5128_ch04ansTE_pp652-661
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652 Additional Answers<br />
31. Maximum value is 11 at x 5;<br />
minimum value is 5 on the interval [3, 2];<br />
local maximum at (5, 9)<br />
32. Maximum value is 4 on the interval [5, 7];<br />
minimum value is 4 on the interval [2, 1].<br />
33. Maximum value is 5 on the interval [3, );<br />
minimum value is 5 on the interval (, 2].<br />
34. Minimum value is 4 on the interval [1, 3].<br />
35. crit. pt. derivative extremum value<br />
x 4 5 0 local max 1 2<br />
10 1/3 1.034<br />
25<br />
x 0 undefined local min 0<br />
36.<br />
crit. pt. derivative extremum value<br />
x 1 0 minimum 3<br />
x 0 undefined local max 0<br />
x 1 0 minimum 3<br />
37.<br />
crit. pt. derivative extremum value<br />
x 2 undefined local max 0<br />
x 2 0 minimum 2<br />
x 2 0 maximum 2<br />
x 2 undefined local min 0<br />
38. crit. pt. derivative extremum value<br />
x 0 0 minimum 0<br />
x 1 2<br />
0 local max 1 44<br />
15 1/2 4.462<br />
5<br />
125<br />
x 3 undefined minimum 0<br />
39. crit. pt. derivative extremum value<br />
x 1 undefined minimum 2<br />
40. crit. pt. derivative extremum value<br />
x 0 undefined local min 3<br />
x 1 0 local max 4<br />
41. crit. pt. derivative extremum value<br />
x 1 0 maximum 5<br />
x 1 undefined local min 1<br />
x 3 0 maximum 5<br />
42.<br />
crit. pt. derivative extremum value<br />
x 1 0 local max 4<br />
x 3.155 0 local max 3.079<br />
46. False. Consider the graph below.<br />
y<br />
53. (a) f (x) 3ax 2 2bx c is a quadratic, so it<br />
can have 0, 1, or 2 zeros, which would be the<br />
critical points of f. Examples:<br />
[–3, 3] by [–5, 5]<br />
The function f(x) x 3 3x has two critical points at x 1 and x 1.<br />
[–3, 3] by [–5, 5]<br />
The function f(x) x 3 1 has one critical point at x 0.<br />
[–3, 3] by [–5, 5]<br />
The function f(x) x 3 x has no critical points.<br />
54. (a) By definition of local maximum, there is an open interval containing c<br />
where f(x) f(c), so f(x) f(c) 0.<br />
(b) Since f (c) exists, the limit exists. Because x → c ,(x c) 0, and<br />
the sign of the quotient must be negative (or zero). This means the<br />
limit is nonpositive.<br />
(c) Since f (c) exists, the limit exists. Because as x → c ,(x c) 0, and<br />
the sign of the quotient must be positive (or zero). This means the limit<br />
is nonnegative.<br />
(d) Assuming that f(c) exists, the one-sided limits in (b) and (c) above<br />
must exist and be equal. Since one is nonpositive and one is nonnegative,<br />
the only possible common value is 0.<br />
(e) There will be an open interval containing c where f (x) f (c) 0. The<br />
difference quotient for the left-hand derivative will have to be negative<br />
(or zero), and the difference quotient for the right-hand derivative will<br />
have to be positive (or zero). Taking the limit, the left-hand derivative<br />
will be nonpositive, and the right-hand derivative will be nonnegative.<br />
Therefore, the only possible value for f (c) is 0.<br />
55. (a)<br />
x<br />
[–0.1, 0.6] by [–1.5, 1.5]<br />
f(0) 0 is not a local extreme value because in any open interval containing<br />
x 0, there are infinitely many points where f (x) 1 and<br />
where f(x) 1.<br />
(b) One possible answer, on the interval [0, 1]:<br />
f (x) <br />
(1 x) cos 1<br />
, 0 x 1<br />
1 x<br />
0, x 1<br />
This function has no local extreme value at x 1. Note that it is continuous<br />
on [0, 1].
