Continuous and differentiability

[Joe20]

Hello,

I have attached the question and the steps worked out. I am not sure if my steps are correctly. Need advise on that.
Next, I am not sure how to show f''(0) exist or not. Thanks in advance!

 

[Opalg]

You have shown that $f$ is differentiable at $0$ and that $f'(0) = 0$. For $x\ne0$ you can use the usual rules for differentiation, to see that $f'(x) = 2x$ if $x>0$ and $f'(x) = -2x$ if $x<0$. Therefore \[f'(x) = \begin{cases}2x&\text{when }x\geqslant0,\\-2x&\text{when }x<0.\end{cases}\] Now you have to decide whether that function is differentiable at $x=0$.

 

[math771]

Hi there,

Thank you for sharing your question and steps. It's always helpful to see the work that has been done so far. As for whether your steps are correct, I would suggest double-checking your calculations and making sure you have followed all the necessary rules and formulas. You could also ask a classmate or your teacher for feedback to ensure accuracy.

Regarding f''(0), one way to show its existence is by using the definition of a derivative. You could show that the limit of the difference quotient as x approaches 0 exists, which would prove the existence of the second derivative at 0. Alternatively, you could also check for differentiability at 0 using the first and second derivative tests.

I hope this helps. Good luck with your problem!