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Chapter 4 38

# Chapter 4 38 - _’ 300 CHAPTER 4 APPLICATIONS OF...

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Unformatted text preview: ____’_/ 300 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 72. P(m) : m4 —I— 61:3 + \$2 : P'(m) = 4m3 + 301:2 + 23: => P”(a:) = 123:2 + 6cm + 2. The graph of P”(:L') is a parabola. If P”(w) has two roots. then it changes sign twice and so has two inﬂection points. This happens when the discriminant of P”(a:) is positive. that is (60)2 — 4- 12 - 2 > 0 4:) 36::2 — 96 > 0 (1) [cl > M m 1.63. If 36c2 — 96 = 0 4:) c : i235. P"(a:) is 0 at one point. but there is still no inﬂection point since P' ':n( ) never changes sign and if 36c2 — 96 < 0 4:} [cl < 443 then PH(\$ ) never changes sign. and so there is no inﬂection point. 52125- 0.5 7125 0:6 C:3 0:1.8 0.5 71 0 5:315 020 02—2 3 For large positive 0, of f becomes ﬂatter for ac < 0, and eventually the dip rises above the m-axis. and then disappears entirely. the graph along with the inﬂection points. As (3 continues to decrease. the dip and the inﬂection points reappear. to the right of the origin. 73. By hypothesis 9 : f ' is differentiable on an open interval containing ct Since (0, f (6)) is a point of inﬂection. the :1: = c. Hence, by the First Derivative Test, f / has a local concavity changes at a: : c, so f”(m) changes signs at a: 2 c. Thus, by Fermat’s Theorem f"(c) : 0. 74. f(a:) : 904 => f’(a:) : 43:3 :> f”(a:) : 122:2 => f"(0) : 0. Form < 0. f”(:1:) > 0, so f is CU on is also CU on (0. 00). Since f does not change concavity at 0. (O7 0) is not an extremum at (-oo,0); form > 0, f"(ac) > 0. so f inﬂection point. x/wj,wehavethatg(w):m\/—x_2— => g'(m):\/;2-+\/;:2\/'w_2:2lml :> 75. Using the fact that [9:] : _ 2 g"(m) : 2:1:(552) 1/2 =1 < 0fora: < Oandg ISBI "(a:) > 0 for :3 > 0. so (0., 0) is an inﬂection point. But g"(0) does not exist. ...
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