Practice M151B Problems for Final Exam
Calculators will not be allowed on the exam. Unjustified answers will not receive credit. On the exam you will be given the following identities: n(n + 1) ; k= 2 k=1
n
n(n + 1)(2n + 1) k = ; 6 k=1
2
n
n
k3 =
k=1
n(n + 1) 2
2
.
1. Compute each of the following limits: 1a.
x2
lim -
x2
x . + 3x - 10
1
1b.
x0
lim xesin( x ) .
1c.
x sin x . x0 (1 - ex )2 lim
x
1d. lim [(x + 1)1/3 - x1/3 ]. ab. Show that
1e. The geometric mean of two positive real numbers a and b is defined as ab = lim (
x
a1/x + b1/x x ) . 2
2a. Find a value for c that makes the given function continuous at all points. f (x) =
sin x , x
c,
x=0 . x=0
2b. Determine whether or not your function from (2a) is differentiable at x = 0. If it is differentiable at this point, compute its derivative there. 3. Find an equation for the line that is tangent to the graph of f (x) = at the point x = 0. 4. Suppose the angle of elevation of the Sun is decreasing at a rate of .25 rad/hr. How fast is the shadow cast by a 400 ft tall building increasing when the angle of elevation of the Sun is ? 6 1 xex 1 + x2
5. Suppose f (x) is continuous on the interval [a, b] and differentiable on the interval (a, b). Show that if f (x) = x for all x (a, b), then there exists some value c (0, 1) so that f (b) - f (a) = c(b - a). 6. Let f (x) = x1/3 (x + 3)2/3 , 6a. Locate the critical points of f and determine the intervals on which f is increasing and the intervals on which f is decreasing. 6b. Locate the possible inflection points for f and determine the intervals on which f is concave up and the intervals on which it is concave down. 6c. Evaluate f at the critical points and at the possible inflection points, and determine the boundary behavior of f by computing limits as x . - < x < .
6d. Use your information from Parts a-c to sketch a graph of this function.
7. Find the side-lengths that maximize the area of an isosceles triangle with given perimeter P = 10. (An isosceles triangle is a triangle with two sidelengths equal.) 8. Find all fixed points for the recursion equation 1 3 an+1 = an + . 4 an
1 Sketch a graph of the function f (a) = 3 a+ a , and use the method of cobwebbing to determine 4 1 whether or not one of these fixed points will be achieved from the starting value a0 = 2 .
9. Find all fixed points for the recursion equation xt+1 = 1 + 2 xt
and determine whether or not each is asymptotically stable or unstable. 10. Suppose a function f (x) is continuous on the interval [0, 1] and that you are given the following table of values: x f (x) 1/8 1/2 3/8 1/3 5/8 -1 7/8 -2 Table 1: Values of f (x) for Problem 1. Use an appropriate Riemann sum to approximate 11. Use the method of Riemann sums to evaluate
2 1 0
f (x)dx.
x + x2 dx.
1
2
12. Evaluate the following indefinite integrals. 12a. ex cos(ex )dx. 12b. 12b. cos-1 xdx. 13. Evaluate the following definite integrals. 13a.
1 3
x dx 1+x
4
x2 dx. 1 + x3
13b.
0
x sec2 xdx.
16. Suppose the base of a certain solid is the region in the xy-plane between the line y = x and the parabola y = x2 . Find the volume of the solid created if every cross section is a right isosceles triangle with hypotenuse in the xy-plane perpendicular to the x-axis.
15. Find the volume obtained when the region between the graphs of y = ex and y = e-x , x [0, 2], is rotated about the x-axis.
