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

Question

fibonaccichick666 · Answer

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

anonymous · Answer

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

fibonaccichick666 · Answer

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

fibonaccichick666 · Answer

neither would I and that is the issue.

fibonaccichick666 · Answer

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

perl · Answer

you can use the piecewise definition of absolute value

fibonaccichick666 · Answer

No, I have to use Taylor's

perl · Answer

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

fibonaccichick666 · Answer

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

perl · Answer

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