5128_ch04ansTE_pp652-661
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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.