Additional Answers 653<br />
Section 4.2<br />
Exercises 4.2<br />
11. Because the trucker’s average speed was 79.5 mph, and by the Mean Value<br />
Theorem, the trucker must have been going that speed at least once during<br />
the trip.<br />
12. Let f (t) denote the temperature indicated after t seconds. We assume that<br />
f (t) is defined and continuous for 0 t 20. The average rate of change<br />
is 10.6°F/sec. Therefore, by the Mean Value Theorem, f (c) 10.6°F/sec<br />
for some value of c in [0, 20].<br />
13. Because its average speed was approximately 7.667 knots, and by the<br />
Mean Value Theorem, it must have been going that speed at least once<br />
during the trip.<br />
14. The runner’s average speed for the marathon was approximately 11.909 mph.<br />
Therefore, by the Mean Value Theorem, the runner must have been going<br />
that speed at least once during the marathon. Since the initial speed and<br />
final speed are both 0 mph and the runner’s speed is continuous, by the<br />
Intermediate Value Theorem, the runner’s speed must have been 11 mph<br />
at least twice.<br />
25. (a) Local max at (2, 1/4);<br />
local min at (2, 1/4)<br />
(b) On (, 2] and [2, )<br />
(c) On [2, 2]<br />
26. (a) None (b) None<br />
(c) On (, 2), (2, 2), and (2, )<br />
27. (a) Local maximum at (1.126, 0.036);<br />
local minimum at (0.559, 2.639)<br />
(b) On (, 1.126] and [0.559, )<br />
(c) On [1.126, 0.559]<br />
39. Possible answers:<br />
(a)<br />
(b)<br />
(c)<br />
(d)<br />
41. One possible answer:<br />
[–3, 3] by [–15, 15]<br />
42. One possible answer:<br />
[–2, 4] by [–2, 4]<br />
(b)<br />
(c)<br />
[–1, 4] by [0, 3.5]<br />
[–1, 4] by [0, 3.5]<br />
40. Possible answers:<br />
(a)<br />
46. Because the Mean Value Theorem applies to the function y sin x on any<br />
interval, and y cos x is the derivative of sin x. So, between any two zeros<br />
of sin x, its derivative, cos x, must be zero at least once.<br />
47. f(x) must be zero at least once between a and b by the Intermediate Value<br />
Theorem.<br />
Now suppose that f (x) is zero twice between a and b. Then by the Mean<br />
Value Theorem, f (x) would have to be zero at least once between the two<br />
zeros of f(x), but this can’t be true since we are given that f (x) 0 on this<br />
interval.<br />
Therefore, f(x) is zero once and only once between a and b.<br />
48. Let f(x) x 4 3x 1. Then f(x) is continuous and differentiable everywhere.<br />
f (x) 4x 3 3, which is never zero between x 2 and x 1.<br />
Since f (2) 11 and f (1) 1, exercise 47 applies, and f(x) has exactly<br />
one zero between x 2 and x 1.<br />
49. Let f(x) x ln (x 1). Then f(x) is continuous and differentiable<br />
1<br />
everywhere on [0, 3]. f (x) 1 , which is never zero on [0, 3].<br />
x 1<br />
Now f(0) 0, so x 0 is one solution of the equation. If there were a<br />
second solution, f(x) would be zero twice in [0, 3], and by the Mean Value<br />
Theorem, f (x) would have to be zero somewhere between the two zeros<br />
of f(x). But this can’t happen, since f (x) is never zero on [0, 3]. Therefore,<br />
f(x) 0 has exactly one solution in the interval [0, 3].<br />
50. Consider the function k(x) f(x) g(x). k(x) is continuous and differentiable<br />
on [a, b], and since k(a) f(a) g(a) 0 and
654 Additional Answers<br />
k(b) f(b) g(b) 0, by the Mean Value Theorem, there must be a point<br />
c in (a, b) where k(c) 0. But since k(c) f (c) g(c), this means<br />
that f (c) g(c), and c is a point where the graphs of f and g have parallel<br />
or identical tangent lines.<br />
Section 4.3<br />
Exercises 4.3<br />
may vary.<br />
(a) y <br />
(b)<br />
may vary.<br />
2898438<br />
(a) y 1 49.<br />
252<br />
(b)<br />
b a<br />
51. False. For example, the function x 3 is increasing on (1, 1), but f(0) 0.<br />
52. True. In fact, f is increasing on [a, b] by Corollary 1 to the Mean Value<br />
Theorem.<br />
57. (a) Increasing: [2, 1.3] and [1.3, 2];<br />
decreasing: [1.3, 1.3];<br />
local max: x 1.3<br />
local min: x 1.3<br />
(b) Regression equation: y 3x 2 5<br />
(c) f(x) x 3 5x<br />
[–2.5, 2.5] by [–8, 10]<br />
58. (a) Toward: 0 t 2 and 5 t 8;<br />
away: 2 t 5<br />
(b) A local extremum in this problem is a time/place where Priya changes<br />
the direction of her motion.<br />
(c) Regression equation:<br />
y 0.0820x 3 0.9163x 2 2.5126x 3.3779<br />
(d) f(t) 0.2459t 2 1.8324t 2.5126<br />
toward: 0 t 1.81 and 5.64 t 8;<br />
away: 1.81 t 5.64<br />
59. f(b 1 ) f(a)<br />
b 1 a <br />
1<br />
<br />
b a<br />
a b<br />
1 1 1<br />
f (c) c 2, so c 2 and c 2 ab.<br />
a b<br />
Thus, c ab.<br />
60. f(b ) f(a)<br />
b <br />
a<br />
b2 a2<br />
b b a<br />
a<br />
f (c) 2c, so2c b a and c a b<br />
.<br />
2<br />
61. By the Mean Value Theorem, sin b sin a (cos c)(b a) for some c<br />
