Question

Let \(f(x)=|x|^3\) Compute f', f'', for all x, but show that f'''(0) DNE.

Answer 1

So here is my dilemma, this is in the Taylor polynomial section, but I cannot figure out how to apply it.

Answer 2

I wouldn't do anything involving Taylor series...

Answer 3

I have \(f'(x)=3|x|^2=3x^2\) and \(f''(x)=6|x|\) and \(f'''(x)=6 ~~x>0; -6~~~x<0\), and at x=0 it is not differentiable since |x| is not diff at x=0

Answer 4

neither would I and that is the issue.

Answer 5

It is in this section so I must have to use it to prove it

Answer 6

you can use the piecewise definition of absolute value

Answer 7

No, I have to use Taylor's

Answer 8

for derivative part, you dont need taylor . thats just derivative

Answer 9

I am aware of the derivatives, however I must rigorously prove f'''(0)=DNE