
6, 7, 8, 9, 10, 20, 50, and 100. Then use a graph of f to support
your guess. 3x2
2x2 30. lim x 2 lim (sx 2 x2
lim x
xl 2 7. lim ■ 2x 3
2x2 5 xl x3
x2 x
1 x3 ; 33. (a) Estimate the value of ; 5. Guess the value of the limit fx 28. lim x5 xl 0 261 24. lim cos x 25. lim s x 1 ❙❙❙❙ 16. lim
xl 18. lim
xl 3x
x t2
t 3 lim f x t 22. lim ( x 1 and x x 1 1
1 43. y x 2) f2 0, lim f x x l3 3 and horizontal asymptote y 2x ) ■ x
x 1. x
■ 2 x 46. y 1
■ ■ ■ ■ ■ 2x 2
x2 1
1 44. y x 45. y sx 2 , x l0 43–46  Find the horizontal asymptotes of the curve and use
them, together with concavity and intervals of increase and
decrease, to sketch the curve. s9x 6 x
x3 1
6x 2 lim f x 42. Find a formula for a function that has vertical asymptotes 2
2 , x l3 x2
9x 2 1
s xl xl xl 2 3y 2
5y 2 4y 20. lim (sx 4 3x) conditions:
lim f x
0, 5
4 sx
■ ■ 2 1
■ ■ ■ 5E04(pp 262271) ❙❙❙❙ 262 1/17/06 2:48 PM Page 262 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 47–50  Find the limits as x l and as x l
. Use this
information, together with intercepts, to give a rough sketch of
the graph as in Example 11.
47. y x2 x 48. y 2 x31 49. y x 4 50. y
■ 1
■ 21 5 x xx
■ ■ Then use these sketches to ﬁnd the following limits.
(a) lim x n
(b) lim x n
x l0 x fx
fx
fx
fx
52. f 2 fx
fx
fx 4 3
2
■ 4x
x 5 2 ■ 0, f 2
0 if 0 x
0 if 0 x
0 if 1 x
f x for all ■ ■ ■ ■ ■ ■ 1, f 0
2, f x
1 or if x
4, lim x l
x lim x l
■ ■ ■ ■ ■ ■ ■ (b) What happens to the concentration as t l ? ; 61. Use a graph to ﬁnd a number N such that
6 x 2 5x 3
2x 2 1 ,
0,
x ■ ■ xl ■ values as x l and as x l
.
(a) Describe and compare the end behavior of the functions
Px
3x 5 5x 3 2 x and Q x
3x 5 by graphing both
functions in the viewing rectangles 2, 2 by 2, 2 and
10, 10 by 10,000, 10,000 .
(b) Two functions are said to have the same end behavior if
their ratio approaches 1 as x l . Show that P and Q have
the same end behavior. N 2 illustrate Deﬁnition 5 by ﬁnding values of N that correspond to
0.5 and
0.1. ; 63. For the limit
lim xl s4x 2 1
x1 2 illustrate Deﬁnition 6 by ﬁnding values of N that correspond to
0.5 and
0.1. ; 64. For the limit
lim xl 2x
sx 1
1 illustrate Deﬁnition 7 by ﬁnding a value of N that corresponds
to M 100.
0.0001?
(b) Taking r 2 in Theorem 4, we have the statement
lim x l 1 x 2
0. Prove this directly using Deﬁnition 5. Px
Qx if the degree of P is (a) less than the degree of Q and
(b) greater than the degree of Q. (ii) n
(iv) n x 65. (a) How large do we have to take x so that 1 x 2 57. Let P and Q be polynomials. Find following ﬁve cases:
(i) n 0
(iii) n 0, n even
(v) n 0, n even whenever s4x 2 1
x1 lim ; 56. By the end behavior of a function we mean the behavior of its 58. Make a rough sketch of the curve y 0.2 ; 62. For the limit sin x
.
x
(b) Graph f x
sin x x. How many times does the graph
cross the asymptote? lim 3 2 xl xl 30 t
200 t Ct 55. (a) Use the Squeeze Theorem to evaluate lim ; 5. 0 for x 0, lim x l t x
,
, lim x l 0 t x
, lim x l 0 t x
■ 3x
x2 30 g of salt per liter of water is pumped into the tank at a
rate of 25 L min. Show that the concentration of salt after
t minutes (in grams per liter) is 0,
0 if x 2,
4,
fx
1, f1
0, lim x l 2 f x
, lim x l 2 f x
, lim x l f x
, lim x l f x
lim x l 0 f x
fx
0 for x 2, f x
0 for x 0 and for 0 ■ for all x 4x 2 fx 60. (a) A tank contains 5000 L of pure water. Brine that contains 0, f 0
1, f x
0 if 0 x 2,
0 if x 2, f x
0 if 0 x 4,
0 if x 4, lim x l f x
0,
f x for all x 0, t x
tx 1
x 53. f 1 54. t 0 xl 59. Find lim x l f x if x 51–54  Sketch the graph of a function that satisﬁes all of the
given conditions.
51. f 2 (d) lim x n xl x3 3 x l0 (c) lim x n x n (n an integer) for the
0, n odd
0, n odd 66. (a) How large do we have to take x so that 1 s x (b) Taking r 1
2 0.0001?
in Theorem 4, we have the statement
lim xl 1
sx 0 Prove this directly using Deﬁnition 5.
67. Use Deﬁnition 6 to prove that lim x l 1x 68. Prove, using Deﬁnition 7, that lim x l x 3 0.
. 5E04(pp 262271) 1/17/06 2:48 PM Page 263 S ECTION 4.5 SUMMARY OF CURVE SKETCHING 69. Prove that 263 70. Formulate a precise deﬁnition of lim f x lim f x lim f 1 t xl lim f x xl tl0 lim f 1 t and Then use your deﬁnition to prove that
xl x3 lim 1 tl0 xl if these limits exist.  4.5 ❙❙❙❙ if these limits exist. Summary of Curve Sketching
So far we have been concerned with some particular aspects of curve sketching: domain,
range, and symmetry in Chapter 1; limits, continuity, and asymptotes in Chapter 2; derivatives and tangents in Chapter 3; and extreme values, intervals of increase and decrease, concavity, points of inﬂection, and horizontal asymptotes in this chapter. It is now time to put all
of this information together to sketch graphs that reveal the important features of functions.
You may ask: What is wrong with just using a calculator to plot points and then joining
these points with a smooth curve? To see the pitfalls of this approach, suppose you have
used a calculator to produce the table of values and corresponding points in Figure 1.
y x fx x 22
7
2
4
2
3 1
2
3
4
5
6 20 fx
7
10
11
10
8
8 5
4
3
2
1
0 15
10
_5 _4 _3 _2 _1 5
0 1 2 3 4 5 6 x FIGURE 1 You might then join these points to produce the curve shown in Figure 2, but the cor  rect graph might be the one shown in Figure 3. You can see the drawbacks of the method
of plotting points. Certain essential features of the graph may be missed, such as the maximum and minimum values between 2 and 1 or between 2 and 5. If you just plot points,
you don’t know when to stop. (How far should you plot to the left or right?) But the use
of calculus ensures that all the important aspects of the curve are illustrated.
y y 20 20 15 15 10 10 5 5 _2 _2
0 FIGURE 2 2 4 x 0 FIGURE 3 2 4 x 5E04(pp 262271) 264 ❙❙❙❙ 1/17/06 2:48 PM Page 264 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION You might respond: Yes, but what about graphing calculators and computers? Don’t
they plot such a huge number of points that the sort of uncertainty demonstrated by Figures 2 and 3 is unlikely to happen?
It’s true that modern technology is capable of producing very accurate graphs. But even
the best graphing devices have to be used intelligently. We saw in Section 1.4 that it is
extremely important to choose an appropriate viewing rectangle to avoid getting a misleading graph. (See especially Examples 1, 3, 4, and 5 in that section.) The use of calculus
enables us to discover the most interesting aspects of graphs and in many cases to calculate maximum and minimum points and inﬂection points exactly instead of approximately.
For instance, Figure 4 shows the graph of f x
8 x 3 21x 2 18 x 2. At ﬁrst
glance it seems reasonable: It has the same shape as cubic curves like y x 3, and it
appears to have no maximum or minimum point. But if you compute the derivative, you
will see that there is a maximum when x 0.75 and a minimum when x 1. Indeed, if
we zoom in to this portion of the graph, we see that behavior exhibited in Figure 5. Without
calculus, we could easily have overlooked it.
30 _2 y=8˛21≈+18x+2 8 4
_10 F IGURE 4 y=8˛21≈+18x+2
0 2
6 FIGURE 5 In the next section we will graph functions by using the interaction between calculus
and graphing devices. In this section we draw graphs by ﬁrst considering the following
information. We don’t assume that you have a graphing device, but if you do have one you
should use it as a check on your work. Guidelines for Sketching a Curve
y 0 x (a) Even function: reﬂectional symmetry C. Symmetry y 0 (b) Odd function: rotational symmetry
FIGURE 6 The following checklist is intended as a guide to sketching a curve y f x by hand. Not
every item is relevant to every function. (For instance, a given curve might not have an
asymptote or possess symmetry.) But the guidelines provide all the information you need
to make a sketch that displays the most important aspects of the function.
A. Domain It’s often useful to start by determining the domain D of f , that is, the set of values of x for which f x is deﬁned.
B. Intercepts The yintercept is f 0 and this tells us where the curve intersects the yaxis.
To ﬁnd the xintercepts, we set y 0 and solve for x. (You can omit this step if the
equation is difﬁcult to solve.) x (i) If f x
f x for all x in D, that is, the equation of the curve is unchanged
when x is replaced by x, then f is an even function and the curve is symmetric about
the yaxis. This means that our work is cut in half. If we know what the curve looks like
for x 0, then we need only reﬂect about the yaxis to obtain the complete curve [see
Figure 6(a)]. Here are some examples: y x 2, y x 4, y
x , and y cos x.
(ii) If f x
f x for all x in D, then f is an odd function and the curve is
symmetric about the origin. Again we can obtain the complete curve if we know what 5E04(pp 262271) 1/17/06 2:48 PM Page 265 ❙❙❙❙ S ECTION 4.5 SUMMARY OF CURVE SKETCHING 265 it looks like for x 0. [Rotate 180° about the origin; see Figure 6(b).] Some simple
examples of odd functions are y x, y x 3, y x 5, and y sin x.
(iii) If f x p
f x for all x in D, where p is a positive constant, then f is called
a periodic function and the smallest such number p is called the period. For instance,
y sin x has period 2 and y tan x has period . If we know what the graph looks
like in an interval of length p, then we can use translation to sketch the entire graph (see
Figure 7).
y FIGURE 7 Periodic function:
translational symmetry ap 0 a a+p a+2p x D. Asymptotes (i) Horizontal Asymptotes. Recall from Section 4.4 that if either lim x l f x
L
or lim x l f x
L, then the line y L is a horizontal asymptote of the curve
y f x . If it turns out that lim x l f x
(or
), then we do not have an asymptote to the right, but that is still useful information for sketching the curve.
(ii) Vertical Asymptotes. Recall from Section 2.2 that the line x a is a vertical
asymptote if at least one of the following statements is true:
1 lim f x xla lim f x xla E. F. G.
In Module 4.5 you can practice using
information about f , f , and asymptotes to determine the shape of the
graph of f . H. lim f x xla lim f x xla (For rational functions you can locate the vertical asymptotes by equating the denominator to 0 after canceling any common factors. But for other functions this method does
not apply.) Furthermore, in sketching the curve it is very useful to know exactly which
of the statements in (1) is true. If f a is not deﬁned but a is an endpoint of the domain
of f , then you should compute lim x l a f x or lim x l a f x , whether or not this limit
is inﬁnite.
(iii) Slant Asymptotes. These are discussed at the end of this section.
Intervals of Increase or Decrease Use the I/D Test. Compute f x and ﬁnd the intervals
on which f x is positive ( f is increasing) and the intervals on which f x is negative
( f is decreasing).
Local Maximum and Minimum Values Find the critical numbers of f [the numbers c where
fc
0 or f c does not exist]. Then use the First Derivative Test. If f changes from
positive to negative at a critical number c, then f c is a local maximum. If f changes
from negative to positive at c, then f c is a local minimum. Although it is usually
preferable to use the First Derivative Test, you can use the Second Derivative Test if c
is a critical number such that f c
0. Then f c
0 implies that f c is a local
minimum, whereas f c
0 implies that f c is a local maximum.
Concavity and Points of Inflection Compute f x and use the Concavity Test. The curve is
concave upward where f x
0 and concave downward where f x
0. Inﬂection
points occur where the direction of concavity changes.
Sketch the Curve Using the information in items A–G, draw the graph. Sketch the asymptotes as dashed lines. Plot the intercepts, maximum and minimum points, and inﬂection 5E04(pp 262271) 266 ❙❙❙❙ 1/17/06 2:48 PM Page 266 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION points. Then make the curve pass through these points, rising and falling according to E,
with concavity according to G, and approaching the asymptotes. If additional accuracy is
desired near any point, you can compute the value of the derivative there. The tangent indicates the direction in which the curve proceeds.
2x 2 EXAMPLE 1 Use the guidelines to sketch the curve y x 2 1 . A. The domain is x x2 1 0 xx 1 , 1 1, 1 1, B. The x and yintercepts are both 0.
C. Since f
x
f x , the function f is even. The curve is symmetric about the yaxis. D. lim xl x 2 lim 1 xl 2
1 x2 1 2 Therefore, the line y 2 is a horizontal asymptote.
Since the denominator is 0 when x
1, we compute the following limits: y y=2 lim x l1 0 x lim x=_1 2x 2 xl 1 x=1 2x 2
x2 lim 1 x l1 2x 2
x2 lim 1 xl 1 2x 2
x2 1 2x 2
x2 1 Therefore, the lines x 1 and x
1 are vertical asymptotes. This information
about limits and asymptotes enables us to draw the preliminary sketch in Figure 8,
showing the parts of the curve near the asymptotes. FIGURE 8 Preliminary sketch
 We have shown the curve approaching its
horizontal asymptote from above in Figure 8.
This is conﬁrmed by the intervals of increase and
decrease. E. fx 4x x 2 2x 2 2x 1
x 2 1 4x
x
1 2 2 2 Since f x
0 when x 0 x
1 and f x
0 when x 0 x 1 , f is
increasing on
, 1 and 1, 0 and decreasing on 0, 1 and 1, .
F. The only critical number is x
0. Since f changes from positive to negative at 0,
f0
0 is a local maximum by the First Derivative Test.
y G.
y=2 4 x2 fx
Since 12 x 2 4 1 2 4x 2 x 2
x2 1 4 1 2x 12 x 2 4
x2 1 3 0 for all x, we have 0
x x=_1 x=1 FIGURE 9 Finished sketch of y= 2≈
≈1 fx 0 &? x2 1 0 &? x 1 and f x
0 &? x
1. Thus, the curve is concave upward on the intervals
, 1 and 1,
and concave downward on 1, 1 . It has no point of inﬂection
since 1 and 1 are not in the domain of f .
H. Using the information in E–G, we ﬁnish the sketch in Figure 9. 5E04(pp 262271) 1/17/06 2:49 PM Page 267 S ECTION 4.5 SUMMARY OF CURVE SKETCHING Domain
xx 1 0
xx
The x and yintercepts are both 0.
Symmetry: None
Since 1 lim xl 1, x2
sx 1 there is no horizontal asymptote. Since s x
positive, we have
lim xl 1 and so the line x
E. G. y= FIGURE 10 0 1 l 0 as x l 1 and f x is always x2
sx 1 1 is a vertical asymptote.
2x sx fx 1 x 2 1 (2 s x
x1 1) x 3x 4
2x 132 We see that f x
0 when x 0 (notice that 4 is not in the domain of f ), so the
3
only critical number is 0. Since f x
0 when 1 x 0 and f x
0 when
x 0, f is decreasing on 1, 0 and increasing on 0, .
F. Since f 0
0 and f changes from negative to positive at 0, f 0
0 is a local
(and absolute) minimum by the First Derivative Test. y x=_1 267 x2
.
sx 1 EXAMPLE 2 Sketch the graph of f x
A.
B.
C.
D. ❙❙❙❙ ≈
œ„„„„
x+1
x fx 2x 1 32 6x 4
4x 3x 2
13 4x 3 x 1 12 3x 2 8x
4x 1 8
52 Note that the denominator is always positive. The numerator is the quadratic
3x 2 8 x 8, which is always positive because its discriminant is b 2 4ac
32,
which is negative, and the coefﬁcient of x 2 is positive. Thus, f x
0 for all x in the
domain of f , which means that f is concave upward on 1,
and there is no point
of inﬂection.
H. The curve is sketched in Figure 10.
EXAMPLE 3 Sketch the graph of f x
A. The domain is .
B. The yintercept is f 0 2 cos x sin 2 x 2 cos x sin 2 x. 2. The xintercepts occur when
2 cos x 2 sin x cos x 2 cos x 1 sin x 0 that is, when cos x 0 or sin x
1. Thus, in the interval 0, 2 , the xintercepts
are 2 and 3 2.
C. f is neither even nor odd, but f x
2
f x for all x and so f is periodic and has
period 2 . Thus, in what follows we need to consider only 0 x 2 and then
extend the curve by translation in H.
D. Asymptotes: None 5E04(pp 262271) ❙❙❙❙ 268 1/17/06 2:49 PM Page 268 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION E. fx 2 sin x 2 cos 2 x 2 2 sin2x 21 2 sin2x 2 2 sin x 1 sin x 2 sin x sin x 1 1 Thus, f x
0 when sin x 1 or sin x
1, so in 0, 2 we have x
6,
2
5 6, and 3 2. In determining the sign of f x in the following chart, we use the
fact that sin x 1 0 for all x.
