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Chapter 4 146 - 408 CHAPTER 4 APPLICATIONS OF...

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Unformatted text preview: 408 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 34. y = f(x) : ln(:z:2 — 1) A. D = (—00, —1) U (1, 00) B. No y—intercept; m—intercepts :tx/i C. Symmetric about the y—axis D. lim ln(a:2 — 1) 2 oo. lim ln(:c2 — 1) : —oo. lim ln(ac2 — 1) 2 —00. so ac = 1 m—>:too 1—.1+ z—p—1* 2 andzv = —1areVA. E. y = f(x) = ln(av2 * 1) => f’(w) : x2 i1 > 0f0rw > 1 and f’(a:) < 0 for a: < —1. so f is increasing 0n (1, 00) and decreasing on (~00, —1). Note that the domain off is |x| > 1. F. No extreme value 2 1 . G. f”($) = —2-($2—+1)2 < 0. so f 1s CD on (700, —1) and (1,00). m _ No IP 102—1 I w3(2m)— 3:2A1 3zc2 3gw2 35-f(:c): m3 :» for i ) 7 964 => 1 ,, m4(—2a:) — (3 , m2)4m3 , 23:2 g 12 f (33) # $8 ’ :65 f’ Estimates: From the graphs of f' and f ”. it appears that f is increasing on -5 5 (—1.730) and (0,1.73) and decreasing on (—00, -1.73) and (1.73, 00); '- #02 f has a local maximum of about f(1.73) : 0.38 and a local minimum of 0-2 about f(—1.7) : —0.38; f is CU on (—2.45, 0) and (2.45, 00). and CD f” . . . ’8 w A 8 on (—00, —2.45) and (0, 2.45); and f has Inflectlon pomts at about (—2.45, 70.34) and (2.45, 0.34). 3 1 $2 70.2 Exact: Now f’(:c) : 4 is positive for 0 < 3:2 < 3. that is. f is 15 m increasing on (—\/§, 0) and (0, (/3); and f’ (:c) is negative (and so f is _5 w h 5 decreasing) on (700, ,fi) and (x/goo). fI (an) : 0 when as 2 ::\/3. f ' goes from positive to negative at w : \/§ so f has a local maximum of 1 5 2 7 -0.25 f (x/g ) : Egg—)3: : 37?; and since f is odd. we know that maxima on the interval (0, oo) correspond to minima on (700, 0). so f has a local 2 2 — 12 . . . minimum of f(7\/§) : gig. Also, f" (at) : igs— 1s posrtive (so 1 3 —0.4 f is CU) on (—x/ES, 0) and (x/é, co), and negative (so f is CD) on (—00. —\/6) and (O, x/é). There are [P at (x/é 53%) and («4%)- ...
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