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Chapter 4 35

# Chapter 4 35 - SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE...

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Unformatted text preview: SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH 297 60. At ﬁrst the depth increases slowly because the base of the mug depth of coffee is wide. But as the mug narrows. the coffee rises more quickly. _ . height Thus. the depth (1 increases at an increasing rate and its graph is of mug concave upward. The rate of increase of d has a maximum where the mug is narrowest; that is. when the mug is half full. It is there that the inﬂection point (IP) occurs. Then the rate of increase of d starts to decrease as the mug widens and the graph time [0 f becomes concave down. ﬁll mug 61. 100 From the graph. we estimate that the most rapid increase in the V percentage of households in the United States with at least one VCR occurs at about t 2 8. To maximize the ﬁrst derivative. we need to determine the values for which the second derivative is 0. We‘ll use V(t) 2 a ‘ , and substitute a 2 85. b 2 53. and c 2 —0.5 later. 0 t 20 1 + be“ a(bcea) (1 +bect)2 V’(t) 2 2 [by the Reciprocal Rule] and (1 + beCt)2 we“ — e” -2(1-l— bed) - bee“: [(1 + be“)2]2 fabc - ce°t(1 + be‘it)[(1 + be“) — 2bed] iabcze‘iﬂ — beat) (1+ ber't)4 (1 + ber:t)3 V"(t) — abc ~ So V”(t) 2 0 2 1 2 be“ 2 ed 2 l/b. Now graph y 2 670‘“ and y 2 5—13. These graphs intersect at t z 7.94 years, which corresponds to roughly midyear 1988. [Alternatively we could use the rootﬁnder on a 20.5t calculator to solve e 2 5—13.0r, if you have already studied logarithms. you can solve e“ 2 1 / b as follows: ct 2111(1/b) 2 t2 (l/c)ln(1/b) 2 22ln 5—13 2 7.94 years. 62. (a) As Ix] —+ 00. t 2 —£L'2/(20'2) ~> —00, and at —> 0‘ The HA is y 2 0. Since it takes on its maximum value at :c 2 0. so does 6‘. Showing this result using derivatives. we have f(3\$) 2 eTEZ/(m’z) 2> f’(x) 2 e‘x2/(2‘T2) (—m/a2). f’(:c) 2 0 2 m 2 0. Because f' changes from positive to negative at a: 2 0. f(0) 2 1 is a local maximum. For inﬂection points, we ﬁnd II 1 —12 U2 _\$2 .02 _1 f (\$)=—; [3 “2 )'1+£E6 /(2 )(—x/a'2)] :;e*\$2/(202)(1_\$2/U2)I f”(a:)20 2 \$2202 2 x2::a.f”(m) <0 2 \$2<a2 2 ~a<m<a.SofisCDon (2010') and CU on (2007 —a) and (a, 00). IP at (:0, 6‘1/2). (b) Since we have IP at a: 2 :0, the inﬂection points move away from the y—axis as 0 increases. (C) From the graph. we see that as 0 increases, the graph tends to . spread out and there is more area between the curve and the x—axis. ...
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