Suppose g:R→R is an everywhere differentiable function. Show that if g(a)≠0, then |g(x)| is differentiable at x =a. Hint: use the fact that composition of 2 differentiable functions is differentiable. b) Now suppose g(a) =0. Show that |g(x)| is differentiable at x = a iff g'(a) =0 Please, help
i'll try tomorrow
ok!! good night, friend.
This is the second time I post this problem. It seems no one wants to help me. :(
not like that @Loser66 few ppl here only know these stuff dont worry :)
I don't know how to do this, I am just here watching sorry.
@Kainui are you kidding me? it 's calculus.
and I know you are an Ace on the field
That's different, I don't believe in proofs.
lol
hahaha..... it's ok!!
part 1
that it only
part 2, g(a)=0 means we need to check defferiantability at g(a)=0 when g'(a)=0 means its smooth so its deferentiable
i wrote these hints hope it helps you to write neat answer
Suppose g(a) =0, then g is differentiable at x =a \(g'(a) = lim_{x\rightarrow a}\dfrac{g(x)- g(a)}{x-a}= lim_{x\rightarrow a}\dfrac{g(x)}{x-a}\) For other piecewise, we have the same argument, and let them equal.
That is \(lim_{x\rightarrow a}\dfrac{g(x)}{x-a} =lim_{x\rightarrow a}\dfrac{-g(x)}{x-a}\) then?? how to argue to let to g'(a) =0??
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