between a and b. Taking the absolute value of both sides and using<br />
⏐cos c⏐ 1 gives the result.<br />
62. Apply the Mean Value Theorem to f on [a, b].<br />
Since f(b) f (a), f (b ) f(a)<br />
is negative, and hence f (x) must be negative<br />
at some point between a and b.<br />
b<br />
a<br />
63. Let f(x) be a monotonic function defined on an interval D. For any two<br />
values in D, we may let x 1 be the smaller value and let x 2 be the larger<br />
value, so x 1 x 2 . Then either f(x 1 ) f(x 2 ) (if f is increasing), or<br />
f(x 1 ) f(x 2 ) (if f is decreasing), which means f(x 1 ) f(x 2 ). Therefore, f is<br />
one-to-one.<br />
25. (a) v(t) 2t 4 (b) a(t) 2<br />
(c) It begins at position 3 moving in a negative direction. It moves to position<br />
1 when t 2, and then changes direction, moving in a positive<br />
direction thereafter.<br />
26. (a) v(t) 2 2t (b) a(t) 2<br />
(c) It begins at position 6 and moves in the negative direction thereafter.<br />
27. (a) v(t) 3t 2 3 (b) a(t) 6t<br />
(c) It begins at position 3 moving in a negative direction. It moves to position<br />
1 when t 1, and then changes direction, moving in a positive direction<br />
thereafter.<br />
28. (a) v(t) 6t 6t 2 (b) a(t) 6 12t<br />
(c) It begins at position 0. It starts moving in the positive direction until it<br />
reaches position 1 when t 1, and then it changes direction. It moves<br />
in the negative direction thereafter.<br />
31. Some calculators use different logistic regression equations, so answers<br />
12655.179<br />
<br />
1 12.871e 0.0326x<br />
[0, 140] by [–200, 12000]<br />
(c) The regression equation predicts a population of 12,209,870. (This is<br />
remarkably close to the 2000 census number of 12,281,054.)<br />
(d) The second derivative has a zero at about 78, indicating that the population<br />
was growing the fastest in 1898. This corresponds to the inflection<br />
point on the regression curve.<br />
(e) The regression equation predicts a population limit of about<br />
12,655,179.<br />
32. Some calculators use different logistic regression equations, so answers<br />
6.<br />
288<br />
e<br />
0.<br />
851x<br />
[0, 9] by [–3.1 10 6 , 3.2 10 7 ]<br />
(c) The zero of the second derivative is about 4.6, which puts the fastest<br />
growth during 1981. This corresponds to the inflection point of the regression<br />
curve.<br />
(d) The regression curve predicts that cable subscribers will approach a<br />
limit of 28,984,386 12,168,450 subscribers (about 41 million).<br />
(e) For many reasons, the potential market for cable TV subscribers between<br />
1977 and 1985 was limited compared to today. Some reasons are: the<br />
technology has improved, televisions have become cheaper, there are<br />
more cable stations to entice viewers, and many more communities<br />
have access to cable.<br />
33. y3 3x 2 and y6x.<br />
y0 at 1. y(1) 0 and y(1) 0, so there is a local minimum at<br />
(1, 3) and a local maximum at (1, 7).<br />
34. y5x 4 80 and y20x 3 .<br />
y0 at 2. y(2) 0 and y(2) 0, so there is a local maximum at<br />
(2, 228) and a local minimum at (2, 28).
Additional Answers 655<br />
35. y3x 2 6x and y6x 6.<br />
y0 at 2 and 0. y(2) 0 and y(0) 0, so there is a local maximum<br />
at (– 2, 2) and a local minimum at (0, 2).<br />
36. y15x 4 75x 2 60 and y60x 3 150x.<br />
y0 at 1 and 2. y(2) 0, y(1) 0, y(1) 0, and y(2) 0;<br />
so there are local maxima at (2, 4) and (1, 58), and there are local<br />
minima at (1, 18) and (2, 36).<br />
37. y(x 1)e x and y(x 2)e x .<br />
y0 at 1 and y(1) 0 , so there is a local minimum at (1, 1/e).<br />
38. y(1 x)e x and y(x 2)e x .<br />
y0 at 1 and y(1) 0, so there is a local maximum at (1, 1e)<br />
41.<br />
y<br />
y = f′(x)<br />
y = f(x)<br />
P<br />
47. One possible answer:<br />
y<br />
(–2, 8) 10<br />
(0, 4)<br />
–5<br />
(2, 0)<br />
–10<br />
48. One possible answer:<br />
5<br />
x<br />
x<br />
y = f′′(x)<br />
42.<br />
43. No. f must have a horizontal tangent line at that point, but it could be increasing<br />
(or decreasing) on both sides of the point, and there would be no<br />
local extremum.<br />
44. No. f (x) could still be positive (or negative) on both sides of x c, in<br />
which case the concavity of the function wouldn’t change at x c.<br />
45. One possible answer:<br />
y<br />
–5<br />
5<br />
5<br />
x<br />
51. (a) Absolute maximum at (1, 2);<br />
absolute minimum at (3, 2)<br />
(b) None<br />
(c) One possible answer:<br />
2<br />
1<br />
–1<br />
–2<br />
y<br />
1<br />
y = f(x)<br />
52. (a) Absolute maximum at (0, 2);<br />
absolute minimum at (2, 1) and (2, 1)<br />
(b) At (1, 0) and (1, 0)<br />
(c) One possible answer:<br />
2<br />
3<br />
x<br />
–5<br />
46. One possible answer:<br />
(d) f(3) f(3), and 1 f(3) 0.
