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Chapter 4 149

# Chapter 4 149 - CHAPTER 4 REVIEW 411 41 f:z arctan(cos(3...

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Unformatted text preview: CHAPTER 4 REVIEW 411 41. f(:z:) : arctan(cos(3 arcsin 37)). We use a CAS to compute f’ and f". and to graph f. f’. and f": 42. . The family of functions f(.’L‘) : ln(sin a: + C) all have the same From the graph of f’. it appears that the only maximum occurs at w : 0 and there are minima at a; : i087. From the graph of f ". it appears that there are inﬂection points at :5 : i052. f(1:) : 111(2w + av sin at) We use the CAS to calculate 4 2+sinx+atcosac / :— d f (xv) 230 +\$sinm an , 2\$2sincc+4sinx—C082m+m2 +5 f (:c) _ 11:2(cos2 w — 4sinm i 5) From the graphs. it seems that f’ > 0 (and so f is increasing) on approximately the intervals (0, 2.7). (4.5, 8.2) and (10.9. 14.3). It seems that f" changes sign (indicating inﬂection points) at a: z 3.8. 5.7. 10.0 and 12.0. Looking back at the graph of f. this implies that the inﬂection points have approximate coordinates (3.8. 1.7). (5.7. 2.1). (10.0., 2.7), and (12.0, 2.9). eriod and all have maximum values at m : 5 + 27m. Since the P 2 domain of In is (0. 00), f has a graph only if sinm + C > 0 somewhere. Since —1 3 sins: g 1. this happens ifC' > 71, that is. f has no graph ifC' S —1. Similarly. ifC > 1, then sina: —l— C > 0 and f is continuous on (700, 00). As O increases. the graph of f is shifted vertically upward and ﬂattens out. If—l < C S 1. f is deﬁned where sinac + C > 0 4:) sinm > —C 4:) sin’1(—C) < :r < 71' ~ sin—1(~C). Since the period is 27r. the domain of f is (27m l sin’1( C)7 (2n l 1)7r sinFl( C)) n an integer. We exclude the case 0 : 0. since in that case f (as) : O for all as. To ﬁnd the maxima and minima. we differentiate: 2 f(:c) = 0276‘“ :> f'(ac) = 6&6“ch (—2033) + e‘Cm2(1)] : ce’cg”2 (726:22 +1). This is 0 where —20\$2 + 1 = 0 a) as : :1/V2c. So if c > 0. there are two maxima or minima. whose sit—coordinates approach 0 as c increases. The negative root gives a minimum and the positive root gives a maximum. by the First Derivative Test. By substituting back into the equation, we see that f(::1/\/%) : C(il/x/ﬁ) e‘cCﬂ/mﬁ 2 :l: c/2e. So as c increases. the extreme points become more pronounced. Note that if c > 0. then lim f(ac) : 0. If C < O, \$‘?::w then there are no extreme values, and lim f (2v) : Zoo. (EHIZOO ...
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