Unformatted text preview: 9. 10. 11. 12.
13. 14. 15. 16. 17. CHAPTER 4 REVIEW 401 True. The graph of one such function is sketched. False. At any point (a. f(a)). we know that f’(a) < 0. So since the tangent line at (a. f(a)) is not horizontal. it
must cross the x—axis—at x : I). say. But since f”(m) > 0 for all m. the graph off must lie above all ofits tangents; in particular. f(b) > 0. But this is a contradiction. since we are given that f(ac) < 0 for all 2:.
True. Let 361 < 332 where $1.12 6 I. Then f(m1) < ﬂu) and g(9c1) < 9(222) (since f andg are increasing on
I) 50 (f+9)($1) : f(m1) +9($1) < f($2) +9($2) = (f+9)($2)
False. f(m) : w and g(m) : 2.73 are both increasing on (0. 1). but f(:c) — g(ac) : ,1; is not increasing on (0.1).
False. Take f(.’E) : ac and g(:c) : cc — 1. Then both f and g are increasing 0n (0, 1). But f(w)g(a:) : 93(93 7 1)
is not increasing on (0, 1).
True. Let :31 < m2 where CL‘1,$2 E I. Then 0 < f(ac1) < ﬂu) and 0 < 9(131) < g(:02) (since f and g are both
positive and increasing). Hence. f(m1) 9(351) < f(a:2)g($1) < f(mg) g($2). So fg is increasing on I. True. Letzcl,:c2 E I and $1 < 1:2. Then f(:v1) < f(m2) (f is increasing) :> 1 1
> f (3:1) f (332)
False. The most general antiderivative is F(a:) : —1/m + 01 for a: < 0 and F(m) : —1/ac + 02 for w > 0 (f is positive) => g(ac1) > g(:c2) => g(x) : 1/f(:c) is decreasing on I. (see Example 1 in Section 4.10). True. By the Mean Value Theorem. there exists a number 0 in (0, 1) such that f(1)— n0) : f’(c)(1 — 0) : f’(c) Since f’(c) is f(1) , W) 75 0. so m) 7s f(0) 3: lim 3;
18. False. lim — 2 “fag : — = 0. not 1.
z—>O 61 11m 61 1
z—>0
EXERCISES
1. fps) = 10 + 27x — x3. 0 g as g 4. f’(m) = 27 — 3m2 : —3(a:2 — 9) : —3(m + 3)(a: — 3) : 0 only when at : 3 (since —3 is not in the domain). f’(a:) > 0 form < 3 and f’(m) < 0 for a: > 3. so f(3) : 64 is a local maximum value. Checking the endpoints, we ﬁnd f(0) : 10 and f(4) : 54. Thus. f(0) = 10 is the absolute
minimum value and f(3) : 64 is the absolute maximum value. f(ac):$—\/E.0§wg4.f'(:17)=1—1/(2\/E):0 4:) 2\/—=1 => m=711.f’(ar)doesn0texist (i) m : O. f’(m) < 0f0r0 < at < i and f’(m) > 0fori < x < 4. so fﬁ) : ~i isalocal and absolute
minimum value. f(0) : 0 and f(4) : 2. so f(4) : 2 is the absolute maximum value.
2
+1 1 — 2 1 i 2
ﬁx): “3 —2Sm£0.f’(:n)*(m “3 I” “1“) 1 I :0 a 332+$+1’ (aﬁ—Eac—l—l)2 T(:c2+m+1)2
a: : —1 (since 1 is not in the domain). f’(w) < 0 for *2 < a: < —1 and f'(a:) > 0for —1 < :1: < 0. so f(—1) : 71 is a local and absolute minimum value. f(72) = ~§ and f(0) : 0. so f(0) : 0 is an absolute
maximum value. ...
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 Spring '10
 Ban
 Calculus

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