656 Additional Answers<br />
53.<br />
y<br />
4<br />
3<br />
2<br />
1<br />
y = f(x)<br />
By the First Derivative test, f has relative minima at x 1 and x 4<br />
and relative maxima at x 2.<br />
(b) f (x) 6(x2 ) <br />
(x<br />
2 2 2x(6x)<br />
2) 2 6(x<br />
2 2)<br />
(x 2 .<br />
2) 2 Since f changes sign at<br />
x 2, there are points of inflection at x 2 .<br />
(c) Comparing the two relative maxima, we see that f(2) 3 ln 6 4<br />
and f(2) 3ln 6 4. The absolute maximum is 3 ln 6 4.<br />
54.<br />
–1<br />
1<br />
2<br />
3<br />
4<br />
5<br />
6<br />
x<br />
Section 4.4<br />
Exercises 4.4<br />
18. (a) V(x) 2x 3 25x 2 75x<br />
(b) Domain: (0, 5)<br />
61. (a) In Exercise 7, a 4 and b 21,<br />
b<br />
so 7 ,which is the x-value where the point of inflection occurs.<br />
3 a 4<br />
The local extrema are at x 2 and x 3 ,which are symmetric<br />
2<br />
about x 7 4 .<br />
b<br />
(b) In exercise 2, a 2 and b 6, so 1, which is the x-value<br />
3 a<br />
where the point of inflection occurs. The local extrema are at x 0 and<br />
x 2, which are symmetric about x 1.<br />
(c) f (x) 3ax 2 2bx c and f (x) 6ax 2b.<br />
The point of inflection will occur where f (x) 0, which is at<br />
b<br />
x .<br />
3 a<br />
If there are local extrema, they will occur at the zeros of f (x). since<br />
f (x) is quadratic, its graph is a parabola and any zeros will be<br />
symmetric about the vertex which will also be where f (x) 0.<br />
abce 62. (a) f (x) b x<br />
(e bx<br />
,<br />
a) 2 so the sign of f (x) is the<br />
same as the sign of the product abc.<br />
(b) f (x) ab2 ceb<br />
x ( e bx a)<br />
( e bx<br />
.<br />
a)<br />
3 Since a 0, this changes sign<br />
when x ln a<br />
due to the e bx a factor in the numerator, and<br />
b<br />
there is a point of inflection at that location.<br />
63. (a) Since f (x) is quadratic it must have 0, 1, or 2 zeros. If f (x) has 0 or 1<br />
zeros, it will not change sign and the concavity of f(x) will not change,<br />
so there is no point of inflection. If f (x) has 2 zeros, it will change<br />
sign twice, and f(x) will have 2 points of inflection.<br />
(b) f(x) has two points of inflection if and only if 3b 2 8ac.<br />
Quick Quiz (Sections 4.1–4.3)<br />
6x<br />
4. (a) f(x) x 2 2 2(x 1)( x 2)<br />
2<br />
x 2 . Critical numbers are x 1<br />
2<br />
and x 2. The sign graph of f is shown below.<br />
<br />
<br />
<br />
–2 1 2 4<br />
x<br />
(c) Maximum volume 66.02 in 3 when x 1.96 in.<br />
(d) V (x) 6x 2 50x 75, so the critical point is at x 25 57<br />
,<br />
6<br />
which confirms the result in part (c).<br />
19. (a) V(x) 2x(24 2x)(18 2x)<br />
(b) Domain: (0, 9)<br />
[0, 9] by [–400, 1600]<br />
(c) Maximum volume 1309.95 in 3 when x 3.39 in.<br />
(d) V(x) 24x 2 336x 864, so the critical point is at x 7 13,<br />
which confirms the result in part (c).<br />
(e) x 2 in. or x 5 in.<br />
(f) The dimensions of the resulting box are 2x in., (24 2x) in.,<br />
and (18 2x) in. Each of these measurements must be positive, so that<br />
gives the domain of (0, 9).<br />
22. Dimensions: Radius 10<br />
2 3 20<br />
8.16 cm, height 11.55 cm;<br />
3<br />
4000<br />
maximum volume 2418.40 cm 3<br />
33<br />
23. Set r(x) c(x): 4x 1/2 4x. The only positive critical value is x 1, so<br />
profit is maximized at a production level of 1000 units. Note that<br />
(r c)(x) 2(x) 3/2 4 0 for all positive x, so the Second Derivative<br />
Test confirms the maximum.<br />
24. Set r(x) c(x): 2x/(x 2 1) 2 (x 1) 2 . We solve this equation graphically<br />
to find that x 0.294 or x 1.525. The graph of y r(x) c(x)<br />
shows a minimum at x 0.294 and a maximum at x 1.525, so profit is<br />
maximized at a production level of about 1,525 units.<br />
25. Set c(x) c(x)/x:3x 2 20x 30 x 2 10x 30. The only positive<br />
solution is x 5, so average cost is minimized at a production level of<br />
d 2<br />
5000 units. Note that d x2 c( x) 2 0 for all positive x, so the Second<br />
x <br />
Derivative Test confirms the minimum.<br />
26. Set c(x) c(x) /x: xe x e x 4x e x 2x. The only positive solution is<br />
x ln 2, so average cost is minimized at a production level of 1000 ln 2,<br />
d 2<br />
which is about 693 units. Note that d x2 <br />
c( x)<br />
<br />
x <br />
e x 0 for all positive x,<br />
so the Second Derivative Test confirms the minimum.
