Unformatted text preview: 298 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 63. f(zc) 2 aa:3 + bw2 + c2: + d 2 f’(av) = 3am2 + me +c. We are
given that f(1) = 0 and f(2) = 3. so f(1) : (1+ b+ c+d = Oand
f(—2) : —8a+4b—2c+d:3. Alsof’(1) :3a+2b+c=0and
f’(—2) 2 12a — 4b —I— c : 0 by Fermat’s Theorem. Solving these four
equations. we geta = g. 2 % c = a: = g. so the function is f(x):§1;(2x3 +3zr2 — 12x+7. 2 I a: a: z ‘ 
64. f(1) : awe!” => f (m) : a[:veb 2 2bm —I— ab 2 1] : me" 2 (bez + 1). For f(2) : 1 to be amaxrmum value. we must have f'(2) = O. f(2) : 1 => 1 = 2116“ and f’(2) : O i 0 : (8b+ 1)ae4b. So 8b+1:0[a7é0] => b:—%andnow1=2ae_1/2 => a=\/E/2. 65. Suppose that f is differentiable on an interval I and f ’(x) > O for all a: in I except a: : c. To show that f is increasing on I. let :01. :32 be two numbers in I with 121 < :82. Case I 331 < 932 < e. Let J be the interval {ac E I I an < c}. By applying the Increasing/Decreasing Test to f on J. we see that f is increasing on J. so f(zrl) < f(ac2)‘
Case 2 c < :121 < 302. Apply the Increasing/Decreasing Test to f on K : {cc 6 I  a: > c}. Case 3 331 < m2 = c. Apply the proof of the Increasing/Decreasing Test. using the Mean Value Theorem (MVT) 0n the interval [351. m2] and noting that the MVT does not require f to be differentiable at
the endpoints of [$1. 372].
Case 4 c = :31 < :22. Same proof as in Case 3. Case 5 301 < c < 332. By Cases 3 and 4. f is increasing on [m1.c] and on [c.rc2]. so f(ml) < f(c) < f(ac2). In all cases. we have shown that f(ml) < f(aJ2). Since m1. 902 were any numbers in I with 2:1 < .132. we have shown
that f is increasing on I. 66. (a) We will make use of the converse of the Concavity Test (along with the stated assumptions); that is, if f
is concave upward on I, then f" > 0 on I. If f and g are CU on I. then f” > 0 and g" > 0 on I, so(f+g)"=f”+g” >OonI : f+gisCUonI. (b)SincefispositiveandCUonI.f>Oandf”>00nI.Sog(cc)=[f(:1:)]2 => g’:2ff' =>
g”:2f’f’+2ff”:2(f’)2+2ff”>0 => gisCUonI. 67. (a) Since f and g are positive. increasing. and CU on I with f" and 9” never equal to O. we have f > 0,
' f’ 2 0.f” > 0.9 > 0.9' 2 0.9” > OonI. Then (fg)' : f’g+fg’ :>
My)” = fl'g + 2f'g' + fg” Z f”g + fg” > 0 on I => fg is CU on I. (b) In part (a). if f and g are both decreasing instead of increasing. then f ' g 0 and g' g 0 on I . so we still have
2f’g’ 2 O on I. Thus, (fg)" : f"g + 2f'g’ —I— fg" Z f"g+ fg" > O on I => fg is CU on I as in part (a). ...
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 Spring '10
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 Calculus

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