Interval f fx increasing on 0, 0 x 6 x 5 6
6 decreasing on 6 5 6 x 3 2 increasing on 5 6, 3 3 2 x 2 increasing on 3 2, 2 6, 5 6
2 F. From the chart in E the First Derivative Test says that f 6
3 s3 2 is a local
maximum and f 5 6
3 s3 2 is a local minimum, but f has no maximum or
minimum at 3 2, only a horizontal tangent.
fx G. 2 cos x 4 sin 2 x 2 cos x 1 4 sin x 1
Thus, f x
0 when cos x 0 (so x
2 or 3 2) and when sin x
4.
From Figure 11 we see that there are two values of x between 0 and 2 for which
1
sin x
x
0 on
2, 1 and 3 2, 2 ,
2. Then f
4 . Let’s call them 1 and
so f is concave upward there. Also f x
0 on 0, 2 , 1, 3 2 , and 2 , 2 ,
so f is concave downward there. Inﬂection points occur when x
2, 1, 3 2,
and 2. y y=Ã x
arcsin 1
4 1 2 2 arcsin 1
4 å¡ 0 å™ 1
_4 2π x F IGURE 11 H. The graph of the function restricted to 0 x 2 is shown in Figure 12. Then it is
extended, using periodicity, to the complete graph in Figure 13. y y
π ”6 , 3œ3
„
2 ’ 2 2 å¡
0 π
6 π
2 5π π
6 3π
2 ”
FIGURE 12 å™
2π x _ 3π
2 _π
2 0 π
2 3π
2 5π
2 y=2 Ł x+Ã 2x 5π
3œ3
„
, _ 2 ’
6 FIGURE 13 7π
2 9π
2 x 5E04(pp 262271) 1/17/06 2:50 PM Page 269 ❙❙❙❙ S ECTION 4.5 SUMMARY OF CURVE SKETCHING 269 Slant Asymptotes
y Some curves have asymptotes that are oblique, that is, neither horizontal nor vertical. If
y=ƒ lim f x mx xl ƒ(mx+b) 0 then the line y m x b is called a slant asymptote because the vertical distance
between the curve y f x and the line y m x b approaches 0, as in Figure 14. (A
similar situation exists if we let x l
.) For rational functions, slant asymptotes occur
when the degree of the numerator is one more than the degree of the denominator. In such
a case the equation of the slant asymptote can be found by long division as in the following example. y=mx+b
0 b x FIGURE 14 x3 EXAMPLE 4 Sketch the graph of f x
A.
B.
C.
D. x 2 1 . ,.
The domain is
The x and yintercepts are both 0.
f x , f is odd and its graph is symmetric about the origin.
Since f x
Since x 2 1 is never 0, there is no vertical asymptote. Since f x l as x l and
fxl
as x l
, there is no horizontal asymptote. But long division gives
x3 fx fx So the line y x 2 x x 1 x 2 1
1
x x x x2 1 1
x2 1 l0 as xl x is a slant asymptote. E. 3x 2 x 2 fx
Since f x F. Although f 0 x 3 2x 1
x 2 1 x2 x2 3
x2 1 2 2 0 for all x (except 0), f is increasing on
,.
0, f does not change sign at 0, so there is no local maximum or minimum.
G.
y ˛
y=
≈+1 ”œ„ , 3 ”_œ„ , _
3 3œ3
„
’
4 inflection
points 0 when x Interval x x 0 6x x 2 1 2 x Since f x 3œ3
„
’
4 x 4x 3 fx 2 0 or x
3 x4
1 3x 2 2 x2 1 2x 4 2x 3
x2 s3, we set up the following chart:
x2 x2 1 3 f fx
CU on ( s3 s3 ) , s3 x 0 CD on ( s3, 0) 0 x s3 CU on (0, s3 ) x s3 CD on (s3, y=x The points of inﬂection are ( s3,
FIGURE 15 x2
13 3 s3 4), 0, 0 , and (s3, 3 s3 4). H. The graph of f is sketched in Figure 15. ) 5E04(pp 262271) ❙❙❙❙ 270 1/17/06  1. y Exercises 3 9x x 7. y 2x5 5x2
x
x1
1
x2 9
x
x2 9
x1
x2 13. y 15. y 4x x 3 1 x 2 2 20 x 3 3 x 5
x
x 12
x
x2 9
x2
x2 9
x2 2
x4 14. y 16. y x 21. y sx 2 1 x2 27. y x x
3x1 29. y x 3
s x2 y L 42. Coulomb’s Law states that the force of attraction between two charged particles is directly proportional to the product of the
charges and inversely proportional to the square of the distance
between them. The ﬁgure shows particles with charge 1 located
at positions 0 and 2 on a coordinate line and a particle with
charge 1 at a position x between them. It follows from
Coulomb’s Law that the net force acting on the middle particle
is
k
k
Fx
0x2
x2
x 22 1
1
x
x
5 xs2 x2 where k is a positive constant. Sketch the graph of the net force
function. What does the graph say about the force?
+1 sx 2
x5 3 28. y 3 sin x 32. y sin x 33. y 35. y
36. y cos x 37. y sin 2 x ■ ■ 2 sin x, x 1 sin x x
x x 2 3 ■ 38. y 2 sin x sin x
cos x
■ sin x 40. y
■ ■ ■ 2
■ x cos x
sin x
■ 2x 3 44. y ■ ■ ■ WL 3
x
12EI W L2 2
x
24EI 2x 2 5x
2x 1 49. xy
■ ■ walls. If a constant load W is distributed evenly along its
length, the beam takes the shape of the deﬂection curve
W
x4
24EI 1
1 ■ 47. y 41. The ﬁgure shows a beam of length L embedded in concrete y x x
5x 4
x3 46. y
■ ■ x2 ■ 2 x
2x x2
x2
■ 3 x
2
■ ■ ■ 47–52  Use the guidelines of this section to sketch the curve. In
guideline D ﬁnd an equation of the slant asymptote. 2 sin x ■ 4x 3 2x 2 5
2x 2 x 3 2 2
0 x2
x 45. y 3 tan x, 2 39. y 43. y tan x 2x
1
2 2 Find an equation of the slant asymptote. Do not sketch  2 x tan x, 34. y 43–46 +1 x the curve. 1 31. y 1
5x 2 3 _1 0 x 26. y 3 W 0 sx 30. y 9x 3 x 24. y 1 s1 2 sx 22. y x x
sx 2 x3
x3 20. y 3 x s5 25. y xx 12. y 3 18. y 19. y 23. y x x 10. y x2 17. y 8x 4 8. y 2 3 5. y 11. y 4. y 2 6. y 15x
4 6x 2 2. y x 2 9. y where E and I are positive constants. (E is Young’s modulus of
elasticity and I is the moment of inertia of a crosssection of
the beam.) Sketch the graph of the deﬂection curve. Use the guidelines of this section to sketch the curve.
x 3. y Page 270 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION  4.5
1–40 2:50 PM x2 1 x2
x 48. y
50. xy x2 52. y 4 12
2 x
x x 1 ■ 2x 3 51. y
■ x2
x ■ 2 ■ 1
1
■ ■ 53. Show that the curve y y 2 x and y ■ ■ ■ 1
1
■ 3
2 ■ ■ s 4 x 2 9 has two slant asymptotes:
2 x. Use this fact to help sketch the curve. ■ 5E04(pp 262271) 1/17/06 2:50 PM Page 271 S ECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS 54. Show that the curve y yx
curve. 2 and y b a x and y
asymptotes of the hyperbola x 2 a 2 56. Let f x x3 lim  4.6 x 4 1 x in the
same manner as in Exercise 56. Then use your results to help
sketch the graph of f . 57. Discuss the asymptotic behavior of f x b a x are slant
y2 b2
1. 1 x. Show that
xl fx x2 271 This shows that the graph of f approaches the graph of y x 2,
and we say that the curve y f x is asymptotic to the
parabola y x 2. Use this fact to help sketch the graph of f . sx 2 4 x has two slant asymptotes:
x 2. Use this fact to help sketch the 55. Show that the lines y ❙❙❙❙ cos x 1 x 2 to sketch
its graph without going through the curvesketching procedure
of this section. 58. Use the asymptotic behavior of f x 0 Graphing with Calculus and Calculators  If you have not already read Section 1.4, you
should do so now. In particular, it explains how
to avoid some of the pitfalls of graphing devices
by choosing appropriate viewing rectangles. The method we used to sketch curves in the preceding section was a culmination of much
of our study of differential calculus. The graph was the ﬁnal object that we produced. In
this section our point of view is completely different. Here we start with a graph produced
by a graphing calculator or computer and then we reﬁne it. We use calculus to make sure
that we reveal all the important aspects of the curve. And with the use of graphing devices
we can tackle curves that would be far too complicated to consider without technology.
The theme is the interaction between calculus and calculators.
2 x 6 3x 5 3x 3 2 x 2. Use the graphs of f
and f to estimate all maximum and minimum points and intervals of concavity.
EXAMPLE 1 Graph the polynomial f x SOLUTION If we specify a domain but not a range, many graphing devices will deduce a
suitable range from the values computed. Figure 1 shows the plot from one such device
if we specify that 5 x 5. Although this viewing rectangle is useful for showing
that the asymptotic behavior (or end behavior) is the same as for y 2 x 6, it is obviously
hiding some ﬁner detail. So we change to the viewing rectangle 3, 2 by 50, 100
shown in Figure 2.
100 41,000
y=ƒ
y=ƒ _5 _3 2 5
_50 _1000 FIGURE 2 FIGURE 1 From this graph it appears that there is an absolute minimum value of about 15.33
when x
1.62 (by using the cursor) and f is decreasing on
, 1.62 and increasing on 1.62, . Also there appears to be a horizontal tangent at the origin and inﬂection points when x 0 and when x is somewhere between 2 and 1.
Now let’s try to conﬁrm these impressions using calculus. We differentiate and get
fx 12 x 5 15x 4 9x 2 4x fx 60 x 4 60 x 3 18 x 4 5E04(pp 272281) 272 ❙❙❙❙ 1/17/06 2:54 PM Page 272 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION When we graph f in Figure 3 we see that f x changes from negative to positive when
x
1.62; this conﬁrms (by the First Derivative Test) the minimum value that we
found earlier. But, perhaps to our surprise, we also notice that f x changes from positive to negative when x 0 and from negative to positive when x 0.35. This means
that f has a local maximum at 0 and a local minimum when x 0.35, but these were
hidden in Figure 2. Indeed, if we now zoom in toward the origin in Figure 4, we see
what we missed before: a local maximum value of 0 when x 0 and a local minimum
value of about 0.1 when x 0.35.
20 1
y=ƒ y=fª(x)
_1
_3 1 2
_5 _1 FIGURE 3
10
_3 2
y=f·(x) _30 FIGURE 5 FIGURE 4 What about concavity and inﬂection points? From Figures 2 and 4 there appear to be
inﬂection points when x is a little to the left of 1 and when x is a little to the right of 0.
But it’s difﬁcult to determine inﬂection points from the graph of f , so we graph the second derivative f in Figure 5. We see that f changes from positive to negative when
x
1.23 and from negative to positive when x 0.19. So, correct to two decimal
places, f is concave upward on
and concave downward on
, 1.23 and 0.19,
1.23, 0.19 . The inﬂection points are 1.23, 10.18 and 0.19, 0.05 .
We have discovered that no single graph reveals all the important features of this
polynomial. But Figures 2 and 4, when taken together, do provide an accurate picture.
EXAMPLE 2 Draw the graph of the function fx x2 7x
x2 3 in a viewing rectangle that contains all the important features of the function. Estimate
the maximum and minimum values and the intervals of concavity. Then use calculus to
ﬁnd these quantities exactly.
SOLUTION Figure 6, produced by a computer with automatic scaling, is a disaster. Some
graphing calculators use 10, 10 by 10, 10 as the default viewing rectangle, so
let’s try it. We get the graph shown in Figure 7; it’s a major improvement.
3 10 10!* y=ƒ
_10 y=ƒ _5 10 5
_10 FIGURE 6 FIGURE 7 5E04(pp 272281) 1/17/06 2:54 PM Page 273 S ECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS ❙❙❙❙ 273 The yaxis appears to be a vertical asymptote and indeed it is because
lim x2 7x
x2 xl0 3 Figure 7 also allows us to estimate the xintercepts: about 0.5 and 6.5. The
exact values are obtained by using the quadratic formula to solve the equation
x 2 7x 3 0; we get x ( 7 s37 ) 2.
To get a better look at horizontal asymptotes, we change to the viewing rectangle
20, 20 by 5, 10 in Figure 8. It appears that y 1 is the horizontal asymptote and
this is easily conﬁrmed: 10
y=ƒ
y=1
_20 lim 20 x2 7x
x2 xl _5 3 lim xl 7
x 1 3
x2 1 To estimate the minimum value we zoom in to the viewing rectangle 3, 0 by
4, 2 in Figure 9. The cursor indicates that the absolute minimum value is about 3.1
when x
0.9, and we see that the function decreases on
, 0.9 and 0,
and
increases on 0.9, 0 . The exact values are obtained by differentiating: FIGURE 8
2 _3 0 y=ƒ
_4 FIGURE 9 fx 7
x2 6
x3 7x 6
x 3 6
6
This shows that f x
0 when x
0 when 7 x 0 and f x
7 and when
6
37
x 0. The exact minimum value is f ( 7 )
3.08.
12
Figure 9 also shows that an inﬂection point occurs somewhere between x
1 and
x
2. We could estimate it much more accurately using the graph of the second derivative, but in this case it’s just as easy to ﬁnd exact values. Since fx 14
x3 18
x4 2 7x 9
x 4 9
0 when x
0 . So f is concave upward on ( 9 , 0) and
we see that f x
7x
7
9
0,
and concave downward on ( , 7 ). The inﬂection point is ( 9 , 71 ).
7
27
The analysis using the ﬁrst two derivatives shows that Figures 7 and 8 display all the
major aspects of the curve. E XAMPLE 3 Graph the function f x
10 y=ƒ
_10 10 SOLUTION Drawing on our experience with a rational function in Example 2, let’s start by
graphing f in the viewing rectangle 10, 10 by 10, 10 . From Figure 10 we have
the feeling that we are going to have to zoom in to see some ﬁner detail and also zoom
out to see the larger picture. But, as a guide to intelligent zooming, let’s ﬁrst take a close
look at the expression for f x . Because of the factors x 2 2 and x 4 4 in the
denominator, we expect x 2 and x 4 to be the vertical asymptotes. Indeed _10 FIGURE 10 x2 x 1 3
.
x 22x 44 lim
x l2 x2 x 1 3
x 22x 4 4 and lim xl4 x2 x 1 3
x 22x 4 4 5E04(pp 272281) 274 ❙❙❙❙ 1/17/06 2:54 PM Page 274 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION To ﬁnd the horizontal asymptotes we divide numerator and denominator by x 6 :
x2 x 1 3
x 22x 4 1
1
x
4 2 2
x 1 1
x
1 3 4
x 4 l0 as xl so the xaxis is the horizontal asymptote.
It is also very useful to consider the behavior of the graph near the xintercepts using
an analysis like that in Example 11 in Section 4.4. Since x 2 is positive, f x does not
change sign at 0 and so its graph doesn’t cross the xaxis at 0. But, because of the factor
x 1 3, the graph does cross the xaxis at 1 and has a horizontal tangent there.
Putting all this information together, but without using derivatives, we see that the curve
has to look something like the one in Figure 11.
y _1 FIGURE 11 1 2 3 4 x Now that we know what to look for, we zoom in (several times) to produce the graphs
in Figures 12 and 13 and zoom out (several times) to get Figure 14.
0.05 0.0001 500
y=ƒ y=
_100 1 _1.5 0.5 y=ƒ
_0.05 FIGURE 12 _0.0001 FIGURE 13 _1 _10 10 FIGURE 14 We can read from these graphs that the absolute minimum is about 0.02 and occurs
when x
20. There is also a local maximum 0.00002 when x
0.3 and a local
minimum 211 when x 2.5. These graphs also show three inﬂection points near 35,
5, and 1 and two between 1 and 0. To estimate the inﬂection points closely we
would need to graph f , but to compute f by hand is an unreasonable chore. If you
have a computer algebra system, then it’s easy to do (see Exercise 15).
We have seen that, for this particular function, three graphs (Figures 12, 13, and 14)
are necessary to convey all the useful information. The only way to display all these
features of the function on a single graph is to draw it by hand. Despite the exaggerations and distortions, Figure 11 does manage to summarize the essential nature of the
function.
EXAMPLE 4 Graph the function f x
, locate all maxisin x sin 2 x . For 0 x
mum and minimum values, intervals of increase and decrease, and inﬂection points correct to one decimal place. 5E04(pp 272281) 1/17/06 2:55 PM Page 275 SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS  The family of functions
fx
sin x sin c x
where c is a constant, occurs in applications to
frequency modulation (FM) synthesis. A sine
wave is modulated by a wave with a different
frequency sin c x . The case where c 2 is
studied in Example 4. Exercise 19 explores
another special case.
1.1 ❙❙❙❙ 275 S OLUTION We ﬁrst note that f is periodic with period 2 . Also, f is odd and f x
1
for all x. So the choice of a viewing rectangle is not a problem for this function: We start
with 0,
by 1.1, 1.1 . (See Figure 15.) It appears that there are three local maximum values and two local minimum values in that window. To conﬁrm this and locate
them more accurately, we calculate that fx cos x sin 2 x 1 2 cos 2 x and graph both f and f in Figure 16. Using zoomin and the First Derivative Test, we
ﬁnd the following values to one decimal place.
