MIT18_014F10_pset7sols

MIT18_014F10_pset7sols - 18.014 Problem Set 7 Solutions...

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18.014 Problem Set 7 Solutions Total: 12 points Problem 1: Given a function g continuous everywhere such that g (1) = 5 and ± 1 ± x g ( t ) dt = 2, let f ( x ) = 1 ( x t ) 2 g ( t ) dt . Prove that 0 2 0 x x f ( x ) = x g ( t ) dt tg ( t ) dt, 0 0 then compute f (1) and f (1). Solution (4 points) We Frst observe that ² ³ 1 2 x x x 2 f ( x ) = x g ( t ) dt 2 x tg ( t ) dt + t g ( t ) dt . 2 0 0 0 This follows simply from the fact that ( x t ) 2 = x 2 2 xt + t 2 , x is a scalar and the linearity of the integral. Now we differentiate. Here the Frst term uses the product rule and the fundamental theorem of calculus, the second term uses the same, and the third term requires only the fundamental theorem of calculus: ² ³ d 2 x x x 2 dx x 0 g ( t ) dt 2 x 0 tg ( t ) dt + 0 t g ( t ) dt x x = x 2 g ( x ) + 2 x g ( t ) dt 2 tg ( t ) dt 2 x 2 g ( x ) + x 2 g ( x ) 0 0 x x = 2 x g ( t ) dt 2 tg ( t ) dt. 0 0 Dividing the result by 2 answers the Frst part of the question. Now, observe by the ±undamental Theorem of Calculus (and the product rule) x x f ( x ) = g ( t ) dt + xg ( x ) xg ( x ) = g ( t ) dt. 0 0 Thus, f”(1)=2. Moreover, f ( x ) = g ( x ) so f (1) = g (1) = 5. Problem 2: (a) Let f ( x ) = e x 1 x for all x . Prove that f ( x ) 0 if x 0 and f ( x ) 0 if x 0. Use this fact to deduce the inequalities e x > 1 + x, e x > 1 x, 1
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for all x > 0. Integrate these inequalities to derive the following further inequalities, all valid for x > 0; x (b) e > 1 + x + x 2 / 2! , e x < 1 x + x 2 / 2!. (c)
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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_pset7sols - 18.014 Problem Set 7 Solutions...

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