MIT18_014F10_pset5sols

MIT18_014F10_pset5sols - 18.014 Problem Set 5 Solutions...

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Unformatted text preview: 18.014 Problem Set 5 Solutions Total: 24 points Problem 1: Let f ( x ) = x 4 + 2 x 2 + 1 for ≤ 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 ≤ x < y ≤ 10. Since y 4 − x 4 = ( y − x )( y 3 + y 2 x + yx 2 + x 3 ) > , we have y 4 > x 4 . Similarly, y 2 > x 2 since y 2 − x 2 = ( y − x )( y + x ) > . 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 ≤ 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 ≤ 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 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 (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 = x < 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 ....
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This note was uploaded on 01/18/2012 for the course MATH 18.014 taught by Professor Christinebreiner during the Fall '10 term at MIT.

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MIT18_014F10_pset5sols - 18.014 Problem Set 5 Solutions...

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