Intervals of increase: π 0, 0.6 , 1.0, 1.6 , 2.1, 2.5 Intervals of decrease: 0 0.6, 1.0 , 1.6, 2.1 , 2.5, Local maximum values: f 0.6 FIGURE 15 1, f 1.6 Local minimum values: _1.1 0.94, f 2.1 f 1.0 1, f 2.5 1 0.94 The second derivative is 1.2
y=ƒ
0 fx
π 1 2 cos 2 x 2 sin x sin 2 x 4 sin 2 x cos x sin 2 x Graphing both f and f in Figure 17, we obtain the following approximate values: y=f ª(x) Concave upward on:
Concave downward on: FIGURE 16 0, 0.8 , 1.3, 1.8 , 2.3, Inflection points: _1.2 0.8, 1.3 , 1.8, 2.3 0, 0 , 0.8, 0.97 , 1.3, 0.97 , 1.8, 0.97 , 2.3, 0.97 1.2 1.2
f 0 π _2π 2π f·
_1.2 FIGURE 17 _1.2 FIGURE 18 Having checked that Figure 15 does indeed represent f accurately for 0 x
we can state that the extended graph in Figure 18 represents f accurately for
2
x 2. , Our ﬁnal example is concerned with families of functions. As discussed in Section 1.4,
this means that the functions in the family are related to each other by a formula that
contains one or more arbitrary constants. Each value of the constant gives rise to a member of the family and the idea is to see how the graph of the function changes as the constant changes. 5E04(pp 272281) 276 ❙❙❙❙ 1/17/06 2:55 PM Page 276 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION _5 4 lim xl _2 c=2 y=
2 1
≈+2x2 lim 4 x 1
2x 2 0 c for any value of c, they all have the xaxis as a horizontal asymptote. A vertical asymptote will occur when x 2 2 x c 0. Solving this quadratic equation, we get
x
1 s1 c. When c 1, there is no vertical asymptote (as in Figure 19). When
c 1, the graph has a single vertical asymptote x
1 because
xl 1 _5 c vary as c varies? 2x SOLUTION The graphs in Figures 19 and 20 (the special cases c
2 and c
2) show
two very differentlooking curves. Before drawing any more graphs, let’s see what members of this family have in common. Since 1
y=
≈+2x+2 F IGURE 19 1 x2 EXAMPLE 5 How does the graph of f x 2 x 1
2x 2 lim 1 xl 1 1
x When c 1, there are two vertical asymptotes: x
Now we compute the derivative:
fx 2x
x2 1
1 2 s1 c (as in Figure 20). 2
2x c 2 _2 FIGURE 20 c=_2 See an animation of Figure 21.
Resources / Module 5
/ Max and Min
/ Families of Functions c=_1
FIGURE 21 This shows that f x
0 when x
1 (if c 1), f x
0 when x
1, and
fx
0 when x
1. For c 1, this means that f increases on
, 1 and
decreases on 1, . For c 1, there is an absolute maximum value
f1
1 c 1 . For c 1, f 1
1 c 1 is a local maximum value and the
intervals of increase and decrease are interrupted at the vertical asymptotes.
Figure 21 is a “slide show” displaying ﬁve members of the family, all graphed in the
viewing rectangle 5, 4 by 2, 2 . As predicted, c 1 is the value at which a transition takes place from two vertical asymptotes to one, and then to none. As c increases
from 1, we see that the maximum point becomes lower; this is explained by the fact that
1 c 1 l 0 as c l . As c decreases from 1, the vertical asymptotes become more
widely separated because the distance between them is 2 s1 c, which becomes large
as c l
. Again, the maximum point approaches the xaxis because 1 c 1 l 0
as c l
. c=0 c=1 c=2 c=3 The family of functions ƒ=1/(≈+2x+c) There is clearly no inﬂection point when c
fx 2 3x 2
x2 1. For c
6x
2x 4
c 1 we calculate that
c 3 and deduce that inﬂection points occur when x
1 s3 c 1 3. So the inﬂection
points become more spread out as c increases and this seems plausible from the last two
parts of Figure 21. 5E04(pp 272281) 1/17/06 2:56 PM Page 277 SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS  4.6
1–8  Produce graphs of f that reveal all the important aspects of
the curve. In particular, you should use graphs of f and f to estimate the intervals of increase and decrease, extreme values, intervals of concavity, and inﬂection points. 4x 4 2. f x x6 3. f x 3 4. f x
5. f x
6. f x sx
x 32 x 3 2 95 x
125 x 3 x 5 x3 sin x
x 4x ■ 17–18  Use a computer algebra system to graph f and to ﬁnd f
and f . Use graphs of these derivatives to estimate the intervals of
increase and decrease, extreme values, intervals of concavity, and
inﬂection points of f . sin 2 x
,
sx 2 1 18. f x 5 cos x 8. f x graphs to estimate the intervals of increase and decrease and
concavity of f . 2x 1
4
sx 4 x 1 3 tan x
2 16. If f is the function of Exercise 14, ﬁnd f and f and use their 17. f x 2x 2 2
x
x2
x
x 2 4x 1
x CAS CAS 29 2 x ■ 75 x 4 3x 7. f x ■ 89 x 2 15 x 5 4 ■ 4 7 cos x, x 10. f x 8x 3 x
■ ■ ■ ■ ■ ■ ■ ■ ■ x ■ 2
■ x
■ 2
■ ■ ■ ■ ■ ■ 13–14  Sketch the graph by hand using asymptotes and intercepts, but not derivatives. Then use your sketch as a guide to producing graphs (with a graphing device) that display the major
features of the curve. Use these graphs to estimate the maximum
and minimum values. x 13. f x CAS ■ ■ ■ ■ ■ ■ ■ 4x 3
xx 1  Describe how the graph of f varies as c varies. Graph
several members of the family to illustrate the trends that you discover. In particular, you should investigate how maximum and
minimum points and inﬂection points move when c changes. You
should also identify any transitional values of c at which the basic
shape of the curve changes. 20. f x x3 22. f x x 2sc 2 1 23. f x x2 cx 2
cx
c 2x 2 1 24. f x 2 x4 21. f x cx 1 x 22 cx 2 4 10x x 1 4
x 23x 1 14. f x
■ ■ 20–25 2 2 sin x, ■ ■ 20 x s9
x 10 11 x
x2 12. f x ■ 3 tions f x
sin x sin c x that occur in FM synthesis. Here
we investigate the function with c 3. Start by graphing f in
the viewing rectangle 0,
by 1.2, 1.2 . How many local
maximum points do you see? The graph has more than are visible to the naked eye. To discover the hidden maximum and
minimum points you will need to examine the graph of f very
carefully. In fact, it helps to look at the graph of f at the same
time. Find all the maximum and minimum values and inﬂection
points. Then graph f in the viewing rectangle 2 , 2 by
1.2, 1.2 and comment on symmetry. 2
■ 3x 2 11. f x ■ x 19. In Example 4 we considered a member of the family of func 2
■ x2 ■ 0 4 9–12  Produce graphs of f that reveal all the important aspects of
the curve. Estimate the intervals of increase and decrease, extreme
values, intervals of concavity, and inﬂection points, and use calculus to ﬁnd these quantities exactly. 9. f x 277 Exercises ; 1. f x ❙❙❙❙ ■ ■ ■ 25. f x
■ ■ cx
■ sin x
■ ■ ■ ■ ■ ■ ■ ■ 2 ■ ■ ■ ■ ■ ■ 15. If f is the function considered in Example 3, use a computer algebra system to calculate f and then graph it to conﬁrm
that all the maximum and minimum values are as given in the
example. Calculate f and use it to estimate the intervals of
concavity and inﬂection points. ■ 26. Investigate the family of curves given by the equation fx
x 4 c x 2 x. Start by determining the transitional
value of c at which the number of inﬂection points changes.
Then graph several members of the family to see what shapes
are possible. There is another transitional value of c at which
the number of critical numbers changes. Try to discover it
graphically. Then prove what you have discovered. ■ 5E04(pp 272281) 278 ❙❙❙❙ 1/17/06 2:56 PM Page 278 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 27. (a) Investigate the family of polynomials given by the equation fx
c x 4 2 x 2 1. For what values of c does the curve
have minimum points?
(b) Show that the minimum and maximum points of every
curve in the family lie on the parabola y 1 x 2. Illustrate by graphing this parabola and several members of the
family.  4.7 28. (a) Investigate the family of polynomials given by the equation fx
2 x 3 c x 2 2 x. For what values of c does the
curve have maximum and minimum points?
(b) Show that the minimum and maximum points of every
curve in the family lie on the curve y x x 3. Illustrate
by graphing this curve and several members of the family. Optimization Problems
The methods we have learned in this chapter for ﬁnding extreme values have practical
applications in many areas of life. A businessperson wants to minimize costs and maximize proﬁts. A traveler wants to minimize transportation time. Fermat’s Principle in optics
states that light follows the path that takes the least time. In this section and the next we
solve such problems as maximizing areas, volumes, and proﬁts and minimizing distances,
times, and costs.
In solving such practical problems the greatest challenge is often to convert the word
problem into a mathematical optimization problem by setting up the function that is to be
maximized or minimized. Let’s recall the problemsolving principles discussed on page 58
and adapt them to this situation:
Steps in Solving Optimization Problems
1. Understand the Problem The ﬁrst step is to read the problem carefully until it is clearly 2.
3. 4.
5. 6. understood. Ask yourself: What is the unknown? What are the given quantities?
What are the given conditions?
Draw a Diagram In most problems it is useful to draw a diagram and identify the
given and required quantities on the diagram.
Introduce Notation Assign a symbol to the quantity that is to be maximized or minimized (let’s call it Q for now). Also select symbols a, b, c, . . . , x, y for other
unknown quantities and label the diagram with these symbols. It may help to use
initials as suggestive symbols—for example, A for area, h for height, t for time.
Express Q in terms of some of the other symbols from Step 3.
If Q has been expressed as a function of more than one variable in Step 4, use the
given information to ﬁnd relationships (in the form of equations) among these
variables. Then use these equations to eliminate all but one of the variables in the
expression for Q. Thus, Q will be expressed as a function of one variable x, say,
Q f x . Write the domain of this function.
Use the methods of Sections 4.1 and 4.3 to ﬁnd the absolute maximum or minimum value of f . In particular, if the domain of f is a closed interval, then the
Closed Interval Method in Section 4.1 can be used. EXAMPLE 1 A farmer has 2400 ft of fencing and wants to fence off a rectangular ﬁeld that
borders a straight river. He needs no fence along the river. What are the dimensions of
the ﬁeld that has the largest area?
 Understand the problem
 Analogy: Try special cases
 Draw diagrams SOLUTION In order to get a feeling for what is happening in this problem, let’s experiment
with some special cases. Figure 1 (not to scale) shows three possible ways of laying out
the 2400 ft of fencing. We see that when we try shallow, wide ﬁelds or deep, narrow 5E04(pp 272281) 1/17/06 2:56 PM Page 279 S ECTION 4.7 OPTIMIZATION PROBLEMS ❙❙❙❙ 279 ﬁelds, we get relatively small areas. It seems plausible that there is some intermediate
conﬁguration that produces the largest area.
400 1000
2200 700 100 1000 700 1000 100 Area=700 · 1000=700,000 ft@ Area=100 · 2200=220,000 ft@ Area=1000 · 400=400,000 ft@ FIGURE 1 Figure 2 illustrates the general case. We wish to maximize the area A of the rectangle.
Let x and y be the depth and width of the rectangle (in feet). Then we express A in terms
of x and y:
A xy  Introduce notation y
x A x We want to express A as a function of just one variable, so we eliminate y by expressing
it in terms of x. To do this we use the given information that the total length of the fencing is 2400 ft. Thus
2 x y 2400
From this equation we have y FIGURE 2 A
Note that x
mize is 0 and x 2 x, which gives x 2400 2x 2400 x 1200 (otherwise A
Ax The derivative is A x
equation 2400 2400 0). So the function that we wish to maxi 2x 2 2400 x 2x 2 0 x 1200 4 x, so to ﬁnd the critical numbers we solve the
2400 4x 0 which gives x 600. The maximum value of A must occur either at this critical number
or at an endpoint of the interval. Since A 0
0, A 600
720,000, and A 1200
0,
the Closed Interval Method gives the maximum value as A 600
720,000.
[Alternatively, we could have observed that A x
4 0 for all x, so A is always
concave downward and the local maximum at x 600 must be an absolute maximum.]
Thus, the rectangular ﬁeld should be 600 ft deep and 1200 ft wide.
EXAMPLE 2 A cylindrical can is to be made to hold 1 L of oil. Find the dimensions that
will minimize the cost of the metal to manufacture the can.
h r
FIGURE 3 SOLUTION Draw the diagram as in Figure 3, where r is the radius and h the height (both in
centimeters). In order to minimize the cost of the metal, we minimize the total surface
area of the cylinder (top, bottom, and sides).
From Figure 4 on page 280 we see that the sides are made from a rectangular sheet
with dimensions 2 r and h. So the surface area is A 2 r2 2 rh 5E04(pp 272281) 280 ❙❙❙❙ 1/17/06 2:56 PM Page 280 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 2πr To eliminate h we use the fact that the volume is given as 1 L, which we take to be
1000 cm3. Thus
r 2h 1000 r h which gives h r 2 . Substitution of this into the expression for A gives 1000 2 r2 A Area 2{πr@} Area (2πr)h 1000
r2 2r 2000
r 2 r2 Therefore, the function that we want to minimize is FIGURE 4 2 r2 Ar 2000
r r 0 To ﬁnd the critical numbers, we differentiate:
y Ar
y=A(r) 1000 0 10 r FIGURE 5 2000
r2 r3 4 500
r2 3
Then A r
0 when r 3 500, so the only critical number is r s500 .
Since the domain of A is 0, , we can’t use the argument of Example 1 concerning
3
endpoints. But we can observe that A r
0 for r s500 and A r
0 for
3
r s500 , so A is decreasing for all r to the left of the critical number and increasing
3
for all r to the right. Thus, r s500 must give rise to an absolute minimum.
[Alternatively, we could argue that A r l as r l 0 and A r l as r l , so
there must be a minimum value of A r , which must occur at the critical number. See
Figure 5.]
3
The value of h corresponding to r s500 is 1000
r2 h
 In the Applied Project on page 288 we investigate the most economical shape for a can by
taking into account other manufacturing costs. 4r 1000
500 23 2 3 500 3
Thus, to minimize the cost of the can, the radius should be s500
should be equal to twice the radius, namely, the diameter. 2r
cm and the height NOTE 1 The argument used in Example 2 to justify the absolute minimum is a variant
of the First Derivative Test (which applies only to local maximum or minimum values) and
is stated here for future reference.
■ Module 4.7 takes you through eight
additional optimization problems, including animations of the physical situations. First Derivative Test for Absolute Extreme Values Suppose that c is a critical number of a
continuous function f deﬁned on an interval.
(a) If f x
0 for all x c and f x
0 for all x c, then f c is the absolute
maximum value of f .
(b) If f x
0 for all x c and f x
0 for all x c, then f c is the absolute
minimum value of f . NOTE 2 An alternative method for solving optimization problems is to use implicit differentiation. Let’s look at Example 2 again to illustrate the method. We work with the same
equations
■ A 2 r2 2 rh r 2h 100 5E04(pp 272281) 1/17/06 2:57 PM Page 281 S ECTION 4.7 OPTIMIZATION PROBLEMS ❙❙❙❙ 281 but instead of eliminating h, we differentiate both equations implicitly with respect to r :
A 4r 2h 2 rh r 2h 2 rh The minimum occurs at a critical number, so we set A
equations
2r
and subtraction gives 2r h rh h 0 0, or h 2h 0, simplify, and arrive at the
rh 0 2r. EXAMPLE 3 Find the point on the parabola y 2
y 0 2 x that is closest to the point 1, 4 . SOLUTION The distance between the point 1, 4 and the point x, y is
¥=2x (1, 4) d 0 1 234 1 d x s( 1 y 2
2 (Alternatively, we could have substituted y
of minimizing d, we minimize its square:
d2 FIGURE 6 2 y 4 2 y 2 2, so the expression for d (See Figure 6.) But if x, y lies on the parabola, then x
becomes (x, y) 1 sx (1 y2
2 fy 1 )2 y 4 2 s 2 x to get d in terms of x alone.) Instead 1)2 y 4 2 (You should convince yourself that the minimum of d occurs at the same point as the
minimum of d 2, but d 2 is easier to work with.) Differentiating, we obtain
fy C D
8 km B FIGURE 7 1) y 2y 4 y3 8 so f y
0 when y 2. Observe that f y
0 when y 2 and f y
0 when
y 2, so by the First Derivative Test for Absolute Extreme Values, the absolute minimum occurs when y 2. (Or we could simply say that because of the geometric nature
of the problem, it’s obvious that there is a closest point but not a farthest point.) The
corresponding value of x is x y 2 2 2. Thus, the point on y 2 2 x closest to 1, 4
is 2, 2 . 3 km
A 2( 1 y 2
2 EXAMPLE 4 A man launches his boat from point A on a bank of a straight river, 3 km
wide, and wants to reach point B, 8 km downstream on the opposite bank, as quickly as
possible (see Figure 7). He could row his boat directly across the river to point C and
then run to B, or he could row directly to B, or he could row to some point D between C
and B and then run to B. If he can row 6 km h and run 8 km h, where should he land to
reach B as soon as possible? (We assume that the speed of the water is negligible compared with the speed at which the man rows.)
SOLUTION If we let x be the distance from C to D, then the running distance
is DB
8 x and the Pythagorean Theorem gives the rowing distance as
AD
s x 2 9. We use the equation time distance
rate 5E04(pp 282291) 282 ❙❙❙❙ 1/17/06 3:22 PM Page 282 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Then the rowing time is s x 2
T as a function of x is 9 6 and the running time is 8 sx 2
6 Tx 9 8 x
8 The domain of this function T is 0, 8 . Notice that if x
he rows directly to B. The derivative of T is
x Tx
Try another problem like this one.
