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If $f(x)=|x|^3$, show that $f^{\prime \prime}(x)$ exists for all real $x$ and find it.
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Verified Answer
When $x \geq 0$, then $f(x)=|x|^3=x^3$
$\therefore \mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^2$ and $\mathrm{f}^{\prime \prime}(\mathrm{x})=6 \mathrm{x}$
which exists for all real values of $x$.
When $x < 0$, then $f^{\prime}(x)=|x|^2=(-x)^3=-x^3$
$\therefore \quad f^{\prime}(x)=-3 x^2$ and $f^{\prime \prime}(x)=-6 x$ which exists for all real values of $x$.
Hence $f^{\prime \prime}(x)= \begin{cases}6 x & , \text { if } x \geq 0 \\ -6 x, & \text { if } x < 0\end{cases}$
$\therefore \mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^2$ and $\mathrm{f}^{\prime \prime}(\mathrm{x})=6 \mathrm{x}$
which exists for all real values of $x$.
When $x < 0$, then $f^{\prime}(x)=|x|^2=(-x)^3=-x^3$
$\therefore \quad f^{\prime}(x)=-3 x^2$ and $f^{\prime \prime}(x)=-6 x$ which exists for all real values of $x$.
Hence $f^{\prime \prime}(x)= \begin{cases}6 x & , \text { if } x \geq 0 \\ -6 x, & \text { if } x < 0\end{cases}$
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