14. Find the area of the region bounded by the graphs of y = x4 and y = 20 - x2 .
Solutions
1a. We have
x2
lim -
x2
where we have observed that x - 2 is negative for x to the left of 2. -|x|e xesin( x ) |x|e. We have
x0
1
x x = lim = -, - (x - 2)(x + 5) + 3x - 10 x2
1b. We apply the Squeeze Theorem in this case, using the inequality
lim -|x|e = lim |x|e = 0,
x0
and so according to the Squeeze Theorem
x0
lim xesin( x ) = 0. 3
1
1c. We apply L'Hospital's Rule twice, sin x + x cos x sin x + x cos x x sin x = lim = lim x )2 x )(-ex ) x0 2(1 - e x0 x0 (1 - e 2e2x - 2ex 2 cos x - x sin x = lim = 1. x0 4e2x - 2ex lim 1d. This limit has the indeterminate form - , so the first thing we do is rearrange it 0 into an expression with the form 0 . We have 1 lim (x + 1)1/3 - x1/3 = lim x1/3 [(1 + )1/3 - 1] x x x 1 1 1 1 (1 + x )1/3 - 1 (1 + x )-2/3 (- x2 ) 3 = lim = lim x x x-1/3 - 1 x-4/3 3 1 -2/3 1 = 0. = lim (1 + ) x x x2/3 1e. We observe that this limit has the general form 1 , and so we can apply L'Hospital's rule. We have
x
lim (
a1/x +b1/x a1/x +b1/x x a1/x + b1/x x ) ) 2 2 = lim ex ln( ) = lim eln( x x 2
= elimx x ln( In order to compute this limit, we write ln( a a1/x + b1/x lim x ln( ) = lim x x 2 = lim
x a1/x
1/x +b1/x
a1/x +b1/x ) 2
.
2 1 x
)
=
1 1 (a1/x ln a + b1/x ln b) = (ln a + ln b), 1/x +b 2 0 as x . The limit is 1 1/2 = e 2 ln(ab) = eln(ab) = ab.
1 x
2 ( 1 a1/x (ln a)(- x12 ) a1/x +b1/x 2 lim x - x12
+ 1 b1/x (ln b)(- x12 )) 2
where in this last step we have used that e 2 (ln a+ln b) 2a. Since
1
sin x = 1, x0 x we can make this function continuous at all points by choosing c = 1. lim
2b. Since the function is separately defined at x = 0, we must proceed from the definition of differentiation. We compute f (0) = lim
sin h -1 f (0 + h) - f (0) = lim h h0 h0 h h cos h - 1 - sin h sin h - h = lim = lim = 0, = lim h0 h0 h0 h2 2h 2
4
where the last two steps both used L'Hospital's rule. We conclude that this function is differentiable at x = 0, and that f (0) = 0. 3. First, f (x) = (1 + x2 )(ex + xex ) - xex (2x) f (0) = 1, (1 + x2 )2
which is the slope of the tangent line. Using f (0) = 0 and the general point-slope form y - f (a) = f (a)(x - a), we conclude y = x. 4. First, observe that what we know is d = -.25 rad/hr and what we want to know is dt where x is the length of the shadow (see the diagram).
dx , dt
We see that the relation between and x is tan = 400 . x
Upon taking a derivative of this equation with respect to t, we obtain sec2 400 dx d =- 2 , dt x dt
1 cos2
6
where we can now fix = , so that sec2 = 6 Combining these observations, we have
=
1
3 4
= 4 , while x = 3
400 tan 6
= 400 3.
x2 d 4 dx =- sec2 = -3(400)(-.25) = +400 dt ft/hr. 400 dt 6 3 5. According to the Mean Value Theorem there exists some value c (a, b) so that f (b) - f (a) = f (c). b-a In this case f (c) = c, and so we conclude f (b) - f (a) = c f (b) - f (a) = c(b - a). b-a 5
6a. The derivative of f (x) is f (x) =
x+1 , x2/3 (x + 3)1/3
from which we find the critical points x = -3, -1, 0. We see that f is increasing on (-, -3] [-1, ) and decreasing on [-3, -1]. 6b. The second derivative of f (x) is 2 , + 3)4/3
f (x) = -
x5/3 (x
6c. Evaluating f at the critical points, possible inflection points, and at the endpoints, we have: f (-3) = 0 f (-1) = - 22/3 f (0) = 0
x-
from which we find that the possible inflection points are x = 0, -3. We see that f is concave up on (-, -3) (-3, 0) and concave down on (0, ).
lim x1/3 (x + 3)2/3 = -
x
lim x1/3 (x + 3)2/3 = + .