Resources / Module 5
/ Max and Min
/ Start of Optimal Lifeguard Thus, using the fact that x
Tx x 8, so the total time 6 sx 2 0 he rows to C and if x 8 1
8 9 0, we have 0 &? x
6 sx 2 &? 16 x 2
&? x 1
8 9
9 x2 &? 4 x 9 3 sx 2 &? 7x 2 81 9 9
s7 The only critical number is x 9 s7. To see whether the minimum occurs at this critical number or at an endpoint of the domain 0, 8 , we evaluate T at all three points: T y=T(x) T0 1 0 2 4 x 6 FIGURE 8 1.5 T 9
s7 s7
8 1 1.33 T8 s73
6 1.42 Since the smallest of these values of T occurs when x 9 s7, the absolute minimum
value of T must occur there. Figure 8 illustrates this calculation by showing the graph
of T .
Thus, the man should land the boat at a point 9 s7 km ( 3.4 km) downstream from
his starting point.
E XAMPLE 5 Find the area of the largest rectangle that can be inscribed in a semicircle of Resources / Module 5
/ Max and Min
/ Start of Max and Min radius r.
SOLUTION 1 Let’s take the semicircle to be the upper half of the circle x 2 y 2 r 2 with
center the origin. Then the word inscribed means that the rectangle has two vertices on
the semicircle and two vertices on the xaxis as shown in Figure 9.
Let x, y be the vertex that lies in the ﬁrst quadrant. Then the rectangle has sides of
lengths 2 x and y, so its area is y (x, y) 2x
_r FIGURE 9 0 A 2 xy y
r x To eliminate y we use the fact that x, y lies on the circle x 2
y sr 2 x 2. Thus
A 2 x sr 2 x2 y2 r 2 and so 5E04(pp 282291) 1/17/06 3:22 PM Page 283 S ECTION 4.7 OPTIMIZATION PROBLEMS The domain of this function is 0
A 2 sr 2 x ❙❙❙❙ 283 r. Its derivative is
2x 2 x2 sr 2 x 2 2 r 2 2x 2
sr 2 x 2 which is 0 when 2 x 2 r 2, that is, x r s2 (since x 0). This value of x gives a
maximum value of A since A 0
0 and A r
0. Therefore, the area of the largest
inscribed rectangle is
A r
s2 2 r
s2 r2 r2
2 r2 SOLUTION 2 A simpler solution is possible if we think of using an angle as a variable. Let be the angle shown in Figure 10. Then the area of the rectangle is
r 2r cos r sin r 2 2 sin cos r 2 sin 2 We know that sin 2 has a maximum value of 1 and it occurs when 2
2. So A
has a maximum value of r 2 and it occurs when
4.
Notice that this trigonometric solution doesn’t involve differentiation. In fact, we
didn’t need to use calculus at all. ¨
r Ł ¨
FIGURE 10  4.7 A r Ã ¨ Exercises 1. Consider the following problem: Find two numbers whose sum is 23 and whose product is a maximum.
(a) Make a table of values, like the following one, so that the
sum of the numbers in the ﬁrst two columns is always 23.
On the basis of the evidence in your table, estimate the
answer to the problem.
First number Second number Product 1
2
3
.
.
. 22
21
20
.
.
. 22
42
60
.
.
. (b) Use calculus to solve the problem and compare with your
answer to part (a).
2. Find two numbers whose difference is 100 and whose product is a minimum.
3. Find two positive numbers whose product is 100 and whose sum is a minimum.
4. Find a positive number such that the sum of the number and its reciprocal is as small as possible.
5. Find the dimensions of a rectangle with perimeter 100 m whose area is as large as possible. 6. Find the dimensions of a rectangle with area 1000 m2 whose perimeter is as small as possible.
7. Consider the following problem: A farmer with 750 ft of fenc ing wants to enclose a rectangular area and then divide it into
four pens with fencing parallel to one side of the rectangle.
What is the largest possible total area of the four pens?
(a) Draw several diagrams illustrating the situation, some with
shallow, wide pens and some with deep, narrow pens. Find
the total areas of these conﬁgurations. Does it appear that
there is a maximum area? If so, estimate it.
(b) Draw a diagram illustrating the general situation. Introduce
notation and label the diagram with your symbols.
(c) Write an expression for the total area.
(d) Use the given information to write an equation that relates
the variables.
(e) Use part (d) to write the total area as a function of one
variable.
(f) Finish solving the problem and compare the answer with
your estimate in part (a).
8. Consider the following problem: A box with an open top is to be constructed from a square piece of cardboard, 3 ft wide, by
cutting out a square from each of the four corners and bending
up the sides. Find the largest volume that such a box can have.
(a) Draw several diagrams to illustrate the situation, some short
boxes with large bases and some tall boxes with small
bases. Find the volumes of several such boxes. Does it
appear that there is a maximum volume? If so, estimate it. 5E04(pp 282291) 284 ❙❙❙❙ 1/17/06 3:22 PM Page 284 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION (b) Draw a diagram illustrating the general situation. Introduce
notation and label the diagram with your symbols.
(c) Write an expression for the volume.
(d) Use the given information to write an equation that relates
the variables.
(e) Use part (d) to write the volume as a function of one
variable.
(f) Finish solving the problem and compare the answer with
your estimate in part (a).
9. A farmer wants to fence an area of 1.5 million square feet in a rectangular ﬁeld and then divide it in half with a fence parallel
to one of the sides of the rectangle. How can he do this so as to
minimize the cost of the fence?
10. A box with a square base and open top must have a volume of 32,000 cm3. Find the dimensions of the box that minimize the
amount of material used.
11. If 1200 cm2 of material is available to make a box with a square base and an open top, ﬁnd the largest possible volume
of the box.
12. A rectangular storage container with an open top is to have a
3 volume of 10 m . The length of its base is twice the width.
Material for the base costs $10 per square meter. Material for
the sides costs $6 per square meter. Find the cost of materials
for the cheapest such container.
13. Do Exercise 12 assuming the container has a lid that is made from the same material as the sides.
14. (a) Show that of all the rectangles with a given area, the one with smallest perimeter is a square.
(b) Show that of all the rectangles with a given perimeter, the
one with greatest area is a square.
15. Find the point on the line y 4x 7 that is closest to the origin. can be inscribed in a circle of radius r.
24. Find the area of the largest rectangle that can be inscribed in a right triangle with legs of lengths 3 cm and 4 cm if two sides of
the rectangle lie along the legs.
25. A right circular cylinder is inscribed in a sphere of radius r. Find the largest possible volume of such a cylinder.
26. A right circular cylinder is inscribed in a cone with height h and base radius r. Find the largest possible volume of such a
cylinder.
27. A right circular cylinder is inscribed in a sphere of radius r. Find the largest possible surface area of such a cylinder.
28. A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus, the diameter of the semicircle is equal
to the width of the rectangle. See Exercise 52 on page 24.) If
the perimeter of the window is 30 ft, ﬁnd the dimensions of
the window so that the greatest possible amount of light is
admitted.
29. The top and bottom margins of a poster are each 6 cm and the side margins are each 4 cm. If the area of printed material on
the poster is ﬁxed at 384 cm2, ﬁnd the dimensions of the poster
with the smallest area.
30. A poster is to have an area of 180 in2 with 1inch margins at the bottom and sides and a 2inch margin at the top. What
dimensions will give the largest printed area?
31. A piece of wire 10 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral
triangle. How should the wire be cut so that the total area
enclosed is (a) a maximum? (b) A minimum?
32. Answer Exercise 31 if one piece is bent into a square and the other into a circle.
33. A cylindrical can without a top is made to contain V cm3 of 16. Find the point on the line 6x point 23. Find the dimensions of the isosceles triangle of largest area that y 9 that is closest to the 3, 1 . 17. Find the points on the ellipse 4 x 2 y2 4 that are farthest away from the point 1, 0 . ; 18. Find, correct to two decimal places, the coordinates of the
point on the curve y tan x that is closest to the point 1, 1 . 19. Find the dimensions of the rectangle of largest area that can be inscribed in a circle of radius r.
20. Find the area of the largest rectangle that can be inscribed in the ellipse x 2 a 2 y2 b2 1. liquid. Find the dimensions that will minimize the cost of the
metal to make the can.
34. A fence 8 ft tall runs parallel to a tall building at a distance of 4 ft from the building. What is the length of the shortest ladder
that will reach from the ground over the fence to the wall of the
building?
35. A coneshaped drinking cup is made from a circular piece of paper of radius R by cutting out a sector and joining the edges
CA and CB. Find the maximum capacity of such a cup.
A B
R 21. Find the dimensions of the rectangle of largest area that can be inscribed in an equilateral triangle of side L if one side of the
rectangle lies on the base of the triangle.
22. Find the dimensions of the rectangle of largest area that has its base on the xaxis and its other two vertices above the xaxis
and lying on the parabola y 8 x 2. C 5E04(pp 282291) 1/17/06 3:23 PM Page 285 S ECTION 4.7 OPTIMIZATION PROBLEMS 36. A coneshaped paper drinking cup is to be made to hold 27 cm3 speed of 20 km h. Another boat has been heading due east at
15 km h and reaches the same dock at 3:00 P.M. At what time
were the two boats closest together? 37. A cone with height h is inscribed in a larger cone with height H so that its vertex is at the center of the base of the larger
cone. Show that the inner cone has maximum volume when
h 1 H.
3 41. Solve the problem in Example 4 if the river is 5 km wide and point B is only 5 km downstream from A.
42. A woman at a point A on the shore of a circular lake with 38. For a ﬁsh swimming at a speed v relative to the water, the
energy expenditure per unit time is proportional to v 3. It is radius 2 mi wants to arrive at the point C diametrically opposite A on the other side of the lake in the shortest possible
time. She can walk at the rate of 4 mi h and row a boat at
2 mi h. How should she proceed? believed that migrating ﬁsh try to minimize the total energy
required to swim a ﬁxed distance. If the ﬁsh are swimming
against a current u u v , then the time required to swim a
distance L is L v u and the total energy E required to
swim the distance is given by
av 3 285 40. A boat leaves a dock at 2:00 P.M. and travels due south at a of water. Find the height and radius of the cup that will use the
smallest amount of paper. Ev ❙❙❙❙ B L
v u where a is the proportionality constant. A (a) Determine the value of v that minimizes E.
(b) Sketch the graph of E. ¨
2 C 2 Note: This result has been veriﬁed experimentally; migrating
ﬁsh swim against a current at a speed 50% greater than the
current speed.
43. The illumination of an object by a light source is directly 39. In a beehive, each cell is a regular hexagonal prism, open at proportional to the strength of the source and inversely proportional to the square of the distance from the source. If two
light sources, one three times as strong as the other, are placed
10 ft apart, where should an object be placed on the line
between the sources so as to receive the least illumination? one end with a trihedral angle at the other end. It is believed
that bees form their cells in such a way as to minimize the surface area for a given volume, thus using the least amount of
wax in cell construction. Examination of these cells has shown
that the measure of the apex angle is amazingly consistent.
Based on the geometry of the cell, it can be shown that the surface area S is given by
S 6sh off the least area from the ﬁrst quadrant. (3s 2s3 2) csc 32
2 s cot where s, the length of the sides of the hexagon, and h, the
height, are constants.
(a) Calculate dS d .
(b) What angle should the bees prefer?
(c) Determine the minimum surface area of the cell (in terms
of s and h).
Note: Actual measurements of the angle in beehives have
been made, and the measures of these angles seldom differ
from the calculated value by more than 2 .
trihedral
angle ¨ rear
of cell 45. Let a and b be positive numbers. Find the length of the shortest line segment that is cut off by the ﬁrst quadrant and passes
through the point a, b .
46. At which points on the curve y 1
tangent line have the largest slope? b
front
of cell 40 x 3 3 x 5 does the 47. Show that of all the isosceles triangles with a given perimeter, the one with the greatest area is equilateral.
CAS 48. The frame for a kite is to be made from six pieces of wood. The four exterior pieces have been cut with the lengths indicated in the ﬁgure. To maximize the area of the kite, how long
should the diagonal pieces be? a h s 44. Find an equation of the line through the point 3, 5 that cuts b a b 5E04(pp 282291) 286 ❙❙❙❙ 1/17/06 3:23 PM Page 286 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION ; 49. A point P needs to be located somewhere on the line AD so
that the total length L of cables linking P to the points A, B,
and C is minimized (see the ﬁgure). Express L as a function of
x
AP and use the graphs of L and dL d x to estimate the
minimum value. the ﬁgure. Show that the shortest length of such a rope occurs
when 1
2.
P
S A
P
5m
Q
2m
B ¨™ ¨¡
R T 3m
53. The upper righthand corner of a piece of paper, 12 in. by 8 in., C D 50. The graph shows the fuel consumption c of a car (measured in
gallons per hour) as a function of the speed v of the car. At very low speeds the engine runs inefﬁciently, so initially c decreases
as the speed increases. But at high speeds the fuel consumption
increases. You can see that c v is minimized for this car when
v
30 mi h. However, for fuel efﬁciency, what must be minimized is not the consumption in gallons per hour but rather the
fuel consumption in gallons per mile. Let’s call this consumption G. Using the graph, estimate the speed at which G has its
minimum value. as in the ﬁgure, is folded over to the bottom edge. How would
you fold it so as to minimize the length of the fold? In other
words, how would you choose x to minimize y ?
12
y x 8 c 54. A steel pipe is being carried down a hallway 9 ft wide. At the end of the hall there is a rightangled turn into a narrower hallway 6 ft wide. What is the length of the longest pipe that can
be carried horizontally around the corner? 0 20 40 6 √ 60 51. Let v1 be the velocity of light in air and v2 the velocity of light in water. According to Fermat’s Principle, a ray of light will
travel from a point A in the air to a point B in the water by a
path ACB that minimizes the time taken. Show that
sin
sin 1 v2 9 v1 2 ¨ 55. An observer stands at a point P, one unit away from a track. where 1 (the angle of incidence) and 2 (the angle of refraction) are as shown. This equation is known as Snell’s Law.
A Two runners start at the point S in the ﬁgure and run along the
track. One runner runs three times as fast as the other. Find the
maximum value of the observer’s angle of sight between the
runners. [Hint: Maximize tan .] ¨¡
P
C
¨
¨™ 1
B 52. Two vertical poles PQ and ST are secured by a rope PRS going from the top of the ﬁrst pole to a point R on the ground
between the poles and then to the top of the second pole as in S 5E04(pp 282291) 1/17/06 3:23 PM Page 287 S ECTION 4.7 OPTIMIZATION PROBLEMS 56. A rain gutter is to be constructed from a metal sheet of width 30 cm by bending up onethird of the sheet on each side
through an angle . How should be chosen so that the gutter
will carry the maximum amount of water? ¨ 10 cm 10 cm 10 cm 57. Where should the point P be chosen on the line segment AB so as to maximize the angle ?
B C L
r4 where L is the length of the blood vessel, r is the radius, and C
is a positive constant determined by the viscosity of the blood.
(Poiseuille established this law experimentally, but it also
follows from Equation 9.4.2.) The ﬁgure shows a main blood
vessel with radius r1 branching at an angle into a smaller vessel with radius r2.
C 2
¨ P 287 pumping the blood. In particular, this energy is reduced when
the resistance of the blood is lowered. One of Poiseuille’s Laws
gives the resistance R of the blood as R
¨ ❙❙❙❙ r™ 3 b vascular
branching
A 5 A r¡ ¨
B 58. A painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer (as in
the ﬁgure). How far from the wall should the observer stand to
get the best view? (In other words, where should the observer
stand so as to maximize the angle subtended at his eye by the
painting?) a (a) Use Poiseuille’s Law to show that the total resistance of the
blood along the path ABC is R C a b cot
r4
1 b csc
r4
2 h
¨ d where a and b are the distances shown in the ﬁgure.
(b) Prove that this resistance is minimized when
cos 59. Find the maximum area of a rectangle that can be circum scribed about a given rectangle with length L and width W . ¨
L
W 60. The blood vascular system consists of blood vessels (arteries, arterioles, capillaries, and veins) that convey blood from the
heart to the organs and back to the heart. This system should
work so as to minimize the energy expended by the heart in r4
2
r4
1 (c) Find the optimal branching angle (correct to the nearest
degree) when the radius of the smaller blood vessel is twothirds the radius of the larger vessel. 5E04(pp 282291) 288 ❙❙❙❙ 1/17/06 3:23 PM Page 288 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 61. Ornithologists have determined that some species of birds tend to avoid ﬂights over large bodies of water during daylight
hours. It is believed that more energy is required to ﬂy over
water than land because air generally rises over land and falls
over water during the day. A bird with these tendencies is
released from an island that is 5 km from the nearest point B
on a straight shoreline, ﬂies to a point C on the shoreline, and
then ﬂies along the shoreline to its nesting area D. Assume that
the bird instinctively chooses a path that will minimize its
energy expenditure. Points B and D are 13 km apart.
(a) In general, if it takes 1.4 times as much energy to ﬂy over
water as land, to what point C should the bird ﬂy in order
to minimize the total energy expended in returning to its
nesting area?
(b) Let W and L denote the energy (in joules) per kilometer
ﬂown over water and land, respectively. What would a large
value of the ratio W L mean in terms of the bird’s ﬂight?
What would a small value mean? Determine the ratio W L
corresponding to the minimum expenditure of energy.
(c) What should the value of W L be in order for the bird to ﬂy
directly to its nesting area D? What should the value of W L be for the bird to ﬂy to B and then along the shore
to D ?
(d) If the ornithologists observe that birds of a certain species
reach the shore at a point 4 km from B, how many times
more energy does it take a bird to ﬂy over water than land? ; 62. Two light sources of identical strength are placed 10 m apart.
An object is to be placed at a point P on a line parallel to the
line joining the light sources and at a distance d meters from it
(see the ﬁgure). We want to locate P on so that the intensity
of illumination is minimized. We need to use the fact that the
intensity of illumination for a single source is directly proportional to the strength of the source and inversely proportional to
the square of the distance from the source.
(a) Find an expression for the intensity I x at the point P.
(b) If d 5 m, use graphs of I x and I x to show that the
intensity is minimized when x 5 m, that is, when P is at
the midpoint of .
