hw22solutions - f c | ≤ ε v-u[Hint Add and substract f...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 521 - Advanced Calculus I Instructor: J. Metcalfe Due: March 31, 2010 Assignment 22 1. Suppose that f : R R is differentiable at c and that f ( c ) = 0. Show that g ( x ) := | f ( x ) | is differentiable at c if and only if f 0 ( c ) = 0. Since f ( c ) = 0, we have lim x c g ( x ) - g ( c ) x - c = lim x c | f ( x ) | x - c = lim x c ± ± ± ± f ( x ) x - c ± ± ± ± | x - c | x - c = lim x c ± ± ± ± f ( x ) - f ( c ) x - c ± ± ± ± | x - c | x - c . We thus see that lim x c + ± ± ± ± f ( x ) - f ( c ) x - c ± ± ± ± | x - c | x - c = | f 0 ( c ) | and lim x c - ± ± ± ± f ( x ) - f ( c ) x - c ± ± ± ± | x - c | x - c = -| f 0 ( c ) | . Thus, if | f 0 ( c ) | 6 = 0, these one sided limits are not the same and the limit as x c does not exist. On the other hand if f 0 ( c ) = 0, these one sided limits are the same and then correspond to the limit as x c . 2. Let f : I R be differentiable at c I . Eastblish the Straddle lemma : Given ε > 0 there exists δ > 0 so that if u,v I satisfy c - δ < u c v < c + δ then we have | f ( v ) - f ( u ) - ( v - u
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ) f ( c ) | ≤ ε ( v-u ). [Hint: Add and substract f ( c )-cf ( c ) to the left side.] Fix ε > 0. Since f is differentiable at c , we have that for any x ∈ I with | x-c | < δ ± ± ± ± f ( x )-f ( c ) x-c-f ( c ) ± ± ± ± < ε/ 2 . That is | f ( x )-f ( c )-f ( c )( x-c ) < ε 2 ( x-c ) . We apply this with x = u so that c-δ < u ≤ c and x = v with c ≤ v < c + δ . Then, | f ( v )-f ( u )-( v-u ) f ( c ) | ≤ | f ( v )-f ( c )-( v-c ) f ( c ) | + | f ( u )-f ( c )-( u-c ) f ( c ) | ≤ ε 2 | v-c | + ε 2 | u-c | ≤ ε | v-u | . 1...
View Full Document

This note was uploaded on 07/15/2010 for the course MATH 521 taught by Professor Metcalfe during the Spring '10 term at University of North Carolina Wilmington.

Ask a homework question - tutors are online