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Unformatted text preview: 76 Section 3.2 21. continued
Check x 27.
3: Since g(3)
lim g(x) x→3 3)2 (4 lim (2x x→3 1 and
1) 2(3) 1 7, the function is not continuous (and hence not differentiable) at x 3. The function is differentiable for all reals except x
2 22. Note that C(x) x,
x 2, xx x
x lim C(0 h→0 3. y NDER (x) actually has two asymptotes for each asymptote of y C(0) lim h→0 hh 0
h tan x. The asymptotes of y occur at x 0. 0:
h)
h tan x Note: Due to the way NDER is defined, the graph of 0
, so it is
0 differentiable for all x except possibly at x
Check x [ 2 , 2 ] by [ 4, 4] dy
dx k 2 NDER (x) 0.001, where k is an integer. A good window for viewing this behavior is [1.566, 1.576] by
lim h h→0 0 [ 1000, 1000]. The function is differentiable for all reals.
28.
23. (a) x 0 is not in their domains, or, they are both
discontinuous at x 0.
1
1
,0
x
x
1
1
For 2 : NDER 2 , 0
x
x (b) For : NDER 1,000,000
[ 2 , 2 ] by [ 20, 20] 0 (c) It returns an incorrect response because even though
these functions are not defined at x
defined at x 0, they are 0.001. The responses differ from each 1
other because 2 is even (which automatically makes
x
1
1
0) and is odd.
NDER 2 , 0
x
x 24. The graph of NDER (x) does not look like the graph of any
basic function.
29. (a) lim f (x) x→1 f (1) x) a(1)2 2 lim (3 x→1 a The relationship is a b(1) b
b 2. (b) Since the function needs to be continuous, we may
assume that a b 2 and f (1) 2.
lim f (1 h→0 [ 5, 5] by [ 10, 10]
dy
x3
dx f (1) f (1 h)
h f (1) lim h→0 lim h→0 lim h→0 dy
dx sin x 26. 3 (1 h) a(1 1 h)2 b(1 h) 2
h
a 2ah ah 2 b bh
h
2ah ah 2 bh (a b
h lim (2a h→0 2a 2 h lim ( 1) h→0 [ 2 , 2 ] by [ 1.5, 1.5] lim h→0
h→0 lim 25. h)
h ah 2
2) b) b Therefore, 2a b
1. Substituting 2 a for b gives
2a (2 a)
1, so a
3.
Then b 2 a 2 ( 3) 5. The values are
a
3 and b 5.
[ 6, 6] by [ 4, 4] dy
dx abs (x) or x 30. The function f (x) does not have the intermediate value
property. Choose some a in ( 1, 0) and b in (0, 1). Then
f (a) 0 and f (b) 1, but f does not take on any value
between 0 and 1. Therefore, by the Intermediate Value
Theorem for Derivatives, f cannot be the derivative of any
function on [ 1, 1]. 77 Section 3.3 31. (a) Note that x
1
x so lim x sin
x→0 1
x sin 0 by the Sandwich Theorem. Therefore, f is continuous at x
f (0 (b) h)
h 1
h sin
h f (0) x 10. (a) f (x) 0. 0 h)
h 1
h h 2 sin g(0) 0 1. 1
h h sin h As noted in part (a), the limit of this as x approaches
zero is 0, so g (0) lim h→0 x(x h lim 1 h→0 f (x) lim x h
h lim h→0 hx(x h)
2 x x2 h) x h h→0 Section 3.3 Exercises (d) No, because the limit in part (c) does not exist.
g(0 lim 1 x h h→0 f (x lim lim interval containing 0). (e) hx
h f (x) h→0 h→0 0 (that is, for h in any open h)
h h)
h
x
(x h)
hx(x h) (b) f (x) 1 and 1 an infinite number of times arbitrarily close to h lim x h→0 1
oscillates
h (c) The limit does not exist because sin f (x h→0 lim 1
h sin h between 9. These are all constant functions, so the graph of each
function is a horizontal line and the derivative of each
function is 0. x, for all x, 2. 0.