(c) If d 10 m, show that the intensity (perhaps surprisingly)
is not minimized at the midpoint.
(d) Somewhere between d 5 m and d 10 m there is a transitional value of d at which the point of minimal illumination abruptly changes. Estimate this value of d by graphical
methods. Then ﬁnd the exact value of d. island x
P
d 5 km
C
B
13 km D
nest 10 m APPLIED PROJECT
The Shape of a Can h r In this project we investigate the most economical shape for a can. We ﬁrst interpret this to mean
that the volume V of a cylindrical can is given and we need to ﬁnd the height h and radius r that
minimize the cost of the metal to make the can (see the ﬁgure). If we disregard any waste metal
in the manufacturing process, then the problem is to minimize the surface area of the cylinder.
We solved this problem in Example 2 in Section 4.7 and we found that h 2r; that is, the height
should be the same as the diameter. But if you go to your cupboard or your supermarket with a
ruler, you will discover that the height is usually greater than the diameter and the ratio h r varies
from 2 up to about 3.8. Let’s see if we can explain this phenomenon.
1. The material for the cans is cut from sheets of metal. The cylindrical sides are formed by bending rectangles; these rectangles are cut from the sheet with little or no waste. But if the
top and bottom discs are cut from squares of side 2r (as in the ﬁgure), this leaves considerable waste metal, which may be recycled but has little or no value to the can makers. If this
is the case, show that the amount of metal used is minimized when Discs cut from squares h
r 8 2.55 5E04(pp 282291) 1/17/06 3:23 PM Page 289 S ECTION 4.8 APPLICATIONS TO BUSINESS AND ECONOMICS ❙❙❙❙ 289 2. A more efﬁcient packing of the discs is obtained by dividing the metal sheet into hexagons and cutting the circular lids and bases from the hexagons (see the ﬁgure). Show that if this
strategy is adopted, then
h
r 4 s3 2.21 3. The values of h r that we found in Problems 1 and 2 are a little closer to the ones that
Discs cut from hexagons actually occur on supermarket shelves, but they still don’t account for everything. If we
look more closely at some real cans, we see that the lid and the base are formed from discs
with radius larger than r that are bent over the ends of the can. If we allow for this we
would increase h r. More signiﬁcantly, in addition to the cost of the metal we need to incorporate the manufacturing of the can into the cost. Let’s assume that most of the expense is
incurred in joining the sides to the rims of the cans. If we cut the discs from hexagons as in
Problem 2, then the total cost is proportional to
4 s3 r 2 2 rh k4 r h where k is the reciprocal of the length that can be joined for the cost of one unit area of
metal. Show that this expression is minimized when
3
sV
k 3 h
r 2
hr hr
4 s3 3
; 4. Plot sV k as a function of x h r and use your graph to argue that when a can is large or
joining is cheap, we should make h r approximately 2.21 (as in Problem 2). But when the
can is small or joining is costly, h r should be substantially larger. 5. Our analysis shows that large cans should be almost square but small cans should be tall and thin. Take a look at the relative shapes of the cans in a supermarket. Is our conclusion usually true in practice? Are there exceptions? Can you suggest reasons why small cans are not
always tall and thin?  4.8 Applications to Business and Economics y inﬂection
point
c (x)=slope
y=C(x)
C (x) 0 x FIGURE 1 Cost function In Section 3.4 we introduced the idea of marginal cost. Recall that if C x , the cost function, is the cost of producing x units of a certain product, then the marginal cost is the rate
of change of C with respect to x. In other words, the marginal cost function is the derivative, C x , of the cost function.
The graph of a typical cost function is shown in Figure 1. The marginal cost C x is the
slope of the tangent to the cost curve at x, C x . Notice that the cost curve is initially concave downward (the marginal cost is decreasing) because of economies of scale (more efﬁcient use of the ﬁxed costs of production). But eventually there is an inﬂection point and
the cost curve becomes concave upward (the marginal cost is increasing), perhaps because
of overtime costs or the inefﬁciencies of a largescale operation.
The average cost function x 1 cx Cx
x represents the cost per unit when x units are produced. We sketch a typical average cost 5E04(pp 282291) 290 ❙❙❙❙ 1/17/06 3:24 PM Page 290 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION y function in Figure 2 by noting that C x x is the slope of the line that joins the origin to
the point x, C x in Figure 1. It appears that there will be an absolute minimum. To ﬁnd
it we locate the critical point of c by using the Quotient Rule to differentiate Equation 1:
y=c(x) xC x cx
Now c x x Cx 0 and this gives Cx 0 0 when x C x Cx
2 Cx
x x FIGURE 2 Average cost function cx Therefore:
If the average cost is a minimum, then
marginal cost average cost
This principle is plausible because if our marginal cost is smaller than our average cost,
then we should produce more, thereby lowering our average cost. Similarly, if our marginal cost is larger than our average cost, then we should produce less in order to lower our
average cost.  See Example 8 in Section 3.4 for an
explanation of why it is reasonable to model
a cost function by a polynomial. EXAMPLE 1 A company estimates that the cost (in dollars) of producing x items is
Cx
2600 2 x 0.001x 2.
(a) Find the cost, average cost, and marginal cost of producing 1000 items, 2000 items,
and 3000 items.
(b) At what production level will the average cost be lowest, and what is this minimum
average cost?
SOLUTION (a) The average cost function is
Cx
x 2600
x Cx cx 2 2 0.001x The marginal cost function is
0.002 x We use these expressions to ﬁll in the following table, giving the cost, average cost, and
marginal cost (in dollars, or dollars per item, rounded to the nearest cent).
x Cx cx Cx 1000
2000
3000 5,600.00
10,600.00
17,600.00 5.60
5.30
5.87 4.00
6.00
8.00 (b) To minimize the average cost we must have
marginal cost
Cx
2 0.002 x average cost
cx
2600
x 2 0.001x 5E04(pp 282291) 1/17/06 3:24 PM Page 291 SECTION 4.8 APPLICATIONS TO BUSINESS AND ECONOMICS  Figure 3 shows the graphs of the marginal
cost function C and average cost function c in
Example 1. Notice that c has its minimum value
when the two graphs intersect. 0.001x 2600
x so x2 2600
0.001 and x c 0 3000 291 This equation simpliﬁes to 10 Cª ❙❙❙❙ 2,600,000 s2,600,000 1612 To see that this production level actually gives a minimum, we note that
cx
5200 x 3 0, so c is concave upward on its entire domain. The minimum
average cost is FIGURE 3 c 1612 2600
1612 2 0.001 1612 $5.22 i tem Now let’s consider marketing. Let p x be the price per unit that the company can
charge if it sells x units. Then p is called the demand function (or price function) and we
would expect it to be a decreasing function of x. If x units are sold and the price per unit
is p x , then the total revenue is
Rx xp x and R is called the revenue function (or sales function). The derivative R of the revenue
function is called the marginal revenue function and is the rate of change of revenue with
respect to the number of units sold.
If x units are sold, then the total proﬁt is
Px Rx Cx and P is called the proﬁt function. The marginal proﬁt function is P , the derivative of
the proﬁt function. In order to maximize proﬁt we look for the critical numbers of P, that
is, the numbers where the marginal proﬁt is 0. But if
Px
then Rx Rx Cx 0 Cx Therefore:
If the profit is a maximum, then
marginal revenue marginal cost
To ensure that this condition gives a maximum, we could use the Second Derivative
Test. Note that
Px
when Rx Rx Cx 0 Cx and this condition says that the rate of increase of marginal revenue is less than the rate of 5E04(pp 292301) 292 ❙❙❙❙ 1/17/06 3:16 PM Page 292 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION increase of marginal cost. Thus, the proﬁt will be a maximum when
Rx Cx and Rx Cx E XAMPLE 2 Determine the production level that will maximize the proﬁt for a company
with cost and demand functions Cx 84 0.01x 2 1.26 x 0.00007x 3 and px 3.5 0.01x SOLUTION The revenue function is Rx xp x 3.5x 0.01x 2 so the marginal revenue function is
Rx 3.5 0.02 x and the marginal cost function is
Cx
 Figure 4 shows the graphs of the revenue
and cost functions in Example 2. The company
makes a proﬁt when R C and the proﬁt is a
maximum when x 103. Notice that the curves
have parallel tangents at this production level
because marginal revenue equals marginal cost. 1.26 0.02 x 0.00021x 2 Thus, marginal revenue is equal to marginal cost when
3.5 0.02 x 1.26 0.02 x 0.00021x 2 Solving, we get 320 x
R 2.24
0.00021 103 To check that this gives a maximum, we compute the second derivatives:
C Rx
0 FIGURE 4 160 0.02 Thus, R x
C x for all x
mize the proﬁt. Cx 0.02 0.00042 x 0. Therefore, a production level of 103 units will maxi EXAMPLE 3 A store has been selling 200 DVD players a week at $350 each. A market
survey indicates that for each $10 rebate offered to buyers, the number of players sold
will increase by 20 a week. Find the demand function and the revenue function. How
large a rebate should the store offer to maximize its revenue?
SOLUTION If x is the number of DVD players sold per week, then the weekly increase in
sales is x 200. For each increase of 20 players sold, the price is decreased by $10. So
1
for each additional player sold, the decrease in price will be 20 10 and the demand
function is px 350 10
20 x 200 450 1
2 x The revenue function is
Rx xp x 450 x 1
2 x2 Since R x
450 x, we see that R x
0 when x 450. This value of x gives an
absolute maximum by the First Derivative Test (or simply by observing that the graph of 5E04(pp 292301) 1/17/06 3:17 PM Page 293 S ECTION 4.8 APPLICATIONS TO BUSINESS AND ECONOMICS ❙❙❙❙ 293 R is a parabola that opens downward). The corresponding price is
p 450
and the rebate is 350 225
offer a rebate of $125.  4.8 1
2 450 450 225 125. Therefore, to maximize revenue the store should Exercises
(b) Sketch a graph of the proﬁt function.
(c) Sketch a graph of the marginal proﬁt function. 1. A manufacturer keeps precise records of the cost C x of making x items and produces the graph of the cost function
shown in the ﬁgure.
(a) Explain why C 0
0.
(b) What is the signiﬁcance of the inﬂection point?
(c) Use the graph of C to sketch the graph of the marginal cost
function. y y=R(x)
y=C (x) C x 0 5–8  For each cost function (given in dollars), ﬁnd (a) the cost,
average cost, and marginal cost at a production level of 1000 units;
(b) the production level that will minimize the average cost; and
(c) the minimum average cost. x 0 5. C x 40,000 300x x2 6. C x 25,000 120x 0.1x 2 7. C x 16,000 200 x 4x 3 ; 8. C x 10,000 340 x 0.3 x 2 2. The graph of a cost function C is given. (a) Draw a careful sketch of the marginal cost function.
(b) Use the geometric interpretation of the average cost c x as
a slope (see Figure 1) to draw a careful sketch of the average cost function.
(c) Estimate the value of x for which c x is a minimum. How
are the average cost and the marginal cost related at that
value of x ?
C
400 ■ ■ ■ 3700 10. C x 339 ■ 4 6 ■ ■ ■ ■ ■ ■ ■  A cost function is given.
(a) Find the average cost and marginal cost functions.
(b) Use graphs of the functions in part (a) to estimate the production level that minimizes the average cost.
(c) Use calculus to ﬁnd the minimum average cost.
(d) Find the minimum value of the marginal cost. 9. C x 2 ■ 0.0001 x 3 ; 9–10 200 0 ■ 2 ■ ■ 0.04 x 2 5x 0.09x 2 25x
■ 0.0003x 3 ■ 0.0004 x 3 ■ ■ ■ ■ ■ ■ ■ x 11–14 3. The average cost of producing x units of a commodity is cx
21.4 0.002 x. Find the marginal cost at a production
level of 1000 units. In practical terms, what is the meaning of
your answer?
4. The ﬁgure shows graphs of the cost and revenue functions reported by a manufacturer.
(a) Identify on the graph the value of x for which the proﬁt is
maximized.  For the given cost and demand functions, ﬁnd the production level that will maximize proﬁt. 11. C x 680 4x 0.01x 2, px 12. C x 680 4x 0.01x 2, px 13. C x
14. C x px
■ ■ 1450 36 x x 2 ■ ■ 12
3 0.001x , 16,000 500x
1700 7x
■ 12 1.6 x
■ 2 x 500 px 60 0.01x 3 0.004 x ,
■ ■ ■ ■ ■ ■ 5E04(pp 292301) 294 ❙❙❙❙ 1/17/06 3:17 PM Page 294 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 15–16  Find the production level at which the marginal cost function starts to increase. 15. C x
16. C x
■ ■ 0.001x 3 0.0002 x
■ ■ 0.3x
3 2 0.25x
■ 6x
2
■ 900
4x
■ 1500
■ ■ ■ ■ ■ ; 17. The cost, in dollars, of producing x yards of a certain fabric
is C x
1200 12 x 0.1x 2 0.0005x 3 and the company
ﬁnds that if it sells x yards, it can charge p x
29 0.00021x
dollars per yard for the fabric.
(a) Graph the cost and revenue functions and use the graphs to
estimate the production level for maximum proﬁt.
(b) Use calculus to ﬁnd the production level for maximum proﬁt.
18. An aircraft manufacturer wants to determine the best selling ; price for a new airplane. The company estimates that the initial
cost of designing the airplane and setting up the factories in
which to build it will be 500 million dollars. The additional
cost of manufacturing each plane can be modeled by the function m x
20 x 5x 3 4 0.01x 2, where x is the number of
aircraft produced and m is the manufacturing cost, in millions
of dollars. The company estimates that if it charges a price p
(in millions of dollars) for each plane, it will be able to sell
xp
320 7.7p planes.
(a) Find the cost, demand, and revenue functions.
(b) Find the production level and the associated selling price of
the aircraft that maximizes proﬁt.
19. A baseball team plays in a stadium that holds 55,000 spectators. With ticket prices at $10, the average attendance had been
27,000. When ticket prices were lowered to $8, the average
attendance rose to 33,000.
(a) Find the demand function, assuming that it is linear.
(b) How should ticket prices be set to maximize revenue?
20. During the summer months Terry makes and sells necklaces on the beach. Last summer he sold the necklaces for $10 each and
his sales averaged 20 per day. When he increased the price by
$1, he found that he lost two sales per day.
(a) Find the demand function, assuming that it is linear.
(b) If the material for each necklace costs Terry $6, what
should the selling price be to maximize his proﬁt?  4.9 21. A manufacturer has been selling 1000 television sets a week at $450 each. A market survey indicates that for each $10 rebate
offered to the buyer, the number of sets sold will increase by
100 per week.
(a) Find the demand function.
(b) How large a rebate should the company offer the buyer in
order to maximize its revenue?
(c) If its weekly cost function is C x
68,000 150 x, how
should the manufacturer set the size of the rebate in order
to maximize its proﬁt?
22. The manager of a 100unit apartment complex knows from experience that all units will be occupied if the rent is $800
per month. A market survey suggests that, on average, one
additional unit will remain vacant for each $10 increase in rent.
What rent should the manager charge to maximize revenue?
23. Store managers want an optimal inventory policy. Overstocking results in excessive storage and interest costs, whereas a small
inventory means added costs for reordering and delivery. A
supermarket manager estimates that a total of 800 cases of
soup will be sold at a steady rate during the coming year and it
costs $4 to store a case for a year. If he places several orders
per year, each consisting of x cases, then there will be an average of 1 x cases in stock at any time and so storage costs for
2
the year are ( 1 x) 4 2 x dollars. He also estimates that the
2
handling cost for each delivery is $100. What is the optimal
reorder quantity x that minimizes costs?
24. Suppose a person has an amount A of spending money deposited to her savings account each month, where it earns
interest at a monthly rate R. Assume she spends the entire
amount throughout the month at a steady rate. When she makes
cash withdrawals from the account, she incurs a transaction
cost T (a combination of bank fees and the cost of her time).
She would save money by making fewer withdrawals, but the
more money she leaves in the account the more interest she
earns. Suppose she makes n cash withdrawals for the same
amount during the month. Then her average cash balance at
any given time is A 2n . (Why?) Find the value of n that minimizes total costs (transaction costs and lost interest), and then
show that the optimal average cash balance is sAT 2R . Newton ’s Method Resources / Module 5
/ Newton’s Method
/ Start of Newton’s Method Suppose that a car dealer offers to sell you a car for $18,000 or for payments of $375 per
month for ﬁve years. You would like to know what monthly interest rate the dealer is, in
effect, charging you. To ﬁnd the answer, you have to solve the equation
1 48 x 1 x 60 1 x 60 1 0 (The details are explained in Exercise 39.) How would you solve such an equation?
For a quadratic equation ax 2 b x c 0 there is a wellknown formula for the roots.
For third and fourthdegree equations there are also formulas for the roots, but they are
extremely complicated. If f is a polynomial of degree 5 or higher, there is no such formula 5E04(pp 292301) 1/17/06 3:17 PM Page 295 SECTION 4.9 NEWTON’S METHOD 0.15 0 0.012 _0.05 FIGURE 1
 Try to solve Equation 1 using the numerical
rootﬁnder on your calculator or computer. Some
machines are not able to solve it. Others are successful but require you to specify a starting point
for the search.
y
{ x ¡, f (x ¡)} y=ƒ
L
0 x™ x ¡ r x f x1 f x1 x Since the xintercept of L is x 2 , we set y
0
If f x 1 { x ¡, f (x ¡)} r
x™ x¡ x1 x x2 f x1
f x1 f x2
f x2 If we keep repeating this process, we obtain a sequence of approximations x 1, x 2, x 3, x 4, . . .
as shown in Figure 3. In general, if the nth approximation is x n and f x n
0, then the
next approximation is given by FIGURE 3
2  Sequences were brieﬂy introduced in
A Preview of Calculus on page 6. A more
thorough discussion starts in Section 12.1. x1 0, we can solve this equation for x 2 : x3 { x ™, f (x ™)} x£ f x1 x2 We use x2 as a second approximation to r.
