MHB Continuous and differentiability

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The discussion revolves around determining the differentiability of a function at a specific point, particularly at x=0. The user has shown that the function is differentiable at this point and that the first derivative f'(0) equals 0. For values of x not equal to 0, the first derivative f'(x) is defined piecewise as 2x for x>0 and -2x for x<0. The main concern is whether the second derivative f''(0) exists, which requires further analysis of the behavior of f'(x) around x=0. Clarification on the existence of f''(0) is needed to conclude the differentiability of the function at that point.
Joe20
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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!
 

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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 $x=0$.
 

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