Written Homework 6 Solutions

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1 Written Homework 6 Solutions Section Explin in terms of liner pproximtions or differentils why the pproximtion is resonble: 1.01) Solution: First strt by finding the liner pproximtion of f x) = x 6 bout 1. Lx) = f ) + f ) x ) = f 1) + f 1) x 1) = 1) ) 5 x 1) = 1 + 6x 1) Now to plug in the vlue to pproximte, L1.01) = ) = ) = 1.06 The pproximtion is resonble becuse the liner pproximtion is good pproximtion s long s x is resonbly smll. In this cse x = 0.1.

2 Section 4.1 x 54. Find the bsolute mximum nd bsolute minimum vlues of f x) = on the intervl [0,] x 2 x+1 Solution: First find the vlue t the endpoints: f 0) = = 0 f ) = = 7 Just to be sure the function does not diverge on the intervl check where the roots of the denomintor re. But note the following: Ax 2 + Bx + C = 1 x 2 + 1) x + 1 B 2 4AC = = < 0 So the roots re imginry nd the function does not diverge on the intervl. Now to find the criticl points of the function. f x) = x2 x + 1) x2x 1) x x 2 x + 1) 2 = x 2 x + 1) 2 It is cler tht f x) = 0 for x = ±1. Since -1 is not in the domin we cn throw this vlue out. Clculte the vlue of f t the criticl points: 1 f 1) = = 1 We now compre ll of the vlues we hve clculted from endpoints nd criticl points): Absolute mx is 1, occurs t x = 1. Absolute min is 0, occurs t x = Find the bsolute mximum nd bsolute minimum vlues of f t) = t8 t) on the intervl [0,8] Solution: Begin by checking the vlue of the endpoints. f 0) = 08 0) = 0 f 8) = 88 8) = 2 0 = 0 Now to find the criticl points. Find the roots of the derivtive. f t) = t 2/ 8 t) t 1/ Test this vlue t 2/ 8 t) t 1/ = 0 t 2/ 8 t) f 2) = 28 2) = 6 2 > 0. = t 1/ 8 t) = t t = 2 We now compre ll of the vlues we hve clculted from endpoints nd criticl points): Absolute mx is 6 2, occurs t t = 2. Absolute min is 0, occurs t x = 0 nd x = 8. Pge 2 of 5

3 6. If nd b re positive numbers, find the mximum vlues of f x) = x 1 x) b, 0 x 1. Solution: Begin by checking the endpoints. Now to find the criticl points. f 0) = 0 1 0) b = 0 f 1) = 1 1 1) b = 1 0) b = 0 f x) = x 1 1 x) b + x b1 x) b 1 1) = x 1 1 x) b bx 1 x) b 1 = x 1 1 x) b 1 1 x) bx) = x 1 1 x) b 1 + b)x) Now, f x) is defined for ll x in 0, 1). f 0) will be zero if > 1, nd similrly f 1) will be zero if b > 1. But we lredy checked the vlues t the endpoints, so we only need to look for criticl points in the interior of the domin 0, 1). The only zero of f x) in 0, 1) is t x = Now plug in tht vlue. ) f = b + b + ) 1 ) b = b + b + Becuse nd b re positive nd nonzero, this is the mximum. Answer: Mximum of b b +b) +b) occurs t x = +b. +b. ) b b + ) b = b b + b) +b). Pge of 5

4 Problem Show tht the eqution hs exctly one rel root: x + e x = 0. Solution: Let f x) = x +e x. Note tht f x) is continuous for ll x. First use the Intermedite Vlue Theorem to show tht root does exist. For the problem in question let = 1 nd b = 0. Note tht: 1) + e 1 = e < e 0 = 1 > 0 So then by the Intermedite Vlue Theorem there exists vlue c [ 1, 0] such tht f c) = 0. As we know, this gurntees tht t lest one such c to exist but there could be more. Assume there were two roots c 1 nd c 2 such tht c 1 < c 2, f c 1 ) = f c 2 ) = 0. Now we pply Rolle s theorem on the intervl [c 1, c 2 ]. Note tht: f x) = x 2 + e x for ll x [c 1, c 2 ] Hence f x) is continuous nd defined on the intervl c 1, c 2 ). By ssumption, f c 1 ) = 0 = f c 2 ). So ll of the conditions for Rolle s theorem re met. Thus, there is number d c 1, c 2 ) such tht f d) = 0. However, f x) = x 2 + e x > 0 for ll x. In prticulr f d) > 0. This is contrdiction so the ssumption tht two roots exist is wrong nd there must be exctly one root. 24. Suppose tht f x) 5 for ll vlues of x. Show tht 18 f 8) f 2) 0. Solution: Use the Men Vlue Theorem. The problem sttes tht f x) exists nd is between two vlues for ll x. Therefore the function f x) is continuous nd differentible everywhere, prticulrly on [2, 8]. Then there exists c such tht: f f 8) f 2) c) = 8 2 However f x) 5 for ll x. In prticulr, f c) 5. f c) 5 = This proves the required sttement. f 8) f 2) 6 5 = 18 f 8) f 2) Suppose tht f nd g re continuous on [, b] md differentible on, b). Suppose lso tht f ) = g) nd f x) < g x) for < x < b. Prove tht f b) < gb). Solution: Using the hint in the text look t the function hx) = f x) gx). Note if hb) < 0 then the desired result follows. Now pply the Men Vlue Theorem to h. Since f nd g re continuous on [, b] nd differentible on, b) then so is h the derivtive is liner nd the difference of continuous functions is continuous). The conditions of the Men Vlue Theorem re sttisfied for h so then there exists c [, b] such tht: hb) h) = h c) b ) However recll tht since f ) = g). Also note tht h) = f ) g) = 0 h c) = f c) g c) < 0 becuse, by ssumption f x) < g x) for ll x. Finlly, becuse b >, we hve tht b > 0. Then: Thus f b) gb) < 0, so f b) < gb). hb) = h) + h c) b ) = 0 + h c) b ) < 0. Pge 4 of 5

5 Problem A Let f x) be differentible t x =. The lineriztion of f t x = is given by Lx) = f ) + f )x ). In clss we sid tht good liner pproximtion should hve the property tht f x) Lx) 0 fster thn x 0. Show tht this is true. Tht is, show tht lim x f x) Lx) x Solution: Recll tht the liner pproximtion bout is given by: = 0 Lx) = f ) f )x ) Then: lim x f x) Lx) x f x) f ) + f )x )) x x f x) f ) f )x ) x x f x) f ) f ) )x ) x x x ) f x) f ) f ) x x x f x) f ) x = f ) f ) = 0. ) f ) Pge 5 of 5

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