Unformatted text preview: / 278 CHAPTER 4 APPLICATIONS OF DlFFEHENTlATlON 20. f(ac) 21:4 + 4x + C. Suppose that f(rc) = 0 has three distinct real roots (1. b. d where a < b < d. Then NI) 0 2 f’ (C1) : f'(C2). so f’(:c) = 0 must have at least two real solutions. However I l f(b) = f(d) : 0. By Rolle‘s Theorem there are numbers C1 and C2 with a < C1 < b and b < 62 < d and 0 : fllx) = 4173 + 4 = 4(173 +1): 4(m + 1) (m2 # at + 1) has as its only real solution x : —1. Thus. f(ac) can have at most {W0 real {0018. 21. (a) Suppose that a cubic polynomial P(ac) has roots 0.1 < (12 < (13 < (14. so P(a1) = P(a2) : P(u3) = P(a4).
By Rolle‘s Theorem there are numbers C1. C2. C3 with (11 < 01 < a2. a2 < C2 < (13 and a3 < 03 < a4 and
P'(C1) : P’(Cg) : P'(C3) : 0. Thus. the second—degree polynomial P’(:c) has three distinct real roots. which is impossible. (b) We prove by induction that a polynomial of degree n has at most n real roots. This is certainly true for n : 1.
Suppose that the result is true for all polynomials of degree n and let P (av) be a polynomial of degree n + 1.
Suppose that P(at) has more than n + 1 real roots. say (11 < (12 < a3 <    < an“ < an+2. Then
P(a1) : P(a2) : ~ .  : P(an__2) : 0. By Rolle's Theorem there are real numbers C1, . . . ,Cn+1 with m < C1 < a2, . .. .an+1 < Cn__1 < an” and P’(C1) : ~ 2 P’ (Cn+1) : 0. Thus. the nth degree
polynomial P’ (CC) has at least it + 1 roots. This contradiction shows that P(m) has at most 71 + 1 real roots. 2. (a) Suppose that f(a) : f(b) : 0 where a < b. By Rolle‘s Theorem applied to f on [(1. b] there is a number C such
thata < C < band f'(C) = 0. (b) Suppose that f(a) : f(b) : f(C) : 0 where a < b < C. By Rolle‘s Theorem applied to f(as) on [(1. b] and
[b. C] there are numbers a < d < b and b < e < C with f’(d) = 0 and f’(e) : 0. By Rolle’s Theorem applied
to f'(a;) on [d. 6] there is a number 9 with d < 9 < e such that f”(g) : 0. (c) Suppose that f is n times differentiable on R and has n + 1 distinct real roots. Then f(”) has at least one real foot. 23. By the Mean Value Theorem. f(4) * f(1) : f'(C)(4 — 1) for some C 6 (1, 4). But for every C E (17 4) we have
f’(C) Z 2. Putting f’(C) Z 2 into the above equation and substituting f(1) : 10. we get
f(4) : f(1) + f’(C)(4 — 1) : 10 —l— 3f’(c) Z 10 + 3 2 : 16. So the smallest possible value of f(4) is 16. 24. If 3 g f’(a:) g 5 for all 1:. then by the Mean Value Theorem. f(8) — f(2) : f'(c)  (8 — 2) for some C in [2, 8]. (f is differentiable for all x. so. in particular. f is differentiable on (2. 8) and continuous on [27 8]. Thus, the
hypotheses of the Mean Value Theorem are satisﬁed.) Since f(8) — f(2) : 6]“ (C) and 3 S f'(C) g 5. it follows
that63 S 6f’(c) g 65 :> 18 S f(8) ~ f(2) g 30. 25. Suppose that such a function f exists. By the Mean Value Theorem there is a number 0 < C < 2 with f’ (C) : w : é. But this is impossible since f'(a:) g 2 < % for all :12. so no such function can exist, 25. Let h : f  9. Then since f and g are continuous on [(1.1)] and differentiable on (a. b). so is h, and thus it satisﬁes the assumptions of the Mean Value Theorem. Therefore. there is a number C with a < C < I) such that h(b) : h(b) # h(a) : h'(C) (b — a). Since h' (C) < 0. h'(C)(b — a) < 0, so f(b) # g(b) : h(b) < 0 and hence
“M < 9(5) ...
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 Spring '10
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 Calculus

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