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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.
\[g'(x)=\frac{f(x)}{|f(x)|}\cdot f'(x)\]so I guess I'm stuck on showing how to deal with the intederminacy of plugging in x=c=0
I guess we must show that\[\frac{f(0)}{|f(0)|}\neq0\]

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Other answers:

can't get l'Hospital to yield anything really...
\[\Large f'(x)=|\frac{f(x)-f(0)}{x-c}-f'(c)|<\epsilon\]
this is the epsilon-delta definition of diff. at c..
sorry should be this\[\Large f'(x)=|\frac{f(x)-f(c)}{x-c}-f'(c)|<\epsilon\]
yeah that looks a bit better...
oh, @mahmit2012 is here, he can likely help.

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