Additional Answers 657<br />
28. (a) At x 1<br />
(b) a b A<br />
0.1 3.72 0.33<br />
0.2 2.86 0.44<br />
0.3 2.36 0.46<br />
0.4 2.02 0.43<br />
0.5 1.76 0.38<br />
0.6 1.55 0.31<br />
0.7 1.38 0.23<br />
0.8 1.23 0.15<br />
0.9 1.11 0.08<br />
1.0 1.00 0.00<br />
(c)<br />
(c)<br />
Changing the value of k changes the maximum strength, but not the<br />
dimensions of the strongest beam. The graphs for different values of k<br />
look the same except that the vertical scale is different.<br />
38. (a) 6 in. wide by 63 in. deep<br />
(b)<br />
[0, 1.1] by [–0.2, 0.6]<br />
(d) Quadratic:<br />
A 0.91a 2 0.54a 0.34<br />
(c)<br />
[0, 12] by [–2000, 8000]<br />
[0, 12] by [–2000, 8000]<br />
[–0.5, 1.5] by [–0.2, 0.6]<br />
Cubic:<br />
A 1.74a 3 3.78a 2 1.86a 0.19<br />
y x 3 (144 x 2 ) 1/2<br />
Changing the value of k changes the maximum stiffness, but not the dimensions<br />
of the stiffest beam. The graphs for different values of k look<br />
the same except that the vertical scale is different.<br />
39. (a) Maximum speed 10 cm/sec;<br />
maximum speed is at t 1 2 , 3 2 , 5 2 , 7 2 seconds;<br />
[–0.5, 1.5] by [–0.2, 0.6]<br />
Quartic:<br />
A 1.92a 4 5.96a 3 6.87a 2 2.71a 0.12<br />
[–0.5, 1.5] by [–0.2, 0.6]<br />
(e) Quadratic: A 0.42;<br />
cubic: A 0.45;<br />
quartic: A 0.46<br />
37. (a) 43 in. wide by 46 in. deep<br />
(b)<br />
position at those times is s 0 cm (rest position);<br />
acceleration at those times is 0 cm/sec 2<br />
(b) The magnitude of the acceleration is greatest<br />
when the cart is at positions s 10 cm;<br />
The speed of the cart is 0 cm/sec at those times.<br />
57. Let P be the foot of the perpendicular from A to the mirror, and Q be the<br />
foot of the perpendicular from B to the mirror. Suppose the light strikes the<br />
mirror at point R on the way from A to B. Let:<br />
a distance from A to P<br />
b distance from B to Q<br />
c distance from P to Q<br />
x distance from P to R<br />
To minimize the time is to minimize the total distance the light travels<br />
going from A to B. The total distance is<br />
D(x) (x 2 a 2 ) 1/2 ((c x) 2 b 2 ) 1/2 .<br />
Then D(x) 0 and D(x) has it minimum when<br />
ac<br />
x c<br />
x , or, a . It follows that<br />
a b a b<br />
bc<br />
c x , or c x c<br />
. This means that the two triangles APR<br />
a b b a b<br />
and BQR are similar, and the two angles must be equal.<br />
59. (a) d v<br />
cr(2r<br />
dr<br />
0 3r) which is zero when<br />
r 2 3 r 0 .<br />
(b)
658 Additional Answers<br />
63. (a) y(0) 0 (b) y(L) 0<br />
(c) y(0) 0, so d 0. y(0) 0, so c 0.<br />
Then y(L) aL 3 bL 2 H and<br />
y(L) 3aL 2 2bL 0.<br />
H H<br />
Solving, a 2 L 3 and b 3 L 2, which gives the equation shown.<br />
64. (a) V(x) 3 2 a x<br />
<br />
2<br />
2 a2 <br />
a<br />
2 <br />
x<br />
2<br />
2<br />
(b) When a 4: r 4 6<br />
, h 4 3<br />
;<br />
3 3<br />
when a 5: r 5 6<br />
, h 5 3<br />
;<br />
3 3<br />
when a 6: r 26, h 23;<br />
when a 8: r 8 6<br />
, h 8 3<br />
<br />
3 3<br />
(c) h<br />
r 2<br />
a<br />
65. (a) The x- and y-intercepts of the line through R and T are x and f (x)<br />
a xf(x) respectively.<br />
The area of the triangle is the product of these two values.<br />
(b) Domain: (0, 10)<br />
2. (a) L(x) 4 5 x 9 5 <br />
(b) Differs from the true value in absolute value by less than 10 3<br />
3. (a) L(x) 2<br />
(b) Differs from the true value in absolute value by less than 10 2<br />
4. (a) L(x) x<br />
(b) Differs from the true value in absolute value by less than 10 2<br />
5. (a) L(x) x <br />
(b) Differs from the true value in absolute value by less than 10 3<br />
6. (a) L(x) x 2 <br />
(b) Differs from the true value in absolute value by less than 10 3<br />
