Question

I have question..

Answer 1

Suppose that f:R to 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'(c)=0.

Answer 2

\[g'(x)=\frac{f(x)}{|f(x)|}\cdot f'(x)\]so I guess I'm stuck on showing how to deal with the intederminacy of plugging in x=c=0

Answer 3

I guess we must show that\[\frac{f(0)}{|f(0)|}\neq0\]

Answer 4

can't get l'Hospital to yield anything really...

Answer 5

\[\Large f'(x)=|\frac{f(x)-f(0)}{x-c}-f'(c)|<\epsilon\]

Answer 6

this is the epsilon-delta definition of diff. at c..

Answer 7

sorry should be this\[\Large f'(x)=|\frac{f(x)-f(c)}{x-c}-f'(c)|<\epsilon\]

Answer 8

yeah that looks a bit better...

Answer 9

oh, @mahmit2012 is here, he can likely help.