4.4 Review Questions - Solutions to Review Problems for the...

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Solutions to Review Problems for the Final ExamSolutions to Maxima, minima, convexity, concavity1.f(x) =23x3-x2-4x+ 1,f(x) = 2x2-2x-4,f(x) = 4x-2(a)f(x) = 0 when 0 =x2-x-2 = (x+ 1)(x-2), i.e.,x=-1,x= 2.(b)f(-1) =-6<0, maximum;f(2) = 6>0 minimum.(c) increasing forx <-1,x >2, (d) convex forx >1/2.2.f(x) = 2x+ 18/x,f(x) = 2-18/x2,f(x) = 36/x3(a)f(x) = 0 when 2 = 18/x2, i.e.,x= 3,x=-3(b)f(-3) =-36/27<0, maximum;f(3) = 36/27>0, minimum(c) increasing forx <-3,x >3; (d) convex forx >03.f(x) =xx2+1,f(x) =(x2+1)-x·2x(x2+1)2=1-x2(x2+1)2(a)f(x) = 0 whenx2= 1, i.e.,x=-1,x= 1; (c) increasing for-1< x <1;(b)f(x) =-2x(x2+1)2-(1-x2)·2(x2+1)(2x)(x2+1)4f(-1) = 2/22>0, minimum;f(1) =-2/22<0, maximum(d)f(x) =-2x(x2+1)(3-x2)(x2+1)4, convex for-3< x <0,x >34.f(x) =-2x-1/2-3 lnx+ 4x,f(x) =x-3/2-3/x+ 2x-1/2,(a)f(x) =x-1/2(x-1-3x-1/2+ 2) = 0 whenx-1/2= 1,2 orx= 1, x= 1/4.(b)f(x) =-(3/2)x-5/2+ 3x-2-x-3/2,f(1) =-(3/2) + 3-1>0 minimumf(1/4) =-(3/2)(32) + 3·16-8 =-8<0 maximum(c) increasing forx <1/4 andx >1; (d) convex whenx-3x1/2+ (3/2)<0, that is whenr1<x < r2whereri= (3±9-6)/25.f(x) =x(x-1)4,f(x) = (x-1)4+ 4x(x-1)3= (5x-1)(x-1)3f(x) = 0 whenx= 1/5,x= 1.fis increasing forx <1/5, decreasing on (1/5,1), andincreasing forx >1 so 1/5 is a maximum and 1 is a minimum.6.(a)h(t) =v(t) = 10-10tandh(0) = 15, soh(t) = 15 + 10t-5t2.(b) Highest point is reached when 0 =v(t) = 10-10t, i.e., timet= 1. Maximum heighth(1) = 15 + 10-5 = 20(c) 0 = 15 + 10t-5t2=-5(t2-2t-3t) =-5(t+ 1)(t-3), sot= 3.

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Term
Spring
Professor
DURRETT
Tags
Calculus, Natural logarithm, dx

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