Question

need help question in comments box

Answer 1

@misty1212

Answer 2

@freckles

Answer 3

@kohai

Answer 4

hint: let's think about a couple of graphs
f(x)=x^2 looks like
and we see we have f'(x)=0 at x=0
f(x)=x^3 looks like
try to make it come to a little rest at x=0
but anyways here f'(x)=0 also at x=0
like we can look at actually finding the derivative to verify what I'm saying about these two graphs here:

f(x)=x^2
f'(x)=2x
2x=0 when x=0

f(x)=x^3
f'(x)=3x^2
3x^2=0 when x=0



Now think about f(x)=|x|

There is a sharp turn at x=0
We say here f'(x) does not exist at x=0

We can also show this by finding the left and right derivatives as x approaches 0
and see if they match
from right hand side f'=1 and from left hand side f'=-1
1 doesn't equal -1
so f' will not exist at x=0