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Chapter 4 141

# Chapter 4 141 - 15 16 17 18 19 CHAPTER 4 REVIEW 403 f(0 = 0...

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Unformatted text preview: 15. 16. 17. 18. 19. CHAPTER 4 REVIEW 403 f(0) = 0. f/(—2) = f’(1)= f’(9) : 0. lim f(x) = 0. 13—400 lim f(ac) z —oo. f’(m) < 0 on (—00. —2). (1.6). and (9.00). z—v6 f’(m) > 0 0n (—2. 1) and (6. 9). f”(:c) > 0 on (—00. 0) and (12, oo). f”(2:) < 0 on (0,6) and (6. 12) ForO < a: < 1. f'(x) = 2x. so f(a:) : 1'2 +0. Since f(0) = 0. f(w) = 2:2 on [0.1]. Forl < m < 3. f’(m) = —1. so f(w)=—x+D.1:f(1):il—1—D => D:2,so f(n:):2ia;.Form>3.f/(m)=1.sof(zc):m—l—E. —1:f(3):3+E :> E:—4.s0f(1:):x—4.Sincef is even. its graph is symmetric about the y—axis. f is odd. f’(m) < 0f0r0 < w < 2. f’(m) > 0f0ra: > 2. f”(a:) > 0 forO < m < 3. f”(a:) < 0 form > 3. lim\$_.oo f(w) : —2 (a) Using the Test for Monotonic Functions we know that f is increasing on (—2. 0) and (4. 00) because f' > 0 on (—2, 0) and (4. 00). and that f is decreasing on (—00. —2) and (0.4) because f’ < 0 on (~00. ,2) and (0. 4). (b) Using the First Derivative Test. we know that f has a local maximum at m = 0 because f’ changes from positive to negative at m = 0. and that f has a local minimum at a: : 4 because 1“ changes from negative to positive atm=4. (C) (d) possible graph of f *V y : f(:c) : 2 — 2% ~ x3 A. D : R B. y—intercept: f(0) = 2. The m—intercept (approximately 0.770917) can be found using Newton‘s Method. C. No symmetry D. No asymptote E. f/(zz) : ~2 i 33:2 : —(3£L‘2 + 2) < 0. so f is decreasing on R. E No extreme value G. f”(x) : —61: < 0 on (07 00) and f”(x) > 0 on (—00, 0). so f is CD on (0. 00) and CU on (—00.0). There is an IP at (0. 2). ...
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