6d. Your plot should look something like this:
7. Let y denote the length of the sides of equal length, and let x denote the length of the side between them. Then the perimeter is 1 10 = 2y + x y = 5 - x. 2 6
By the Pythagorean Theorem, the height of such a triangle is h = area to be maximized is
1 y 2 - 4 x2 , and so the
1 1 1 1 1 1 A = x y 2 - x2 A(x) = x (5 - x)2 - x2 = x 25 - 5x, 2 4 2 2 4 2
0 x 5.
(The upper limit of 5 is clear both because a value of x larger than this would put a negative number under the radical, and because the single side cannot be more than half the perimeter.) In order to maximize A(x), we compute
25 - 15 x 2 4 . A (x) = 25 - 5x
The critical values are x = 10 , 5, where we observe that x = 5 is also a boundary value. 3 Checking A(x) at the critical and boundary values, we find A(0) = 0 A( 5 10 )= 3 3 A(5) = 0. 25 25 = 3 3 3
10 3
We conclude that the maximum area is 3253 and the side-lengths are x = 1 5 - 2 ( 10 ) = 10 . That is, an equilateral triangle. 3 3
and y =
8. The fixed points solve
1 1 1 3 a = a + a = a2 = 4. 4 a 4 a We conclude that the fixed points are 2. In order to use cobwebbing, we must sketch a graph of the function 3 1 f (a) = a + . 4 a First, setting 1 3 f (a) = - 2 = 0, 4 a 2 2 we find that the critical points are a = 3 , 0. The function is increasing on (-, - 3 ] 2 2 2 [ 3 , ) and decreasing on [- 3 , 3 ]. Next, f (a) = 2 , a3
and so the only possible point of inflection is a = 0. The function is concave down on (-, 0)
7
and concave up on (0, ). Finally, 1 3 lim ( a + ) = - a- 4 a 2 f (- ) = - 3 3 1 3 lim ( a + ) = - x0- 4 a 3 1 lim+ ( a + ) = + x0 4 a 2 f( ) = 3 3 3 1 lim ( a + ) = a- 4 a The plot of this function and the cobwebbing are depicted below. We conclude
n
lim an = 2.
9. In order to find the fixed points, we solve 2 x=1+ , x which becomes (upon multiplication by x) x2 - x - 2 = (x - 2)(x + 1) = 0,
2 and the fixed points are x = -1, 2. In order to check for stability we set f (x) = 1 + x , and compute 2 f (x) = - 2 . x
8
We have f (-1) = - 2 -1 is unstable 1 f (2) = - 2 is stable. 2 10. Since no value for f (x) is given at either x = 0 or at x = 1, we cannot take a Riemann sum with left or right endpoints. We see, however, that the values of x are precisely the midpoints 1 1 of the subintervals in the partition P = [0, 4 , 2 , 3 , 1]. The most reasonable Riemann sum is 4
4
f (ck )xk ,
k=1
where the ck are the interval midpoints. That is, 1 13 1 1 f (ck )xk = ( + - 1 - 2) = - . 2 3 4 24 k=1 11. In this case x =
n b-a n 4
=
2-1 n
1 = n , and we use right endpoints xk = 1 + kx. We have
An =
k=1 n
[(1 + kx) + (1 + kx)2 ]x
k k2 1 k ) + (1 + 2 + 2 )] = [(1 + n n n n k=1 1 = n 1 1+ 2 n k=1
n
1 k+ n k=1
n
2 1+ 2 n k=1
n
1 k+ 3 n k=1
n
n
k2
k=1
1 n(n + 1) 2 n(n + 1) 1 n(n + 1)(2n + 1) = 1+...