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MIT18_014F10_pset5sols

MIT18_014F10_pset5sols - 18.014 Problem Set 5 Solutions...

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18.014 Problem Set 5 Solutions Total: 24 points Problem 1: Let f ( x ) = x 4 + 2 x 2 + 1 for 0 x 10. (a) Show f is strictly increasing; what is the domain of its inverse function g ? (b) Find an expression for g , using radicals. Solution (4 points) (a) Let 0 x < y 10. Since y 4 x 4 = ( y x )( y 3 + y 2 x + yx 2 + x 3 ) > 0 , we have y 4 > x 4 . Similarly, y 2 > x 2 since y 2 x 2 = ( y x )( y + x ) > 0 . Summing, we get f ( y ) = y 4 + 2 y 2 + 1 > x 4 + 2 x 2 + 1 and f is strictly increasing. Since f (0) = 1, f (10) = 10 , 201, and f is strictly increasing, the domain of its inverse function g is { x | 1 x 10 , 201 } . (b) Observe g ( x ) = x 1. It’s a good exercise to check that f ( g ( x )) = g ( f ( x )) = x . Problem 2: (a) Show by example that the conclusion of the extreme value theorem can fail if f is only continuous on [ a, b ) and bounded on [ a, b ]. (b) Let f ( x ) = x for 0 x < 1; let f (1) = 5. Show that the conclusion of the small span theorem fails for the function f ( x ). Solution (4 points) (a) Define f ( x ) = x if 0 x < 1, and define f (1) = 0. Let x [0 , 1]. We claim f ( x ) is not a maximum value of the function f on [0 , 1]. If x = 1, then f (1) = 0, and 0 is not a maximum of f on [0 , 1] since f ( 1 ) = 1 > 0. If 2 2 x = 1, then define 1 + x y = . 2 Note that f ( x ) = x < y = f ( y ); hence, f ( x ) is not a maximum of f on [0 , 1]. We conclude that f has no maximum on [0 , 1]. 1

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(b) Suppose that the conclusion of the small span theorem is true for the function f ( x ) in part (b). Then given = 1, we can find a partition 0 = x 0 < x 1 < < · · · x n 1 < x n = 1 of the interval [0 , 1] such that whenever x i 1 x < y x i , we have | f ( x ) f ( y ) | < 1 . Consider the interval [ x n 1 , x n ] = [ x n 1 , 1], and put y = 1. For any x n 1 x < 1, we have | f ( x ) f ( y ) | = | x 5 | = 5 x > 4 since x < 1 < 5. This is a contradiction. Thus, the conclusion of the small span theorem fails for this function. b Problem 3: Assume f is continuous on [ a, b ]. Assume also that a f ( x ) g ( x ) dx = 0 for every function g that is continuous on [ a, b ]. Prove that f ( x ) = 0 for all x in [ a, b ].
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