7. f(0) 1. Also, f (x) k(1 x) k1 , so f (0) k.<br />
This means the linearization at x 0 is L(x) 1 kx.<br />
19. (a) dy (3x 2 3) dx<br />
(b) dy 0.45 at the given values<br />
2 2x2<br />
20. (a) dy (1 x2) 2 dx<br />
(b) dy 0.024 at the given values<br />
21. (a) dy (2x ln x x) dx<br />
(b) dy 0.01 at the given values<br />
1 2x2<br />
22. (a) dy (1 x2) 1 /2 dx<br />
(b) dy 0.2 at the given values<br />
23. (a) dy (cos x) e sin x dx<br />
(b) dy 0.1 at the given values<br />
24. (a) dy csc 1 3<br />
x <br />
cot 1 3<br />
x <br />
dx<br />
The vertical asymptotes at x 0 and x 10 correspond to horizontal<br />
or vertical tangent lines, which do not form triangles.<br />
(c) Height 15, which is 3 times the y-coordinate of the center of the<br />
ellipse.<br />
(d) Part (a) remains unchanged.<br />
The domain is (0, C) and the graph is similar.<br />
The minimum area occurs when x 2 3C 2<br />
. From this, it follows that<br />
4<br />
the triangle has minimum area when its height is 3B.<br />
Section 4.5<br />
Quick Review 4.5<br />
9.<br />
10.<br />
[0, 10] by [–100, 1000]<br />
[0, π] by [–0.2, 1.3]<br />
(b) dy 0.205525 at the given values<br />
25. (a) dy (x <br />
dx1) 2<br />
(b) dy 0.01 at the given values<br />
x y<br />
26. (a) dy 2 2 x<br />
dx<br />
(b) dy 0.0375 at the given values<br />
53. Since V 4 3 r3 , we have dV 4r 2 dr 4r 2 1<br />
16<br />
r 2<br />
.<br />
4<br />
The volume error in each case is simply r 2<br />
in 3 .<br />
4<br />
Sphere Type True Radius Tape Error Radius Error Volume Error<br />
Orange 2" 1/8" 1/16 " 1 in 3<br />
Melon 4" 1/8" 1/16 " 4 in 3<br />
Beach Ball 7" 1/8" 1/16 " 12.25 in 3<br />
54. Since A 4r 2 1 r<br />
, we have dA 8rdr 8r <br />
16 <br />
. 2<br />
The surface area error in each case is simply 2<br />
r in 2 .<br />
Sphere Type True Radius Tape Error Radius Error S. Area Error<br />
Orange 2" 1/8" 1/16 " 1 in 2<br />
Melon 4" 1/8" 1/16 " 2 in 2<br />
Beach Ball 7" 1/8" 1/16 " 3.5 in 2<br />
1<br />
64. If x 1 h, then f(x 1 ) 2h 1/2 and x 2 h h 1/2<br />
<br />
1<br />
<br />
h 2h h. If x 1 h,<br />
2h 1/2<br />
1<br />
then f(x 1 ) 2h 1/2 and x 2 h 2h h.<br />
Exercises 4.5<br />
1. (a) L(x) 10x 13<br />
(b) Differs from the true value in absolute value by less than 10 1
Additional Answers 659<br />
65. x 2 2, x 3 4, x 4 8, and x 5 16;<br />
⏐x n ⏐ 2 n1 .<br />
[–10, 10] by [–3, 3]<br />
66. (a) b 0 f(a), b 1 f (a), and b 2 f (a)<br />
.<br />
2<br />
(b) 1 x x 2<br />
(c) As one zooms in, the two graphs quickly become indistinguishable.<br />
They appear to be identical.<br />
(d) The quadratic approximation is<br />
1 (x 1) (x 1) 2 .<br />
As one zooms in, the two graphs quickly become indistinguishable.<br />
They appear to be identical.<br />
(e) The quadratic approximation is<br />
x x2<br />
1 . 2 8<br />
As one zooms in, the two graphs quickly become indistinguishable.<br />
They appear to be identical.<br />
(f) For f :1 x ;<br />
for g:1 (x 1) 2 x;<br />
x<br />
for h:1 2<br />
67. Finding a zero of sin x by Newton’s method would use the recursive formula<br />
sin(<br />
xn)<br />
x n1 x n x c os(<br />
xn)<br />
n tan x n , and that is exactly what the calculator<br />
would be doing. Any zero of sin x would be a multiple of p.<br />
69. lim tan x<br />
lim sin x/ cos x<br />
<br />
x→0 x x→0 x<br />
lim<br />
x→0 1<br />
sin x<br />
cos x x<br />
lim 1<br />
<br />
x→0 cos x lim sin x<br />
<br />
x→0 x<br />
(1)(1) 1.<br />
70. g(a) c, so if E(a) 0, then g(a) f(a) and<br />
c f(a). Then<br />
E(x) f(x) g(x) f(x) f(a) m(x a).<br />
E(x)<br />
Thus, f (x) f (a)<br />
m.<br />
x a x a<br />
lim f(x ) f(a)<br />
f (a), so if the limit of<br />
x→a x a<br />
E(x)<br />
is zero, then m f (a) and g(x) L(x).<br />
x a<br />
Section 4.6<br />
Exercises 4.6<br />
