Answer
The number for which the function $ f\left( x \right)=\left\{ \begin{align}
& 5x\text{ if }x<4 \\
& \text{21 if }x=4 \\
& {{x}^{2}}+4\text{ if }x>4
\end{align} \right.$ is discontinuous is $ a=4$
Work Step by Step
Consider the function $ f\left( x \right)=\left\{ \begin{align}
& 5x\text{ if }x<4 \\
& \text{21 if }x=4 \\
& {{x}^{2}}+4\text{ if }x>4
\end{align} \right.$,
Check the discontinuity of the function at $ x=4$
Find the value of $ f\left( x \right)$ at $ x=4$,
From the definition of the function,
$ f\left( 4 \right)=21$
The function is defined at the point $ x=4$.
Now find the value of $\,\underset{x\to 4}{\mathop{\lim }}\,f\left( x \right)$,
First find the left-hand limit of $\,f\left( x \right)$,
That is,
$\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f\left( x \right)=5\left( 4 \right)=20$
Now find the right-hand limit of $\,f\left( x \right)$,
That is,
$\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f\left( x \right)={{4}^{2}}+4=20$
Since the left-hand limit and right-hand limit are equal, that is $\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f\left( x \right)=20=\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f\left( x \right)$
Thus $\,\underset{x\to 4}{\mathop{\lim }}\,f\left( x \right)=20$
From the above steps, $\,\underset{x\to 4}{\mathop{\lim }}\,f\left( x \right)\ne f\left( 4 \right)$
Thus, the function does not satisfy the third property of being continuous.
Hence, the function $ f\left( x \right)=\left\{ \begin{align}
& 5x\text{ if }x<4 \\
& \text{21 if }x=4 \\
& {{x}^{2}}+4\text{ if }x>4
\end{align} \right.$ is discontinuous at $ x=4$.
Thus, the number for which the function $ f\left( x \right)=\left\{ \begin{align}
& 5x\text{ if }x<4 \\
& \text{21 if }x=4 \\
& {{x}^{2}}+4\text{ if }x>4
\end{align} \right.$ is discontinuous is $ x=4$
