The set of all points where the function f(x) | vedclass

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(A) The function is given by $f(x) = 2x|x|$.We can rewrite this as a piecewise function:$f(x) = \begin{cases} 2x^2, & x \geq 0 \ -2x^2, & x < 0 \end{

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$(-\infty, \infty)$

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(A) The function is given by $f(x) = 2x|x|$.We can rewrite this as a piecewise function:$f(x) = \begin{cases} 2x^2, & x \geq 0 \ -2x^2, & x Since $2x^2$ and $-2x^2$ are polynomials,$f(x)$ is differentiable for all $x 
eq 0$. We only need to check the differentiability at $x = 0$.Left Hand Derivative $(LHD)$ at $x = 0$:$f'(0^-) = \lim_{h ightarrow 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h ightarrow 0^-} \frac{-2h^2 - 0}{h} = \lim_{h ightarrow 0^-} (-2h) = 0$.Right Hand Derivative $(RHD)$ at $x = 0$:$f'(0^+) = \lim_{h ightarrow 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h ightarrow 0^+} \frac{2h^2 - 0}{h} = \lim_{h ightarrow 0^+} (2h) = 0$.Since $LHD = RHD = 0$,the function $f(x)$ is differentiable at $x = 0$.Therefore,$f(x)$ is differentiable for all $x \in (-\infty, \infty)$.

The set of all points where the function $f(x) = 2x|x|$ is differentiable is

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