3. s Section 3.3 Rules for Differentiation
(pp. 112–121) dy
dx
d 2y
dx 2 d
( x 2)
dx
d
( 2x)
dx dy
dx
d 2y
dx 2 d 13
x
dx 3
d2
(x )
dx dy
dx
d 2y
dx 2 d
(2x)
dx
d
(2)
dx dy
dx
d 2y
dx 2 d2
(x )
dx
d
(2x)
dx dy
dx d 13
x
dx 3 d
(3)
dx 2x 0 2x 2
d
(x)
dx
d
(1)
dx x2
2x d
(1)
dx 1
0 2 2x 0 2 0 Quick Review 3.3
1. (x 2
x
2. 2)(x 1 1)
x 2 2x 1
1 2
x
2x 3
2x 2 3x 4
1 5. (x 6. 1 x x 2x 1 3x 4
2x 2 4
2 2)(x
x 3 x
x 5
x2 1)
2x 1 x2 1 x2
x 3x 2 1 3. 3x 2
4. x2 1 x x
x2 x 2x
2 1
x
1 x 1 x 1 4. 21 1 x 5.
2 5x x2 x 2 2x 3
2x 2
2 2x 32
x
2 4
2x 2 x 2 x 1 x 1 2x 2x
2 dy
dx 2 2 21 6. 2 dy
dx 3 x (x 3 1 x 2 ) x 2 x 1 7. dy
dx At x 1.173, 500x 6 1305.
At x 2.394, 500x 6 94,212
After rounding, we have:
At x 1, 500x 6 1305.
At x 2, 500x 6 94,212.
8. (a) f (10) 8. 7 (b) f (0) 7 (c) f (x h) f (x )
(d) lim
x
x→a 9.
7
f (a)
a 7
a lim 0 x→a 0 2x 3x 2 1 2 6x 2x d
(4x 3)
dx 4x 1 0 2x 1 42x 4 d3
(x )
dx 3x 2 2x 6x
d
(2x 2)
dx 0 4x 3 d
(21x 2)
dx d
(15)
dx 21x 2 d
(4x)
dx 4x
12x 2 dy
dx
d 2y
dx 2 d
d
(5x 3)
(3x 5) 15x 2 15x 4
dx
dx
d
d
(15x 2)
(15x 4) 30x 60x 3
dx
dx dy
dx d
(4x 2)
dx 2 dy
dx 2 1 d
(3x 2)
dx d
(7x 3)
dx 21x 2 2x d2
(x )
dx 2 8x
7
lim
x→a x d
(1)
dx d
(x)
dx d4
(x )
dx d 2y
dx 2 0 d
(x)
dx d
(2x)
dx 4x 3
[0, 5] by [ 6, 6] d
(x)
dx d
( 1)
dx 0 1 1 d
(1)
dx d 2y
dx 2 7. d 12
x
dx 2 d2
(x )
dx 0 2 d
d
(x)
(1) 2x
dx
dx
d
(1) 2 0 2
dx 3 8 d
(8x)
dx 0 d
( 8x 3)
dx 8x
d
(8)
dx d
(1)
dx
3 8
24x 4 0 24x 4 78 10. Section 3.3
dy
dx d1 4
x
dx 4
d
(3)
dx x 5 x 5 d 2y
dx 2 x 4 x 6 dy
dx d1 2
x
dx 2 d
(x 1)
dx (x 18.
x 4 3 x 3 d
( x 5)
dx 5x
11. (a) d1 3
x
dx 3 5 4x x 2 x 0 4 d
(x 3 )
dx 1)(2x) (x 2x 2 2x x2 3x 2 2x 1) (x 19. d (x
dx (x dy
dx d
[(x
dx 1)(x d3
(x
dx 3x (x 2 2 x2 d
1) (x
dx 1)
(2x 3 3x 2 dy
dx 2 1)] x 1) h lim f (x 3x
3x x( 1)2 x 2
2 2)(2x h)
h h→0 lim (1) 3) 5x 6) f (x) lim h→0 1 h→0 d
dx (x 2 3) d
( u)
dx u(x) is a function of x. x 2 u(x lim u(x 3
x2 d x2 3
x
dx
3
1
x2 d
(x
dx 3x 1 ) 1 2 3x f (x ) 0 dy
14.