Next we repeat this procedure with x 1 replaced by x 2 , using the tangent line at
x 2 , f x 2 . This gives a third approximation: y x¢ x1 0 and obtain f x1 x2 0 295 (see the note on page 187). Likewise, there is no formula that will enable us to ﬁnd the
exact roots of a transcendental equation such as cos x x.
We can ﬁnd an approximate solution to Equation 1 by plotting the left side of the equation. Using a graphing device, and after experimenting with viewing rectangles, we produce the graph in Figure 1.
We see that in addition to the solution x 0, which doesn’t interest us, there is a solution between 0.007 and 0.008. Zooming in shows that the root is approximately 0.0076. If
we need more accuracy we could zoom in repeatedly, but that becomes tiresome. A faster
alternative is to use a numerical rootﬁnder on a calculator or computer algebra system. If
we do so, we ﬁnd that the root, correct to nine decimal places, is 0.007628603.
How do those numerical rootﬁnders work? They use a variety of methods, but most of
them make some use of Newton’s method, also called the NewtonRaphson method. We
will explain how this method works, partly to show what happens inside a calculator or
computer, and partly as an application of the idea of linear approximation.
The geometry behind Newton’s method is shown in Figure 2, where the root that we are
trying to ﬁnd is labeled r. We start with a ﬁrst approximation x 1, which is obtained by
guessing, or from a rough sketch of the graph of f , or from a computergenerated graph
of f. Consider the tangent line L to the curve y f x at the point x 1, f x 1 and look at
the xintercept of L, labeled x 2. The idea behind Newton’s method is that the tangent line
is close to the curve and so its xintercept, x2 , is close to the xintercept of the curve
(namely, the root r that we are seeking). Because the tangent is a line, we can easily ﬁnd
its xintercept.
To ﬁnd a formula for x2 in terms of x1 we use the fact that the slope of L is f x1 , so its
equation is
y FIGURE 2 ❙❙❙❙ xn xn 1 f xn
f xn If the numbers x n become closer and closer to r as n becomes large, then we say that
the sequence converges to r and we write
lim x n nl r 5E04(pp 292301) 296 ❙❙❙❙ 1/17/06 3:18 PM Page 296 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION  Although the sequence of successive approximations converges to the desired root for
functions of the type illustrated in Figure 3, in certain circumstances the sequence may not
converge. For example, consider the situation shown in Figure 4. You can see that x 2 is a
worse approximation than x 1. This is likely to be the case when f x 1 is close to 0. It might
even happen that an approximation (such as x 3 in Figure 4) falls outside the domain of f .
Then Newton’s method fails and a better initial approximation x 1 should be chosen. See
Exercises 29–32 for speciﬁc examples in which Newton’s method works very slowly or
does not work at all. y x™ 0 x£ x¡ r x EXAMPLE 1 Starting with x 1
equation x 3 2 x 5 0.
FIGURE 4 2, ﬁnd the third approximation x 3 to the root of the SOLUTION We apply Newton’s method with x3 fx 2x 5 and 3x 2 fx 2 Newton himself used this equation to illustrate his method and he chose x 1 2 after
some experimentation because f 1
6, f 2
1, and f 3
16. Equation 2
becomes
3
xn
2xn 5
xn 1 xn
2
3x n
2
With n 1 we have  Figure 5 shows the geometry behind the
ﬁrst step in Newton’s method in Example 1.
Since f 2
10, the tangent line to
y x 3 2 x 5 at 2, 1 has equation
y 10 x 21 so its xintercept is x 2 2.1. x2 x3
1 x1 23 2 2x1 5
3x 2
2
1 22
5
322 2 2.1 1 Then with n
1.8 2.2 x3 x™ x2
2.1 y=10x21
_2 FIGURE 5 2 we obtain
x3
2 2x2 5
3x 2
2
2 2.1 3 2 2.1
3 2.1 2 2 It turns out that this third approximation x 3 5 2.0946 2.0946 is accurate to four decimal places. Suppose that we want to achieve a given accuracy, say to eight decimal places, using
Newton’s method. How do we know when to stop? The rule of thumb that is generally used
is that we can stop when successive approximations x n and x n 1 agree to eight decimal
places. (A precise statement concerning accuracy in Newton’s method will be given in
Exercises 12.12.)
Notice that the procedure in going from n to n 1 is the same for all values of n. (It is
called an iterative process.) This means that Newton’s method is particularly convenient
for use with a programmable calculator or a computer.
6
EXAMPLE 2 Use Newton’s method to ﬁnd s2 correct to eight decimal places.
6
SOLUTION First we observe that ﬁnding s2 is equivalent to ﬁnding the positive root of the equation
x6 2 0 5E04(pp 292301) 1/17/06 3:18 PM Page 297 SECTION 4.9 NEWTON’S METHOD so we take f x
becomes x6 If we choose x 1 297 6 x 5 and Formula 2 (Newton’s method) 2. Then f x
xn ❙❙❙❙ 1 x6 2
n
5
6x n xn 1 as the initial approximation, then we obtain
x2 1.16666667 x3 1.12644368 x4 1.12249707 x5 1.12246205 x6 1.12246205 Since x 5 and x 6 agree to eight decimal places, we conclude that
6
s2 1.12246205 to eight decimal places.
EXAMPLE 3 Find, correct to six decimal places, the root of the equation cos x x. SOLUTION We ﬁrst rewrite the equation in standard form: cos x
Therefore, we let f x
xn
y y=x y=cos x
1 π
2 x π x cos x
1 x. Then f x xn cos x n x n
sin x n 1 0
sin x
xn cos x n
sin x n xn
1 In order to guess a suitable value for x 1 we sketch the graphs of y cos x and y x in
Figure 6. It appears that they intersect at a point whose xcoordinate is somewhat less
than 1, so let’s take x 1 1 as a convenient ﬁrst approximation. Then, remembering to
put our calculator in radian mode, we get
x2 0.75036387 x3 0.73911289 x4 0.73908513 x5 FIGURE 6 1, so Formula 2 becomes 0.73908513 Since x 4 and x 5 agree to six decimal places (eight, in fact), we conclude that the root of
the equation, correct to six decimal places, is 0.739085.
Instead of using the rough sketch in Figure 6 to get a starting approximation for
Newton’s method in Example 3, we could have used the more accurate graph that a calculator or computer provides. Figure 7 suggests that we use x1 0.75 as the initial approximation. Then Newton’s method gives 1 y=cos x x2 0 FIGURE 7 1 0.73911114 x3 y=x 0.73908513 x4 0.73908513 and so we obtain the same answer as before, but with one fewer step. 5E04(pp 292301) 298 ❙❙❙❙ 1/17/06 3:19 PM Page 298 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION You might wonder why we bother at all with Newton’s method if a graphing device is
available. Isn’t it easier to zoom in repeatedly and ﬁnd the roots as we did in Section 1.4?
If only one or two decimal places of accuracy are required, then indeed Newton’s method
is inappropriate and a graphing device sufﬁces. But if six or eight decimal places are
required, then repeated zooming becomes tiresome. It is usually faster and more efﬁcient
to use a computer and Newton’s method in tandem—the graphing device to get started and
Newton’s method to ﬁnish.  4.9 Exercises
6. x 3 ton’s method is used to approximate the root r of the equation
fx
0 with initial approximation x 1 1.
(a) Draw the tangent lines that are used to ﬁnd x 2 and x 3, and
estimate the numerical values of x 2 and x 3.
(b) Would x 1 5 be a better ﬁrst approximation? Explain. x2 1 0, x1 7. x 4 20 0, x1 2 8. x 1. The ﬁgure shows the graph of a function f . Suppose that New 5 ■ 2
■ 0,
■ x1 1 1 ■ ■ ■ ■ ■ ■ ■ ■ ■ ; 9. Use Newton’s method with initial approximation x1 1 to
ﬁnd x 2 , the second approximation to the root of the equation
x 3 x 3 0. Explain how the method works by ﬁrst graphing the function and its tangent line at 1, 1 . y ; 10. Use Newton’s method with initial approximation x1 1 to
ﬁnd x 2 , the second approximation to the root of the equation
x 4 x 1 0. Explain how the method works by ﬁrst graphing the function and its tangent line at 1, 1 . 1
0 r 1 s x 11–12 2. Follow the instructions for Exercise 1(a) but use x 1 9 as the starting approximation for ﬁnding the root s.  Use Newton’s method to approximate the given number
correct to eight decimal places. ■ 3. Suppose the line y 5x 4 is tangent to the curve y f x
when x 3. If Newton’s method is used to locate a root of the
equation f x
0 and the initial approximation is x1 3, ﬁnd
the second approximation x2. 4. For each initial approximation, determine graphically what happens if Newton’s method is used for the function whose
graph is shown.
(a) x1 0
(b) x1 1
(c) x1 3
(d) x1 4
(e) x1 5 7
12. s1000 3
11. s30
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 13–16  Use Newton’s method to approximate the indicated root
of the equation correct to six decimal places. 13. The root of 2 x 3
14. The root of x 4 6x 2
x 3x 4 0 in the interval 1, 2
x2 15. The positive root of sin x
16. The positive root of 2 cos x
■ ■ ■ ■ ■ 0 in the interval 2, 3 1 x4 ■ ■ ■ ■ ■ ■ ■ y 17–22  Use Newton’s method to ﬁnd all roots of the equation
correct to six decimal places. 17. x 4
0 1 3 5 x 1 3
19. s x x2 21. cos x
■ 5–8  Use Newton’s method with the speciﬁed initial approximation x 1 to ﬁnd x 3 , the third approximation to the root of the given
equation. (Give your answer to four decimal places.) 5. x 3 2x 4 0, x1 1 18. x 5 1 ■ 2 3 x2 22. tan x sx ■ 5x 20. sx x ■ ■ ■ ■ ■ ; 23–26 s1
■ x2
■ ■ ■  Use Newton’s method to ﬁnd all the roots of the equation
correct to eight decimal places. Start by drawing a graph to ﬁnd
initial approximations. 23. x 5 x4 5x 3 x2 4x 3 0 5E04(pp 292301) 1/17/06 3:19 PM Page 299 S ECTION 4.9 NEWTON’S METHOD 24. x 2 4
2 25. x s 2
■ 4 x2 ■ x
■ x 299 34. Use Newton’s method to ﬁnd the absolute minimum value of x2
2 ❙❙❙❙ 26. 3 sin x 1 ■ ■ ■ ■ 2
■ ■ ■ ■ a 0 to
derive the following squareroot algorithm (used by the
ancient Babylonians to compute sa ) :
1
xn
2 1 28. (a) Apply Newton’s method to the equation 1 x a 0 to derive the following reciprocal algorithm:
1 2
ax n 2xn x3 point of the curve y cos x correct to six decimal places. 36. Of the inﬁnitely many lines that are tangent to the curve y
sin x and pass through the origin, there is one that has
the largest slope. Use Newton’s method to ﬁnd the slope of that
line correct to six decimal places.
37. A grain silo consists of a cylindrical main section, with height a
xn (b) Use part (a) to compute s1000 correct to six decimal
places. xn sin x correct to six decimal places. 35. Use Newton’s method to ﬁnd the coordinates of the inﬂection 2x ■ 27. (a) Apply Newton’s method to the equation x 2 xn x2 the function f x 1 30 ft, and a hemispherical roof. In order to achieve a total volume of 15,000 ft 3 (including the part inside the roof section),
what would the radius of the silo have to be?
38. In the ﬁgure, the length of the chord AB is 4 cm and the length of the arc AB is 5 cm. Find the central angle , in radians, correct to four decimal places. Then give the answer to the nearest
degree.
5 cm (This algorithm enables a computer to ﬁnd reciprocals
without actually dividing.)
(b) Use part (a) to compute 1 1.6984 correct to six decimal
places. 4 cm A B ¨ 29. Explain why Newton’s method doesn’t work for ﬁnding the root of the equation x 3
tion is chosen to be x 1 3x
1. 6 0 if the initial approxima 30. (a) Use Newton’s method with x 1 ; 1 to ﬁnd the root of the
equation x 3 x 1 correct to six decimal places.
(b) Solve the equation in part (a) using x 1 0.6 as the initial
approximation.
(c) Solve the equation in part (a) using x 1 0.57. (You
deﬁnitely need a programmable calculator for this part.)
(d) Graph f x
x 3 x 1 and its tangent lines at x1 1,
0.6, and 0.57 to explain why Newton’s method is so sensitive to the value of the initial approximation. 39. A car dealer sells a new car for $18,000. He also offers to sell the same car for payments of $375 per month for ﬁve years.
What monthly interest rate is this dealer charging?
To solve this problem you will need to use the formula for
the present value A of an annuity consisting of n equal payments
of size R with interest rate i per time period: 3 tion sx 0 with any initial approximation x 1
your explanation with a sketch. 0. Illustrate 48 x 1
fx 0
0 x 60 1 3x 4 28 x 3 6x 2 correct to two decimal places.. x 60 1 24 x 0 the point 1, 0 . (The unit here is the distance between the
centers of Earth and the Sun, called an astronomical unit:
y
L¢
Earth 2 function f x
28 x
6x
24 x correct to three
3x
decimal places.
(b) Find the absolute minimum value of the function
fx n 40. The ﬁgure shows the Sun located at the origin and Earth at 33. (a) Use Newton’s method to ﬁnd the critical numbers of the
3 i Use Newton’s method to solve this equation. then the root of the equation f x
0 is x 0. Explain why
Newton’s method fails to ﬁnd the root no matter which initial
approximation x 1 0 is used. Illustrate your explanation with
a sketch.
4 1 Replacing i by x, show that 32. If if x
sx
s x if x R
1
i A 31. Explain why Newton’s method fails when applied to the equa 1 x Sun
L∞ L¡ 7
L£ L™ x 5E04(pp 292301) 300 ❙❙❙❙ 1/17/06 3:20 PM Page 300 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 1 AU 1.496 10 8 km.) There are ﬁve locations L 1 , L 2 , L 3 ,
L 4 , and L 5 in this plane of rotation of Earth about the Sun
where a satellite remains motionless with respect to Earth
because the forces acting on the satellite (including the gravitational attractions of Earth and the Sun) balance each other.
These locations are called libration points. (A solar research
satellite has been placed at one of these libration points.)
If m1 is the mass of the Sun, m 2 is the mass of Earth, and
r m 2 m1 m 2 , it turns out that the xcoordinate of L 1 is  4.10 the unique root of the ﬁfthdegree equation
x5 2 r x4 21 px rx 2r x 3 1
r 1 1 r x2 0 and the xcoordinate of L 2 is the root of the equation
2rx 2 px 0
6 Using the value r 3.04042 10 , ﬁnd the locations of the
libration points (a) L 1 and (b) L 2. Antiderivatives
A physicist who knows the velocity of a particle might wish to know its position at a given
time. An engineer who can measure the variable rate at which water is leaking from a tank
wants to know the amount leaked over a certain time period. A biologist who knows the
rate at which a bacteria population is increasing might want to deduce what the size of the
population will be at some future time. In each case, the problem is to ﬁnd a function F
whose derivative is a known function f. If such a function F exists, it is called an antiderivative of f. Resources / Module 6
/ Antiderivatives
/ Start of Antiderivatives Definition A function F is called an antiderivative of f on an interval I if Fx y ˛ y= 3 +3 f x for all x in I . For instance, let f x
x 2. It isn’t difﬁcult to discover an antiderivative of f if we keep
13
the Power Rule in mind. In fact, if F x
x 2 f x . But the function
3 x , then F x
13
2
Gx
100 also satisﬁes G x
x . Therefore, both F and G are antiderivatives
3x
13
of f . Indeed, any function of the form H x
C, where C is a constant, is an anti3x
derivative of f . The question arises: Are there any others?
To answer this question, recall that in Section 4.2 we used the Mean Value Theorem to
prove that if two functions have identical derivatives on an interval, then they must differ
by a constant (Corollary 4.2.7). Thus, if F and G are any two antiderivatives of f , then
Fx ˛ y= 3 +2
˛ y= 3 +1 fx Gx so G x
Fx
C, where C is a constant. We can write this as G x
have the following result. Fx C, so we y= ˛
0 x 3 1 Theorem If F is an antiderivative of f on an interval I , then the most general
antiderivative of f on I is
Fx
C ˛
y= 3 1
˛ y= 3 2 where C is an arbitrary constant.
FIGURE 1 Members of the family of
antiderivatives of ƒ=≈ Going back to the function f x
x 2, we see that the general antiderivative of f is
x 3 C. By assigning speciﬁc values to the constant C, we obtain a family of functions
whose graphs are vertical translates of one another (see Figure 1).
3 EXAMPLE 1 Find the most general antiderivative of each of the following functions. (a) f x sin x (b) f x x n, n 0 (c) f x x 3 5E04(pp 292301) 1/17/06 3:20 PM Page 301 SECTION 4.10 ANTIDERIVATIVES ❙❙❙❙ 301 S OLUTION (a) If F x
cos x, then F x
sin x, so an antiderivative of sin x is
Theorem 1, the most general antiderivative is G x
cos x C.
xn 1
n1 d
dx (b) n 1 xn
n1 xn x n is Thus, the general antiderivative of f x xn 1
n1 Fx
This is valid for n cos x. By 0 because then f x C x n is deﬁned on an interval. (c) If we put n
3 in part (b) we get the particular antiderivative F x
x2
2
3
by the same calculation. But notice that f x
x is not deﬁned at x 0. Thus, Theorem 1 tells us only that the general antiderivative of f is x 2
2
C on any interval
that does not contain 0. So the general antiderivative of f x
1 x 3 is
1
2x 2
1
2x 2 Fx C1 if x 0 C2 if x 0 As in Example 1, every differentiation formula, when read from right to left, gives rise
to an antidifferentiation formula. In Table 2 we list some particular antiderivatives. Each
formula in the table is true because the derivative of the function in the right column
appears in the left column. In particular, the ﬁrst formula says that the antiderivative of a
constant times a function is the constant times the antiderivative of the function. The second formula says that the antiderivative of a sum is the sum of the antiderivatives. (We use
the notation F
f, G
t.)