45. (a) The point being plotted would correspond to a point on the edge of the<br />
wheel as the wheel turns.<br />
(b) One possible answer:<br />
16pt, where t is in seconds.<br />
(c) Assuming counterclockwise motion, the rates are as follows.<br />
p 4 ; d x<br />
71.086 ft/sec<br />
dt<br />
d y<br />
71.086 ft/sec<br />
dt<br />
2 : d x<br />
100.531 ft/sec<br />
dt<br />
d y<br />
0 ft/sec<br />
dt<br />
p: d x<br />
0 ft/sec<br />
dt<br />
d y<br />
100.531 ft/sec<br />
dt<br />
46. (a) One possible answer:<br />
x 30 cos , y 40 30 sin <br />
(b) Assuming counterclockwise motion, the rates are as follows.<br />
When t 5: d x<br />
0 ft/sec<br />
dt<br />
d y<br />
18.850 ft/sec<br />
dt<br />
when t 8: d x<br />
17.927 ft/sec<br />
dt<br />
d y<br />
5.825 ft/sec<br />
dt<br />
Quick Quiz (Sections 4.4–4.6)<br />
4. (a) Since d y<br />
1 1<br />
dx<br />
2 x1/2 , the slope at (25, 5) is . 1 0<br />
1<br />
The linearization is L(x) (x 25) 5, so L(26) 5.1.<br />
1 0<br />
f(<br />
5)<br />
(b) x 2 5 5 52 26<br />
5.1.<br />
f (<br />
5)<br />
2(5)<br />
(c) The appropriate linearization is for the curve y 3 x at the point (27, 3).<br />
Since d y<br />
1 1<br />
dx<br />
3 x2/3 , the slope at (27, 3) is . The linearization is<br />
2 7<br />
1<br />
L(x) (x 27) 3, so L(26) 3 – 1/27 2.963.<br />
2 7<br />
Review Exercises<br />
5. (a) Approximately (, 0.385]<br />
(b) Approximately [0.385, )<br />
(c) None (d) (, )<br />
(e) Local maximum at (0.385, 1.215)<br />
(f) None<br />
6. (a) [1, ) (b) (, 1]<br />
(c) (, )<br />
(d) None<br />
(e) Local minimum at (1, 0)<br />
(f ) None<br />
7. (a) [0, 1) (b) (1, 0]<br />
(c) (1, 1)<br />
(d) None<br />
(e) Local minimum at (0, 1)<br />
(f) None<br />
8. (a) (, 2 1/3 ] (, 0.794]<br />
(b) [2 1/3 ,1) [0.794, 1) and (1, )<br />
(c) (, 2 1/3 ) (, 1.260) and (1, )<br />
(d) (1.260, 1)<br />
(e) Local maximum at<br />
21/3 , 2 3 21/3 (0.794, 0.529)<br />
(f)<br />
21/3 , 1 3 21/3 (1.260, 0.420)<br />
<br />
9. (a) None (b) [1, 1]<br />
(c) (1, 0) (d) (0, 1)<br />
(e) Local maximum at (1, p);<br />
local minimum at (1, 0)<br />
(f)<br />
0, p 2
660 Additional Answers<br />
10. (a) [3, 3]<br />
(e) Local maximum at (0.889, 1.011);<br />
(b) (,3] and [3, )<br />
local minimum at (0, 0)<br />
(c) Approximately (2.584, 0.706) and (3.290, )<br />
(f) <br />
(d) Approximately (, 2.584) and (0.706, 3.290)<br />
2 9 , 0.667 <br />
(e) Local maximum at<br />
16. (a) Approximately (, 0.215]<br />
3, (b) Approximately [0.215, 2) and (2, )<br />
3 1<br />
(1.732, 0.183);<br />
(c) Approximately (2, 3.710)<br />
4<br />
(d) (, 2) and approximately (3.710, )<br />
local minimum at<br />
(e) Local maximum at (0.215, 2.417)<br />
3,<br />
(f) (3.710, 3.420)<br />
(1.732, 0.683)<br />
4<br />
35. .<br />
y<br />
(f) (2.584, 0.573), (0.706, 0.338), and (3.290, 0.161)<br />
2<br />
11. (a) (0, 2] (b) [2, 0)<br />
(c) None (d) (2, 0) and (0, 2)<br />
(e) Local maxima at (2, ln 2) and (2, ln 2)<br />
x<br />
(f) None<br />
–3<br />
3<br />
12. (a) Approximately [0, 0.176], 0.994, p 2 ,<br />
y = f(x)<br />
[2.148, 2.965], <br />
3.834, 3 p <br />
2 , and 5.591, 2p <br />
(b) Approximately [0.176, 0.994], p 2 , 2.148 ,<br />
p<br />
[2.965, 3.834], and 3 , 5.591<br />
2 <br />
(c) Approximately (0.542, 1.266), (1.876, 2.600),<br />
(3.425, 4.281), and (5.144, 6.000)<br />
(d) Approximately (0, 0.542), (1.266, 1.876),<br />
(2.600, 3.425), (4.281, 5.144),<br />
and (6.000, 2p)<br />
(e) Local maxima at (0.176, 1.266), p 2 ,0 <br />
p<br />
and (2.965, 1.266), 3 ,2<br />
2 <br />
, and (2p, 1);<br />
local minima at (0, 1),<br />
(0.994, 0.513),<br />
(2.148, 0.513), (3.834, 1.806),<br />
and (5.591, 1.806)<br />
Note that the local extrema at x 3.834,<br />