dx d
dx 0 5x x2 5x
x2 2 2x d (x
dx
d
(1
dx (3x 5
2 x 1 ) 0 1) 3x d1x
dx 1 x 2
x 2 2x 1
(1 x 2)2 (1 d
x2
dx 1 x 3 (1 (2x
2)2 23. (a) At x 5)(3) 4 2)2 0, d
f (x)
dx 1 d
(uv)
dx u(0)v (0) ( 1)( 3) f (x)
[ f (x)]2 v(0)u (0) 13 x ) 0, (c) At x 0, (5)(2) x)(2x) (d) At x
7(2) x 3)(2x) x 2( 3x 2)
(1 x 3)2 d
(c)
dx v(0)u (0) u(0)v (0)
du
[v(0)]2
dx v
( 1)( 3) (5)(2)
7
( 1)2 (b) At x
2 d x3 1
dx
x3
3
x4 x 2)( 1) (1
(1 x 2)2 du
dx 19
(3x d
(1 5x 1
dx
5
2
x2
x3 3 1)(x 2 x
x3
3 2)(2)
(3x u(x) [ f (x)]2 (5)(2)
d 2x
dx 3x u(x) d
d
(c f (x)) c
f (x) f (x)
dx
dx
d
d
c
f (x) 0 c
f (x)
dx
dx d
1
22.
dx f (x) This is equivalent to the answer in part (a).
dy
13.
dx h)
h h→0 3) h) [ u(x)]
h h)
h h→0 lim (x 2 u(x lim h→0 3) (x) 21. dy
dx d x2
dx x 2 2)
2) f (x) h→0 h 1 x2 17. 1
1)2 x. d
(x)
dx 3
dx
x
dx x(2x) dy
dx x 12 6x 2
(x 2 3x 2)2 1 x2 16. 1) 2 d
dx dy
dx x( 5x 6) (2x 3 3x 2
(x 2 3x 2)2 1)(1) lim 2x x (x 2 15. x 2 2
2 2)(2x 3) (x 2 3x
(x 2 3x 2)2 3x (b) Note that u dy
dx x 1) 1)2 1)(x
1)(x 20. (a) Let f (x) (b) 2 1 dy
(b)
dx 12. (a) ( 1 1 1) d
(x 2)
dx 3 d
[(x 1)(x 2 1)]
dx
d
(x 1) (x 2 1) (x 2
dx (x (x x (x
x 2 1 2 x( x 2x 2 x d
dx 2 d
(x 4)
dx 3x dy
dx 1 1) x 4 2x
(1 x 3)2 0, u(0)v (0) v(0)u (0)
[u(0)]2 dv
dx u
( 1)( 3)
(5)2
d
(7v
dx 2u) 2( 3) 20 7
25 7v (0) 2u (0) (x h)
h x Section 3.3 24. (a) At x 2, d
(uv)
dx (3)(2) (1)( 4) u(2)v (2) v(2)u (2) 2 v(2)u (2) u(2)v (2)
du
[v(2)]2
dx v
(1)( 4) (3)(2)
10
(1)2 (b) At x 2, 29. y (x) 2, [ 3, 3] by [ 10, 30] d
(d) Use the result from part (a) for
(uv).
dx
d
At x 2,
(3u 2v 2uv)
dx
d
3u (2) 2v (2) 2 (uv)
dx 3( 4) 2(2) 30. y (x) 3x 2
y ( 2) 12
The tangent line has slope 12 and passes through ( 2,
so its equation is y 2(2)
xintercept is 3
x
2 3x
3
3(2)2 3 9 y 1
and passes through (2, 3).
9
1
(x
9
1
x
9 2) 12x 3 29
9 31. y (x) (x 2 1)(4) 4x(2x)
(x 2 1)2 At the origin: y (0)
The tangent is y
At (1, 2): y (1) 4x 2 4
(x 2 1)2 4
4x. 0 The tangent is y 2. Graphical support: Graphical support: [ 4.7, 4.7] by [ 3.1, 3.1]
[ 4.7, 4.7] by [ 2.1, 4.1] 28. y (x) 3x 2 1
The slope is 4 when 3x 2 1 4, at x
1. The tangent
at x
1 has slope 4 and passes through ( 1, 2), so its
equation is y 4(x 1) 2, or y 4x 2. The tangent at
x 1 has slope 4 and passes through (1, 2), so its equation
is y 4(x 1) 2, or y 4x 2. The smallest slope
occurs when 3x 2 1 is minimized, so the smallest slope is
1 and occurs at x 0. 32. y (x)
y (2) x 2)(0) 8(2x)
(4 x 2)2 (4 16x
x 2)2 1
2 The tangent has slope
equation is y 1
(x
2 Graphical support: Graphical support: [ 3, 5] by [ 2, 4]
[ 4.7, 4.7] by [ 3.1, 3.1] (4 8), 16. The [ 3, 3] by [ 20, 20] 2 The tangent line has slope 9, so the perpendicular line has y 8, or y 6, so the 3
slope is . (iii)
2 slope 2) Graphical support: 25. y (x) 2x 5
y (3) 2(3) 5 11
The slope is 11. (iii)
26. The given equation is equivalent to y 12(x 4
and the yintercept is 16.
