{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MIT18_014F10_pset7sols

# MIT18_014F10_pset7sols - 18.014 Problem Set 7 Solutions...

This preview shows pages 1–3. Sign up to view the full content.

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 first observe that 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 first 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: 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 first part of the question. Now, observe by the Fundamental 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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) e x > 1 + x + x 2 / 2! + x 3 / 3! , e x < 1 x + x 2 / 2! x 3 / 3!.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 6

MIT18_014F10_pset7sols - 18.014 Problem Set 7 Solutions...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online