x 3 p , and x 5.591 are also absolute extrema.<br />
2<br />
(f) (0.542, 0.437), (1.266, 0.267),<br />
(1.876, 0.267), (2.600, 0.437), (3.425, 0.329), (4.281, 0.120),<br />
(5.144, 0.120), and (6.000, 0.329)<br />
2<br />
13. (a) 0, <br />
(b) (, 0] and <br />
2<br />
3<br />
, <br />
(c) (, 0) (d) (0, )<br />
(e) Local maximum at<br />
2 16<br />
(1.155, 3.079)<br />
3<br />
3<br />
3 1<br />
<br />
, 33 <br />
(f ) None<br />
14. (a) Approximately [0.578, 1.692]<br />
(b) Approximately (, 0.578] and [1.692, )<br />
(c) Approximately (, 1.079)<br />
(d) Approximately (1.079, )<br />
(e) Local maximum at (1.692, 20.517);<br />
local minimum at (0.578, 0.972)<br />
(f) (1.079, 13.601)<br />
15. (a) 0, 8 9 <br />
(b) (, 0] and <br />
8 , )<br />
9<br />
(c) <br />
, 2 9 <br />
(d) 2 9 ,0 <br />
and (0, )<br />
–3<br />
36. (a) Absolute minimum is 2 at x 1;<br />
absolute maximum is 3 at x 3<br />
(b) None<br />
(c)<br />
1633001.59<br />
41. (a) With some rounding, y <br />
1 17.471e 0.06378t<br />
(b) The regression fit looks very good:<br />
[0, 80] by [0, 1600000]<br />
(c) 1,476,128 829,210 2,305,337<br />
(d) About 1885; the graph has an inflection at this point.<br />
(e) The maximum population will be about 2,462,000.<br />
(f) There are many possible causes. Advances in transportation began<br />
drawing the population southward after 1920, and Tennessee was wellsituated<br />
geographically to become a crossroads of river, railroad, and<br />
automobile routes. By the year 2000 there had been numerous other<br />
demographic changes. It should also be pointed out that the census<br />
years in the data (1850 – 1910) include the years of the Civil War and<br />
Reconstruction, so the regression is based on unusual data to begin with.<br />
51. (a) V(x) x(15 – 2x)(5 – x)<br />
(b) 0 x 5<br />
[0, 5] by [10, 70]
Additional Answers <strong>661</strong><br />
(c) Maximum volume 66.019 in 3 when x 1.962 in.<br />
(d) V(x) 6x 2 – 50 x 75 which is zero at x 25 57<br />
1.962.<br />
6<br />
(b)<br />
y<br />
2<br />
70. (a) The only x-value for which f has a relative maximum is x – 2. That<br />
is the only place where the derivative of f goes from positive to<br />
negative.<br />
(b) The only x-value for which f has a relative minimum is x 0. That is<br />
the only place where the derivative of f goes from negative to positive.<br />
(c) The graph of f is concave up on (1, 1) and on (2, 3). Those are the intervals<br />
on which the derivative of f is increasing.<br />
(d)<br />
y<br />
LRAM 1.25<br />
2<br />
x<br />
6. (a)<br />
2<br />
y<br />
–3 3<br />
x<br />
72. (a) V a 2 b, and b 60 2a<br />
15 a 4<br />
2 , so V 15pa2 p a 3 .<br />
2<br />
Thus d V<br />
30pa 3p a 2<br />
3 pa(20 a). The domain of consideration<br />
da<br />
2 2<br />
for a in this problem is (0, 30), so a 20 is the only critical number.<br />
The cylinder of maximum volume is formed when a 20 and b 5.<br />
(b) The sign graph for the derivative d V<br />
3 pa(20 – a) on the interval<br />
da<br />
2<br />
(0, 30) is as follows:<br />
<br />
x<br />
0 20 30<br />
By the First Derivative Test, there is a maximum at x 20.<br />
(b)<br />
2<br />
y<br />
RRAM 1.25<br />
2<br />
x<br />
CHAPTER 5<br />
2<br />
x<br />
Section 5.1<br />
Exercises 5.1<br />
5. (a) y<br />
2<br />
7.<br />
MRAM 1.375<br />
n LRAM n MRAM n RRAM n<br />
10 1.32 1.34 1.32<br />
50 1.3328 1.3336 1.3328<br />
100 1.3332 1.3334 1.3332<br />
500 1.333328 1.333336 1.333328<br />
R<br />
2<br />
x<br />
13.<br />
14.<br />
n<br />
MRAM<br />
10 526.21677<br />
20 524.25327<br />
40 523.76240<br />
80 523.63968<br />
160 523.60900<br />
n error % error<br />
10 2.61799 0.5<br />
20 0.65450 0.125<br />
40 0.16362 3.12 10 2<br />
80 0.04091 7.8 10 3<br />
160 0.01023 2 10 3