Question

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

Answer 1

i'll try tomorrow

Answer 2

ok!! good night, friend.

Answer 3

This is the second time I post this problem. It seems no one wants to help me. :(

Answer 4

not like that @Loser66 few ppl here only know these stuff dont worry :)

Answer 5

I don't know how to do this, I am just here watching sorry.

Answer 6

@Kainui are you kidding me? it 's calculus.

Answer 7

and I know you are an Ace on the field

Answer 8

That's different, I don't believe in proofs.

Answer 9

lol

Answer 10

hahaha..... it's ok!!

Answer 11

part 1

Answer 12

that it only

Answer 13

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

Answer 14

i wrote these hints hope it helps you to write neat answer

Answer 15

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.

Answer 16

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??