3 12 27. y (x)
y (2) 1 Graphical support: u(2)v (2) v(2)u (2)
dv
[u(2)]2
dx u
(3)(2) (1)( 4)
10
(3)2
9 (c) At x 6x 2 6x 12
6(x 2 x 2)
6(x 1)(x 2)
The tangent is parallel to the xaxis when y
0, at x
and at x 2. Since y( 1) 27 and y(2) 0, the two
points where this occurs are ( 1, 27) and (2, 0). 79 1
and passes through (2, 1). Its
2
1
2) 1, or y
x 2.
2 80 33. Section 3.4
an 2
d nRT
V2
dV V nb
d
(V nb) (nRT) dP
dV (c)
(nRT) dV (V
0
(V nR T
nb)2 d
(V
dV nb) nb)2 2an 2V d
(an 2V 2)
dV 3 d
(uv)
dx (V 34. 35. ds
dt
d 2s
dt 2 d
(4.9t 2)
dt
d
(9.8t)
dt dR
dM 2an
V3 nRT
nb)2 CM Exploration 1 M2 37. If the radius of a sphere is changed by a very small amount
r, the change in the volume can be thought of as a very
thin layer with an area given by the surface area, 4 r 2, and
a thickness given by r. Therefore, the change in the
volume can be thought of as (4 r 2)( r), which means that
the change in the volume divided by the change in the
radius is just 4 r 2.
38. Let t(x) be the number of trees and y(x) be the yield per tree
x years from now. Then t(0) d
(ty)
dx 156, y(0) 12, t (0) 13, 1.5. The rate of increase of production is t(0)y (0) y(0)t (0) (156)(1.5) (12)(13) 39. Let m(x) be the number of members and c(x) be the
pavillion cost x years from now. Then m(0)
250, m (0) 6, and c (0) of each member’s share is
(65)(10) (250)(6)
(65)2 dc
dx m Growth Rings on a Tree 1. Figure 3.22 is a better model, as it shows rings of equal
area as opposed to rings of equal width. It is not likely that
a tree could sustain increased growth year after year,
although climate conditions do produce some years of
greater growth than others.
2. Rings of equal area suggest that the tree adds
approximately the same amount of wood to its girth each
year. With access to approximately the same raw materials
from which to make the wood each year, this is how most
trees actually grow.
3. Since change in area is constant, so also is
change in area
. If we denote this latter constant by k, we
2
k
have
r, which means that r varies
change in radius inversely as the change in the radius. In other words, the
change in radius must get smaller when r gets bigger, and
viceversa. Exploration 2 390 bushels of annual production per year. c(0) u du
u
v dv
v
(u du)(v) (u)(v dv)
(v dv)(v)
uv vdu uv udv
v 2 vdv
vdu udv
v2 s Section 3.4 Velocity and Other Rates
of Change (pp. 122–133) 36. If the radius of a circle is changed by a very small amount
r, the change in the area can be thought of as a very thin
strip with length given by the circumference, 2 r, and
width r. Therefore, the change in the area can be thought
of as (2 r)( r), which means that the change in the area
divided by the change in the radius is just 2 r. and y (0) u
v 9.8
M
3
13
M
3 dv
dx u . This is equivalent to the product (d) Because dx is ‘infinitely small,” and this could be
thought of as dividing by zero. 9.8t C
d
M2
2
dM
dC2
M
dM 2 du
dx rule given in the text. (e) d
2 v Modeling Horizontal Motion 1. The particle reverses direction at about t
t 2.06. 0.61 and 65, 10. The rate of change
m(0)c (0) c(0)m (0)
[m(0)]2 0.201 dollars per year. Each member’s share of the cost is decreasing by approximately
20 cents per year.
40. (a) It is insignificant in the limiting case and can be treated
as zero (and removed from the expression).
(b) It was “rejected” because it is incomparably smaller
than the other terms: v du and u dv. 2. When the trace cursor is moving to the right the particle is
moving to the right, and when the cursor is moving to the
left the particle is moving to the left. Again we find the
particle reverses direction at about t 0.61 and t 2.06. ...
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