2 Table of Antidifferentiation Formulas Function Particular antiderivative cf x
 To obtain the most general antiderivative
from the particular ones in Table 2, we have to
add a constant (or constants), as in Example 1. fx
xn n cF x
tx Particular antiderivative cos x Fx sin x sin x Gx cos x 2 sec x tan x sec x tan x xn 1
n1 1 Function sec x EXAMPLE 2 Find all functions t such that tx 4 sin x 2x 5 sx x
SOLUTION We ﬁrst rewrite the given function as follows:
2x 5
sx
x
x
Thus, we want to ﬁnd an antiderivative of
tx 4 sin x tx 4 sin x 4 sin x 2x 4 x 12 2x 4 1
sx 5E04(pp 302310) 302 ❙❙❙❙ 1/17/06 3:10 PM Page 302 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Using the formulas in Table 2 together with Theorem 1, we obtain
tx 4 cos x x5
5 2
2
5 4 cos x x1 2 C 1
2 x5 2sx C In applications of calculus it is very common to have a situation as in Example 2, where
it is required to ﬁnd a function, given knowledge about its derivatives. An equation that
involves the derivatives of a function is called a differential equation. These will be
studied in some detail in Chapter 10, but for the present we can solve some elementary differential equations. The general solution of a differential equation involves an arbitrary
constant (or constants) as in Example 2. However, there may be some extra conditions
given that will determine the constants and therefore uniquely specify the solution.
EXAMPLE 3 Find f if f x x s x and f 1 2. SOLUTION The general antiderivative of x3 2 fx
is x5 2 fx To determine C we use the fact that f 1 2
5 2 8
5 2
5 12 x 2 x C C 2 , so the particular solution is
2x 5 2
5 fx
EXAMPLE 4 Find f if f x5 2 2: f1
Solving for C, we get C 2
5 C 5
2 6x 8 4, f 0 SOLUTION The general antiderivative of f x 4, and f 1 12 x 2 6x 4 is x3
x2
6
4 x C 4 x 3 3x 2
3
2
Using the antidifferentiation rules once more, we ﬁnd that
fx fx 4 12 x4
4 3 x3
3 4 x2
2 Cx D x4 x3 To determine C and D we use the given conditions that f 0
f0
0 D 4, we have D 4. Since
f1
we have C 1 1 2 C 1. 4x C 2x 2 Cx 4 and f 1 4 1. Since 1 3x D 4 3. Therefore, the required function is
fx x4 x3 2x 2 The Geometry of Antiderivatives
If we are given the graph of a function f , it seems reasonable that we should be able to
sketch the graph of an antiderivative F . Suppose, for instance, that we are given that 5E04(pp 302310) 1/17/06 3:10 PM Page 303 S ECTION 4.10 ANTIDERIVATIVES ❙❙❙❙ 303 F0
1. Then we have a place to start, the point 0, 1 , and the direction in which we
move our pencil is given at each stage by the derivative F x
f x . In the next example
we use the principles of this chapter to show how to graph F even when we don’t have a
formula for f . This would be the case, for instance, when f x is determined by experimental data.
y EXAMPLE 5 The graph of a function f is given in Figure 2. Make a rough sketch of an
antiderivative F , given that F 0
2. y=ƒ
0 1 2 3 4 x FIGURE 2
y y=F(x) SOLUTION We are guided by the fact that the slope of y
F x is f x . We start at the
point 0, 2 and draw F as an initially decreasing function since f x is negative when
0 x 1. Notice that f 1
f3
0, so F has horizontal tangents when x 1 and
x 3. For 1 x 3, f x is positive and so F is increasing. We see that F has a local
minimum when x 1 and a local maximum when x 3. For x 3, f x is negative
and so F is decreasing on 3, . Since f x l 0 as x l , the graph of F becomes
ﬂatter as x l . Also notice that F x
f x changes from positive to negative at
x 2 and from negative to positive at x 4, so F has inﬂection points when x 2
and x 4. We use this information to sketch the graph of the antiderivative in Figure 3. 2 s1 x 3 x, sketch the graph of the antiderivative F that satisﬁes
the initial condition F 1
0.
EXAMPLE 6 If f x 1
0 FIGURE 3 1 x SOLUTION We could try all day to think of a formula for an antiderivative of f and still be
unsuccessful. A second possibility would be to draw the graph of f ﬁrst and then use it
to graph F as in Example 5. That would work, but instead let’s create a more accurate
graph by using what is called a direction ﬁeld.
Since f 0
1, the graph of F has slope 1 when x 0. So we draw several short
tangent segments with slope 1, all centered at x 0. We do the same for several other
values of x and the result is shown in Figure 4. It is called a direction ﬁeld because each
segment indicates the direction in which the curve y F x proceeds at that point.
y
4 y
4 3 3 2 2 1 1 _1 1 2 3x _1 _1 1 2 3x _1 FIGURE 4 FIGURE 5 A direction field for ƒ=œ„„„„„x.
1+˛
The slope of the line segments above x=a is f(a). The graph of an antiderivative
follows the direction field. Now we use the direction ﬁeld to sketch the graph of F . Because of the initial condition F 1
0, we start at the point 1, 0 and draw the graph so that it follows the
directions of the tangent segments. The result is pictured in Figure 5. Any other antiderivative would be obtained by shifting the graph of F upward or downward. 5E04(pp 302310) 304 ❙❙❙❙ 1/17/06 3:10 PM Page 304 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION R ectilinear Motion
Antidifferentiation is particularly useful in analyzing the motion of an object moving in a
straight line. Recall that if the object has position function s f t , then the velocity funcs t . This means that the position function is an antiderivative of the veloction is v t
v t , so the velocity function is
ity function. Likewise, the acceleration function is a t
an antiderivative of the acceleration. If the acceleration and the initial values s 0 and v 0
are known, then the position function can be found by antidifferentiating twice.
EXAMPLE 7 A particle moves in a straight line and has acceleration given by
6t 4. Its initial velocity is v 0
6 cm s and its initial displacement is at
s0 9 cm. Find its position function s t . SOLUTION Since v t at 6t vt Note that v 0 4, antidifferentiation gives
6 t2
2 4t C. But we are given that v 0 4t 6, so C 3t 2 vt Since v t 3t 2 C 4t C
6 and 6 s t , s is the antiderivative of v :
st This gives s 0
function is 3 t3
3 4 t2
2 6t D D. We are given that s 0
t3 st t3 2t 2 9, so D
2t 2 6t 6t D 9 and the required position
9 An object near the surface of the earth is subject to a gravitational force that produces
a downward acceleration denoted by t. For motion close to the ground we may assume that
t is constant, its value being about 9.8 m s2 (or 32 ft s2 ).
EXAMPLE 8 A ball is thrown upward with a speed of 48 ft s from the edge of a cliff 432 ft above the ground. Find its height above the ground t seconds later. When does it
reach its maximum height? When does it hit the ground?
SOLUTION The motion is vertical and we choose the positive direction to be upward. At
time t the distance above the ground is s t and the velocity v t is decreasing. Therefore,
the acceleration must be negative and we have at dv
dt 32 Taking antiderivatives, we have
vt 32 t C To determine C we use the given information that v 0
vt 32 t 48 48. This gives 48 0 C, so 5E04(pp 302310) 1/17/06 3:11 PM Page 305 ❙❙❙❙ S ECTION 4.10 ANTIDERIVATIVES The maximum height is reached when v t
antidifferentiate again and obtain
16t 2 st
Using the fact that s 0 48t 432, we have 432 v t , we D 0 16t 2 st
 Figure 6 shows the position function of the
ball in Example 8. The graph corroborates the
conclusions we reached: The ball reaches its
maximum height after 1.5 s and hits the ground
after 6.9 s. 0, that is, after 1.5 s. Since s t D and so 48t 432 The expression for s t is valid until the ball hits the ground. This happens when s t
that is, when
16t 2 48t 305 432 0 27 0, 0 500 t2 or, equivalently, 3t Using the quadratic formula to solve this equation, we get
3 t
8 0 We reject the solution with the minus sign since it gives a negative value for t. Therefore,
the ball hits the ground after 3(1 s13 ) 2 6.9 s. FIGURE 6  4.10 Exercises 1–16  Find the most general antiderivative of the function.
(Check your answer by differentiation.) 1. f x
6x 2 3. f x 1 8x
x3 5. f x 5x 1 4 7. f x 6 sx 9. f x 2. f x 5x 5 3x 7 15. f t
■ 17–36  21. f x
t 4
sx 3 12. f x 5 sin x ■ 3 cos t 16. f sec t tan t
■ x2 3 14. f t u2 ■ ■ 6
■ ■ ■ 3x , f1 f 24 x 2 2x 10, 4
sin cos , f0 3 s t, f4 20, 2 34. f x 20 x 35. f x 2 36. f
18. f x 2 1 45 20. f x 22. f t t 60t 23. f x 1 24. f x 8x 3 6 x, f0 12 x 8
3, f1 st x6 x 12 x
cos x, sin x, f0 4, 4 1 f0 4 7 f2 15 f0 f0 3 f0 f4 1, 1, 3 f1 2,
3, f0 f0 5, 9, 2 f 0 f1 f0 2, 8,
f f1
2 5
0 1, f0 ■ ■ 1 cos x x
2 x3 12 x,
3 t 1 40 x 3, 6x 3 2 2 33. f x ■ f1 sec 2 t, 32. f t 13 Find f .
12 x 2 0, 31. f 7 sec2 6x x 10 30. f x 4 sin t ■ 3 x 4, f1 29. f x
2x 6 2x 2 2 cos t 5x , 28. f x 3
sx 4 4x 3
x6 5 2x 27. f t 8 sx 6 26. f x 5x 3 3x 1.7 8. f x 3 su ■ 17. f x
19. f u4 25. f x 4 x 10 2x 10. t x 4 st ■ x 20 6. f x 6
sx x3 13. h x 7x 3 4 x2 4 4. f x 3 10
x9 11. f u 3 s13
2 ■ ■ ■ ■ ■ ■ ■ ■ ■ 37. Given that the graph of f passes through the point 1, 6 and that the slope of its tangent line at x, f x is 2 x
38. Find a function f such that f x 6 ■ is tangent to the graph of f . 3 1, ﬁnd f 2 . x and the line x y 0 5E04(pp 302310) 306 ❙❙❙❙ 1/17/06 3:12 PM Page 306 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 39–40  The graph of a function f is shown. Which graph is an
antiderivative of f and why? 39. 40. y f  Draw a graph of f and use it to make a rough sketch of
the antiderivative that passes through the origin. y 45. f x
a sin x 2 , 46. f x f b a ; 45–46 1 x4 ■ ■ b
c 47.
■ ■ ■ 4 1 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ sketch of an antiderivative F , given that F 0 48. y ■ ■ ■ ■ y 2 41. The graph of a function is shown in the ﬁgure. Make a rough 10
8 1 0. 6 y 0 1 4 3x 2 2 _1 0
_2 _2
0 ■ 47–48  A direction ﬁeld is given for a function. Use it to draw
the antiderivative F that satisﬁes F 0
2. c ■ x x x ■ 0 4 12 x 8 x 2 ■ ■ 49–50 ■  ﬁes F 0 ■ ■ 49 f x sin x
,
x 50. f x √ x tan x, ■ ■ ■ ■ ■ ■ ■ Use a direction ﬁeld to graph the antiderivative that satis0. 42. The graph of the velocity function of a car is shown in the ﬁg ure. Sketch the graph of the position function. ■ ■ ■ 0 x 2
2 ■ ■ x 2 ■ ■ ■ ■ ■ ■ ■ 51. A function is deﬁned by the following experimental data. Use a
t 0 direction ﬁeld to sketch the graph of its antiderivative if the initial condition is F 0
0.
x 43. The graph of f is shown in the ﬁgure. Sketch the graph of f if f is continuous and f 0 1. 0.2 0.4 0.6 0.8 1.0 1.2 1.4 fx 0 0.2 0.5 0.8 1.0 0.6 0.2 0 1.6
0.1 1 x 2 and use
it to sketch several members of the family of antiderivatives.
(b) Compute the general antiderivative explicitly and sketch
several particular antiderivatives. Compare with your sketch
in part (a). 52. (a) Draw a direction ﬁeld for the function f x y
2 y=fª(x) 1
0
_1 0 1 2 x 53–58  A particle is moving with the given data. Find the position of the particle. 53. v t ; 44. (a) Use a graphing device to graph f x 2 x 3 sx.
(b) Starting with the graph in part (a), sketch a rough graph of
the antiderivative F that satisﬁes F 0
1.
(c) Use the rules of this section to ﬁnd an expression for F x .
(d) Graph F using the expression in part (c). Compare with
your sketch in part (b). sin t 54. v t cos t, s0 1.5 st, s4 10 55. a t t s0 1, v 0 56. a t cos t sin t, s0 57. a t 10 sin t 2, 3 cos t, 0 3 0, v 0
s0 0, 5
s2 12 5E04(pp 302310) 1/17/06 3:12 PM Page 307 S ECTION 4.10 ANTIDERIVATIVES 58. a t
■ ■ 10
■ 3 t 2, ■ s0 ■ 3t ■ 0, s 2
■ ■ ■ ■ ■ 59. A stone is dropped from the upper observation deck (the Space Deck) of the CN Tower, 450 m above the ground.
(a) Find the distance of the stone above ground level at time t.
(b) How long does it take the stone to reach the ground?
(c) With what velocity does it strike the ground?
(d) If the stone is thrown downward with a speed of 5 m s,
how long does it take to reach the ground?
60. Show that for motion in a straight line with constant acceleration a, initial velocity v0 , and initial displacement s0 , the dis placement after time t is
1
2 s at 2 v0 t per second from a point s0 meters above the ground. Show that
vt v2
0 19.6 s t x
1 s x, in grams per centimeter, where x is measured in
centimeters from one end of the rod. Find the mass of the rod.
67. Since raindrops grow as they fall, their surface area increases and therefore the resistance to their falling increases. A raindrop has an initial downward velocity of 10 m s and its
downward acceleration is
a 9
0 0.9t if 0
if t t 10
10 If the raindrop is initially 500 m above the ground, how long
does it take to fall?
68. A car is traveling at 50 mi h when the brakes are fully applied, s0 61. An object is projected upward with initial velocity v0 meters
2 307 66. The linear density of a rod of length 1 m is given by 10 ■ ❙❙❙❙ producing a constant deceleration of 22 ft s2. What is the distance covered before the car comes to a stop?
69. What constant acceleration is required to increase the speed of a car from 30 mi h to 50 mi h in 5 s? s0 70. A car braked with a constant deceleration of 16 ft s2, produc62. Two balls are thrown upward from the edge of the cliff in Example 8. The ﬁrst is thrown with a speed of 48 ft s and the
other is thrown a second later with a speed of 24 ft s. Do the
balls ever pass each other?
63. A stone was dropped off a cliff and hit the ground with a speed of 120 ft s. What is the height of the cliff?
length L and linear density , then the board takes on the shape
of a curve y f x , where
mt L x 1
2 tL x 2 and E and I are positive constants that depend on the material
of the board and t
0 is the acceleration due to gravity.
(a) Find an expression for the shape of the curve.
(b) Use f L to estimate the distance below the horizontal at
the end of the board.
y 0 71. A car is traveling at 100 km h when the driver sees an accident 80 m ahead and slams on the brakes. What constant deceleration is required to stop the car in time to avoid a pileup?
72. A model rocket is ﬁred vertically upward from rest. Its acceler 64. If a diver of mass m stands at the end of a diving board with EI y ing skid marks measuring 200 ft before coming to a stop. How
fast was the car traveling when the brakes were ﬁrst applied? x 65. A company estimates that the marginal cost (in dollars per item) of producing x items is 1.92 0.002 x. If the cost of producing one item is $562, ﬁnd the cost of producing 100 items. ation for the ﬁrst three seconds is a t
60 t, at which time the
fuel is exhausted and it becomes a freely “falling” body. Fourteen seconds later, the rocket’s parachute opens, and the (downward) velocity slows linearly to 18 ft s in 5 s. The rocket
then “ﬂoats” to the ground at that rate.
(a) Determine the position function s and the velocity function
v (for all times t). Sketch the graphs of s and v.
(b) At what time does the rocket reach its maximum height,
and what is that height?
(c) At what time does the rocket land?
73. A highspeed bullet train accelerates and decelerates at the rate of 4 ft s2. Its maximum cruising speed is 90 mi h.
(a) What is the maximum distance the train can travel if it
accelerates from rest until it reaches its cruising speed and
then runs at that speed for 15 minutes?
(b) Suppose that the train starts from rest and must come to
a complete stop in 15 minutes. What is the maximum distance it can travel under these conditions?
(c) Find the minimum time that the train takes to travel
between two consecutive stations that are 45 miles apart.
(d) The trip from one station to the next takes 37.5 minutes.
How far apart are the stations? 5E04(pp 302310)
❙❙❙❙ 308  1/17/06 3:12 PM Page 308 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 4 Review ■ CONCEPT CHECK 1. Explain the difference between an absolute maximum and a local maximum. Illustrate with a sketch. 9. (a) Given an initial approximation x1 to a root of the equation (b) State the Mean Value Theorem and give a geometric
interpretation.
5. (a) State the Increasing /Decreasing Test. (b) State the Concavity Test.
6. (a) State the First Derivative Test. (b) State the Second Derivative Test.
(c) What are the relative advantages and disadvantages of these
tests?
