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cinarBest ResponseYou've already chosen the best response.0
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.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
I guess we must show that\[\frac{f(0)}{f(0)}\neq0\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
can't get l'Hospital to yield anything really...
 one year ago

cinarBest ResponseYou've already chosen the best response.0
\[\Large f'(x)=\frac{f(x)f(0)}{xc}f'(c)<\epsilon\]
 one year ago

cinarBest ResponseYou've already chosen the best response.0
this is the epsilondelta definition of diff. at c..
 one year ago

cinarBest ResponseYou've already chosen the best response.0
sorry should be this\[\Large f'(x)=\frac{f(x)f(c)}{xc}f'(c)<\epsilon\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
yeah that looks a bit better...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
oh, @mahmit2012 is here, he can likely help.
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.0
dw:1354859726253:dw
 one year ago
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