5333-Solns-Assignment8-Sp13 - 5333 Analysis Solutions Assignment 8 Section 26 1(a False For example f(x = 1\/x on(0 1(0 1 is a bounded interval f has

# 5333-Solns-Assignment8-Sp13 - 5333 Analysis Solutions...

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5333: AnalysisSolutions, Assignment # 8Section 261.(a) False. For example,f(x) = 1/xon (0,1). (0,1) is a bounded interval,fhasneither a maximum nor a minimum.(b) False. You needfdifferentiable in order to apply the Mean-Value Theorem.(c) False.For example, letf(x) =x3on(-1,1).Thenfprime(x) = 3x2andfprime(0) = 0,butfhas neither a maximum nor a minimum atx= 0.2.(a) True. This is Corollary 26.7.(b) False. Setf(x) =x2sin(1/x)forxnegationslash= 0andf(0) = 0(see Pb. 6, Section25).fis differentiable butfprimeis not continuous atx= 0.(c) True.Iffis not injective onI,then there must exist a pair of pointsx1, x2, x1< x2,such thatf(x1) =f(x2). Now,fis continuous on[x1, x2]and differentiable on(x1, x2).Therefore, by Rolle’s Theorem, there exists apointc(x1, x2)such thatfprime(c) = 0.3.f(x) =x2-3x+ 4forx[0,2].(a)fprime(x) = 2x-3.fprime(x) = 0atx= 3/2;fprime(x)<0on(0,3/2);fprime(x)>0on(3/2,2). Therefore,fis strictly decreasing on[0,2];fis strictly increasingon[3/2,2].(b)f(3/2) = 7/4is the minimum value off;f(0) = 4is the maximum value off.9.f(x) =x3onI= (-1,1) is strictly increasing onIandfprime(0) = 0;f(x) =x+ sinxis strictly increasing on(-2π,2π) andf(-π) =f(π) = 0;f(x) =-x3onI= (-∞,)is strictly decreasing onIandfprime(0) = 0.13.(a) By the Mean-Value Theorem, there is a pointc1(0,1) such thatfprime(c1) =f(1)-f(0)1-0= 2(b) Sincef(1) =f(2) = 2,there is a pointc2(1,2)such thatfprime(c2) = 0byRolle’s Theorem.(c) Sincefprime(c1) = 2andfprime(c2) = 0,and since0<54<2,there is a pointc3(c1, c2)such thatfprime(c3) =54.1
15.(a) Letepsilon1 >0. Chooseδ=epsilon1/m. Then, for anyx, y(a, b), x < y,we have, bythe Mean-Value Theorem,f(y)-f(x) =fprime(c)(y-x)wherec(x, y).Therefore, if|y-x|< δ,then|f(y)-f(x)|=|fprime(c)| |y-x| ≤m|y-x|< mepsilon1m=epsilon1.andfis uniformly continuous on(a, b).(b) Setf(x) =braceleftBiggx2sin(1/x2),0< x10,x= 0Thenfis continuous on[0,1],andfis differentiable on(0,1). Sincefis continuous on the compact set[0,1], it is uniformly continuous. Nowfprime(x) =x2cos(1/x2)(-2/x3) + 2xsin(1/x2) =-2cos(1/x2)x+ 2xsin(1/x2)which is unbounded on(0,1) – it is unbounded at 0.23.(a)Letepsilon1 >0. Chooseδ= (epsilon1/M)1.Then, if|x-y|< δ,|f(x)-f(y)| ≤M|x-y|α< M δα=Mepsilon1M=epsilon1.Therefore,fis uniformly continuous.(b)α >1. Choose anycI. Then|f(x)-f(c)||x-c|M|x-c|α|x-c|=M|x-c|α-1It follows from this thatlimxcf(x)-f(c)x-c= 0. Therefore,fprime(c) = 0. Sincecisarbitrary,fprime(x) = 0for allxI. Therefore,fis constant onI.(c)f(x) =|x|satisfies the condition onI= (-1,1), for example, andfis notdifferentiable onI;fis not differentiable atx= 0.(d)Supposegis differentiable onIand thatgprimeis bounded;that is, there is anumberMsuch that|gprime(x)| ≤Mfor allxI. Then, for anyx, yI, y > x,g(y)-g(x) =gprime(c)(y-x),wherec(x, y)by the Mean-Value Theorem.We can now conclude that|g(y)-g(x)|=|gprime(c)| |(y-x)| ≤M|y-x|for allx, yI.Thus,gsatisfies a Lipschitz condition of order1.2

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