7. Explain the meaning of each of the following statements. L ■ 0, then f has a local maximum or minimum at c. 2. If f has an absolute minimum value at c, then f c 0. 3. If f is continuous on a, b , then f attains an absolute maxi mum value f c and an absolute minimum value f d at some
numbers c and d in a, b .
1 f 1 , then there is a number
0. 5. If f x 0 for 1 6. If f 2 0, then 2, f 2 is an inﬂection point of the
f x. curve y
7. If f x 0 x t x for 0
1. x 6, then f is decreasing on (1, 6). x 1, then f x 8. There exists a function f such that f 1 fx
fx 0 for all x. 10. (a) What is an antiderivative of a function f ? (b) Suppose F1 and F2 are both antiderivatives of f on an interval I . How are F1 and F2 related? ■ 10. There exists a function f such that f x fx 11. If f and t are increasing on an interval I , then f t is increas ing on I .
12. If f and t are increasing on an interval I , then f t is increas ing on I .
13. If f and t are increasing on an interval I , then f t is increasing on I .
14. If f and t are positive increasing functions on an interval I , then f t is increasing on I .
0 on I , then t x 16. The most general antiderivative of f x 0, and
Fx 0, f x 0, and 1 f x is decreasing on I . t x for
2, f 3 0, f x 0 for all x. 15. If f is increasing and f x 1 for all x. 9. There exists a function f such that f x fx
0, explain geometrically, with a diagram, how the
second approximation x2 in Newton’s method is obtained.
(b) Write an expression for x2 in terms of x1, f x 1 ,
and f x 1 .
(c) Write an expression for x n 1 in terms of x n , f x n , and
f xn .
(d) Under what circumstances is Newton’s method likely to fail
or to work very slowly? TRUEFALSE QUIZ Determine whether the statement is true or false. If it is true, explain why.
If it is false, explain why or give an example that disproves the statement. 1 and f c L. need calculus to graph a function? 4. (a) State Rolle’s Theorem. c such that c f x has the horizontal asymptote y 8. If you have a graphing calculator or computer, why do you (b) Deﬁne a critical number of f . 4. If f is differentiable and f L xl 3. (a) State Fermat’s Theorem. 1. If f c xl (d) The curve y (b) Explain how the Closed Interval Method works. xl (b) lim f x
(c) lim f x 2. (a) What does the Extreme Value Theorem say? (a) lim f x ■ 1
x x 2 is C 0, and
17. If f x exists and is nonzero for all x, then f 1 f 0. 5E04(pp 302310) 1/17/06 3:14 PM Page 309 C HAPTER 4 REVIEW ■ EXERCISES  Find the local and absolute extreme values of the function
on the given interval. 10 2. f x x 3. f x sx,
x
x x2
x2 4. f x
5. f x x 6. f x x 3, 27x y y=f ª(x) 0, 4
_2
0 _1 2, 0 , 2 x 3, 1 2 ■ 7–12 ■  cos x, 17–28 0,  5 ■ ■ 8. lim
tl 9. lim
xl t 1t t 2 2t ■ ■ ■ ■ xl ■ ■ 3x
■ x 2
x3 6x 2 15 x 4 x4 3x 3 3x 2 x 2x ■ 2
t 1 xl x3 1 22. y 1
x x 1 24. y x s1 x x x s2 26. y sx sin2x x 3
sx 2 cos x 4 sin x
12. lim
xl
sx 2 x)
■ 25. y
27. y x2 10. lim x x2 2 3 23. y 1 8 xx
2 1 20. y 1 21. y s4x 2 1
3x 1 11. lim (s 4 x 2 7 x3 17. y
■ Find the limit. 3 6 Use the guidelines of Section 4.5 to sketch the curve. 19. y ■ 3x 4 x 5
7. lim
xl
6x 4 2 x 2 1 ■ 4 0,
2 18. y ■ 3 2, 1 sin 2 x, sin x ■ 0, 4
1 309 (d) Sketch a possible graph of f . 1–6 1. f x ❙❙❙❙ ■ ■ 28. y
■ ■ ■ ■ 4x
■ tan x,
■ 2 ■ ■ x 2 ■ ■ ■ ■ ■ ■ ■ ■ ; 29–32
13–15  Produce graphs of f that reveal all the important aspects
of the curve. Use graphs of f and f to estimate the intervals of
increase and decrease, extreme values, intervals of concavity, and
inﬂection points. In Exercise 29 use calculus to ﬁnd these quantities
exactly. Sketch the graph of a function that satisﬁes the given
conditions.
 13. f 0 0, f
2
f1
f9
0,
lim x l f x
0, lim x l 6 f x
,
fx
0 on
, 2 , 1, 6 , and 9, ,
fx
0 on 2, 1 and 6, 9 ,
fx
0 on
, 0 and 12, ,
fx
0 on 0, 6 and 6, 12 14. f 0 fx
fx ■ ■ 31. f x
32. f x 0, f is continuous and even,
1 if 1
2 x if 0 x 1, f x
1 if x 3 15. f is odd, fx
fx 29. f x fx
0 for x
0 for x ■ ■ 0 for 0 x 2,
2, f x
0 for 0
3, lim x l f x
■ ■ ■ ■ ■ x ■ x2 1
x3 3x 6 5x 5 x4 2 sin x cos x,
■ ■ 3
sx 30. f x 0
■ 5x 3
x
■ 1 2x 2 x 2 2
■ ■ ■ ■ ■ 3,
33. Show that the equation x 101 x 51 x 1 0 has exactly one real root.
x
2 3, 34. Suppose that f is continuous on 0, 4 , f 0 2
■ ■ ■ ■ 16. The ﬁgure shows the graph of the derivative f of a function f . (a) On what intervals is f increasing or decreasing?
(b) For what values of x does f have a local maximum or
minimum?
(c) Sketch the graph of f . fx 5 for all x in 0, 4 . Show that 9 1, and
f4 21. 35. By applying the Mean Value Theorem to the function fx x 1 5 on the interval 32, 33 , show that
2 5
s33 2.0125 36. For what values of the constants a and b is 1, 6 a point of inﬂection of the curve y x3 ax 2 bx 1? ■ 5E04(pp 302310) 310 ❙❙❙❙ 1/17/06 3:14 PM Page 310 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION f x 2 , where f is twice differentiable for all x,
fx
0 for all x 0, and f is concave downward on
and concave upward on 0, .
(a) At what numbers does t have an extreme value?
(b) Discuss the concavity of t. 50. Use Newton’s method to ﬁnd all roots of the equation 37. Let t x 38. Find two positive integers such that the sum of the ﬁrst number and four times the second number is 1000 and the product of
the numbers is as large as possible.
39. Show that the shortest distance from the point x 1, y1 to the x2 sin x ,0 51. Use Newton’s method to ﬁnd the absolute maximum value of the function f t
places. By C cos t 0
53–58 40. Find the point on the hyperbola x y  Find f .
sx 5 55. f t 8 that is closest to the point 3, 0 .
41. Find the smallest possible area of an isosceles triangle that is circumscribed about a circle of radius r.
42. Find the volume of the largest circular cone that can be x sin x, 2 . Use Newton’s method when necessary. x 2t 0 is A x 1 B y1 C
sA2 B 2 t 2 correct to eight decimal t 52. Use the guidelines in Section 4.5 to sketch the curve y 53. f x straight line Ax 1 correct to six decimal places. 3x u 56. f u ■ su 1 58. f x 2x
■ 54. f x 3 sin t, 2 u 57. f x ■ 4
5
sx , 3x
■ 5 f1 3 48 x 2, 6x
3 f0 3 sec2x 8x f0 2 4x
■ 1, 5,
■ f0 f0
■ 2 2,
■ f1
■ 0
■ ■ ■ inscribed in a sphere of radius r.
43. In ABC, D lies on AB, CD AB, AD
BD
4 cm,
5 cm. Where should a point P be chosen on CD
and CD
so that the sum PA
PB
PC is a minimum? 44. Do Exercise 43 when CD 2 cm. K L
C C
L where K and C are known positive constants. What is the
length of the wave that gives the minimum velocity?
46. A metal storage tank with volume V is to be constructed in the shape of a right circular cylinder surmounted by a hemisphere.
What dimensions will require the least amount of metal?
47. A hockey team plays in an arena with a seating capacity of 15,000 spectators. With the ticket price set at $12, average
attendance at a game has been 11,000. A market survey indicates that for each dollar the ticket price is lowered, average
attendance will increase by 1000. How should the owners of
the team set the ticket price to maximize their revenue from
ticket sales? ; 48. A manufacturer determines that the cost of making x units of a
commodity is C x
1800 25x 0.2 x 2 0.001x 3 and the
demand function is p x
48.2 0.03x.
(a) Graph the cost and revenue functions and use the graphs to
estimate the production level for maximum proﬁt.
(b) Use calculus to ﬁnd the production level for maximum
proﬁt.
(c) Estimate the production level that minimizes the average
cost.
49. Use Newton’s method to ﬁnd the root of the equation x5 x4 3x2 3x
six decimal places. 2 , and use that graph to sketch
fx
x 2 sin x 2 , 0 x
the antiderivative F of f that satisﬁes the initial condition
F0
0.
x 4 x 3 c x 2.
In particular you should determine the transitional value of c at
which the number of critical numbers changes and the transitional value at which the number of inﬂection points changes.
Illustrate the various possible shapes with graphs. ; 60. Investigate the family of curves given by f x 45. The velocity of a wave of length L in deep water is
v ; 59. Use a graphing device to draw a graph of the function 0 in the interval 1, 2 correct to 61. A canister is dropped from a helicopter 500 m above the ground. Its parachute does not open, but the canister has been designed
to withstand an impact velocity of 100 m s. Will it burst?
62. In an automobile race along a straight road, car A passed car B twice. Prove that at some time during the race their accelerations were equal. State the assumptions that you make.
63. A light is to be placed atop a pole of height h feet to illuminate a busy trafﬁc circle, which has a radius of 40 ft. The intensity
of illumination I at any point P on the circle is directly proportional to the cosine of the angle (see the ﬁgure) and inversely
proportional to the square of the distance d from the source.
(a) How tall should the light pole be to maximize I ?
(b) Suppose that the light pole is h feet tall and that a woman is
walking away from the base of the pole at the rate of 4 ft s.
At what rate is the intensity of the light at the point on her
back 4 ft above the ground decreasing when she reaches the
outer edge of the trafﬁc circle?
¨
h d
40 5E04(pp 311313) 1/17/06 3:00 PM Page 311 PROBLEMS
PLUS One of the most important principles of problem solving is analogy (see page 58). If you
are having trouble getting started on a problem, it is sometimes helpful to start by solving
a similar, but simpler, problem. The following example illustrates the principle. Cover up
the solution and try solving it yourself ﬁrst.
EXAMPLE If x, y, and z are positive numbers, prove that x2 1 y2 1 z2
x yz 1 8 SOLUTION It may be difﬁcult to get started on this problem. (Some students have tackled
it by multiplying out the numerator, but that just creates a mess.) Let’s try to think of a
similar, simpler problem. When several variables are involved, it’s often helpful to think
of an analogous problem with fewer variables. In the present case we can reduce the
number of variables from three to one and prove the analogous inequality x2 1 1 2 x for x 0 In fact, if we are able to prove (1), then the desired inequality follows because
x2 1 y 2 1 z2
x yz x2 1 y2 1
x z2 1 1
z y 222 8 The key to proving (1) is to recognize that it is a disguised version of a minimum problem. If we let
fx x2 1
x What have we learned from the solution to this
example?
 To solve a problem involving several
variables, it might help to solve a similar
problem with just one variable.
 When trying to prove an inequality, it might
help to think of it as a maximum or minimum
problem. 1 1
x x x x 0 1 x 2 , so f x
0 when x 1. Also, f x
0 for 0
1. Therefore, the absolute minimum value of f is f 1 then f x
1
fx
0 for x
means that Look Back x2 x 1 and
2. This for all positive values of x 2 and, as previously mentioned, the given inequality follows by multiplication.
The inequality in (1) could also be proved without calculus. In fact, if x 0, we have
x2 1 2 &? x2 &? x x 1
1 2 2 x &? x2 2x 1 0 0 Because the last inequality is obviously true, the ﬁrst one is true too. 311 5E04(pp 311313) 1/17/06 3:01 PM P RO B L E M S Page 312 1. Show that sin x s2 for all x. cos x 2. Show that x 2 y 2 4 x2 4 y2 16 for all numbers x and y such that x 3. Let a and b be positive numbers. Show that not both of the numbers a 1 2 and y
b and b 1 2.
a 1 can be greater than 4.
1 x 2 at which the tangent line cuts from the ﬁrst quadrant the triangle with the smallest area. 4. Find the point on the parabola y 5. Find the highest and lowest points on the curve x 2 y2 xy 12. 6. Water is ﬂowing at a constant rate into a spherical tank. Let V t be the volume of water in the tank and H t be the height of the water in the tank at time t.
(a) What are the meanings of V t and H t ? Are these derivatives positive, negative, or
zero?
(b) Is V t positive, negative, or zero? Explain.
(c) Let t1, t 2, and t 3 be the times when the tank is onequarter full, half full, and threequarters
full, respectively. Are the values H t1 , H t 2 , and H t 3 positive, negative, or zero? Why?
7. Find the absolute maximum value of the function 1
x 1 fx 1 8. Find a function f such that f x
1
2 1 1 ,f 0 2 0, and f x 0 for all x, or prove that such a function cannot exist.
y m x b intersects the parabola y x 2 in points A and B (see the ﬁgure). Find
the point P on the arc AOB of the parabola that maximizes the area of the triangle PAB. 9. The line y y=≈
B 10. Sketch the graph of a function f such that f x fx A 0 for x 1, and lim x l fx 0 for all x, f x
0. x 0 for x 1, 11. Determine the values of the number a for which the function f has no critical number:
y=mx+b fx
O P x a2 a 6 cos 2 x 2x cos 1 12. Sketch the region in the plane consisting of all points x, y such that x 2 xy
FIGURE FOR PROBLEM 9 a y x2 y2 13. Let ABC be a triangle with BAC 120 and AB
AC
(a) Express the length of the angle bisector AD in terms of x
(b) Find the largest possible value of AD . C 1.
AB . 14. (a) Let ABC be a triangle with right angle A and hypotenuse a BC . (See the
ﬁgure.) If the inscribed circle touches the hypotenuse at D, show that D CD 1
2 ( BC AC AB ) 1
2 C, express the radius r of the inscribed circle in terms of a and .
(b) If
(c) If a is ﬁxed and varies, ﬁnd the maximum value of r.
A
FIGURE FOR PROBLEM 14 B 15. A triangle with sides a, b, and c varies with time t, but its area never changes. Let be the angle opposite the side of length a and suppose always remains acute.
(a) Express d dt in terms of b, c, , db dt, and dc dt.
(b) Express da dt in terms of the quantities in part (a).
16. ABCD is a square piece of paper with sides of length 1 m. A quartercircle is drawn from B to D with center A. The piece of paper is folded along EF , with E on AB and F on AD, so that A
falls on the quartercircle. Determine the maximum and minimum areas that the triangle AEF
could have. 312 5E04(pp 311313) 1/17/06 3:01 PM Page 313 17. The speeds of sound c1 in an upper layer and c2 in a lower layer of rock and the thickness h of the upper layer can be determined by seismic exploration if the speed of sound in the lower
layer is greater than the speed in the upper layer. A dynamite charge is detonated at a point P
and the transmitted signals are recorded at a point Q, which is a distance D from P. The ﬁrst
signal to arrive at Q travels along the surface and takes T1 seconds. The next signal travels
from P to a point R, from R to S in the lower layer, and then to Q, taking T2 seconds. The third
signal is reﬂected off the lower layer at the midpoint O of RS and takes T3 seconds to reach Q.
(a) Express T1, T2, and T3 in terms of D, h, c1 , c2 , and .
(b) Show that T2 is a minimum when sin
c1 c2.
(c) Suppose that D 1 km, T1 0.26 s, T2 0.32 s, T3 0.34 s. Find c1 , c2 , and h.
P Q D
Speed of sound=c¡ h ¨ ¨
R S O
Speed of sound=c™ Note: Geophysicists use this technique when studying the structure of Earth’s crust, whether
searching for oil or examining fault lines.
18. For what values of c is there a straight line that intersects the curve y cx 3 12 x 2 5x 2 in four distinct points? d
B x4 E C x
r
F D
FIGURE FOR PROBLEM 19 19. One of the problems posed by the Marquis de l’Hospital in his calculus textbook Analyse des Inﬁniment Petits concerns a pulley that is attached to the ceiling of a room at a point C by a
rope of length r. At another point B on the ceiling, at a distance d from C (where d r), a
rope of length is attached and passed through the pulley at F and connected to a weight W .
The weight is released and comes to rest at its equilibrium position D. As l’Hospital argued,
this happens when the distance ED is maximized. Show that when the system reaches equilibrium, the value of x is
r
(r
4d sr 2 8d 2 ) Notice that this expression is independent of both W and .
20. Given a sphere with radius r, ﬁnd the height of a pyramid of minimum volume whose base is a square and whose base and triangular faces are all tangent to the sphere. What if the base of
the pyramid is a regular ngon (a polygon with n equal sides and angles)? (Use the fact that
the volume of a pyramid is 1 Ah, where A is the area of the base.)
3
21. Assume that a snowball melts so that its volume decreases at a rate proportional to its surface area. If it takes three hours for the snowball to decrease to half its original volume, how much
longer will it take for the snowball to melt completely?
22. A hemispherical bubble is placed on a spherical bubble of radius 1. A smaller hemispherical FIGURE FOR PROBLEM 22 bubble is then placed on the ﬁrst one. This process is continued until n chambers, including
the sphere, are formed. (The ﬁgure shows the case n 4.) Use mathematical induction to
prove that the maximum height of any bubble tower with n chambers is 1 sn. 313 ...
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This note was uploaded on 02/04/2010 for the course M 56435 taught by Professor Hamrick during the Fall '09 term at University of Texas at Austin.
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