Let $f:\mathbb{R}\to \mathbb{R}$ be a function defined by $f\left( x \right)=\max \left\{ x,{{x}^{3}} \right\}$. The set of all points where $f\left( x \right)$ is not differentiable is
A. $\left\{ -1,1 \right\}$
B. $\left\{ -1,0 \right\}$
C. $\left\{ 0,1 \right\}$
D. $\left\{ -1,0,1 \right\}$

Hint: We have to find the continuity and the differentiability of the given piecewise function at certain points. We use the points and respective functions to find the maximum value that will be available to operate. We take the final conclusion depending on the equality of the theorem $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)$ for differentiability where \[{{f}^{'}}\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}\].

Complete step-by-step answer:
We have to show the continuity and the differentiability of the given function at certain intervals.
If the function is not continuous at certain points, then the function will never be differentiable at those points.
We check the continuity for the points $\left\{ -1,0,1 \right\}$.
$f\left( x \right)=\max \left\{ x,{{x}^{3}} \right\}=\left\{ \begin{align}
  & x\text{ }x\in \left( -\infty ,-1 \right) \\
 & {{x}^{3}}\text{ }x\in \left[ -1,0 \right] \\
 & x\text{ }x\in \left( 0,1 \right) \\
 & {{x}^{3}}\text{ }x\in \left[ 1,\infty \right) \\
\end{align} \right.$
For the function if the condition $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)$ satisfies then it will be differentiable where \[{{f}^{'}}\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}\].
Here ${{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left[ f\left( x \right) \right]$.
We check differentiability at three different points of $x=\left\{ -1,0,1 \right\}$.
At $x=-1$, we get
\[\begin{align}
  & \underset{x\to {{\left( -1 \right)}^{-}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)={{\left[ \dfrac{d}{dx}\left( x \right) \right]}_{x=\left( -1 \right)}}=1 \\
 & \underset{x\to {{\left( -1 \right)}^{+}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)={{\left[ \dfrac{d}{dx}\left( {{x}^{3}} \right) \right]}_{x=\left( -1 \right)}}={{\left[ 3{{x}^{2}} \right]}_{x=\left( -1 \right)}}=3 \\
\end{align}\]
Therefore, $f\left( x \right)$ is not differentiable at $x=-1$ as \[\underset{x\to {{\left( -1 \right)}^{-}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)\ne \underset{x\to {{\left( -1 \right)}^{+}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)\].
At $x=0$, we get
\[\begin{align}
  & \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)={{\left[ \dfrac{d}{dx}\left( {{x}^{3}} \right) \right]}_{x=0}}={{\left[ 3{{x}^{2}} \right]}_{x=0}}=0 \\
 & \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)={{\left[ \dfrac{d}{dx}\left( x \right) \right]}_{x=0}}=1 \\
\end{align}\]
Therefore, $f\left( x \right)$ is not differentiable at $x=0$ as \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)\ne \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)\].
At $x=1$, we get
\[\begin{align}
  & \underset{x\to {{1}^{-}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)={{\left[ \dfrac{d}{dx}\left( x \right) \right]}_{x=1}}=1 \\
 & \underset{x\to {{1}^{+}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)={{\left[ \dfrac{d}{dx}\left( {{x}^{3}} \right) \right]}_{x=1}}={{\left[ 3{{x}^{2}} \right]}_{x=1}}=3 \\
\end{align}\]
Therefore, $f\left( x \right)$ is not differentiable at $x=1$ as \[\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)\ne \underset{x\to {{1}^{+}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)\].
Therefore, $f\left( x \right)$ is not differentiable at points $\left\{ -1,0,1 \right\}$. The correct option is D.
So, the correct answer is “Option D”.

Note: This type of differentiability checking is called differentiability of piecewise function. A piecewise function is differentiable at a point if both of the pieces have derivatives at that point, and the derivatives are equal at that point.