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anonymous 3 years ago I have question..

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1. anonymous

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.

2. TuringTest

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

3. TuringTest

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

4. TuringTest

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

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. TuringTest

yeah that looks a bit better...

9. TuringTest

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

10. anonymous

|dw:1354859726253:dw|

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