Answer
The number for which the function $ f\left( x \right)=\left\{ \begin{align}
& 7x\text{ if }x<6 \\
& \text{41 if }x=6 \\
& {{x}^{2}}+6\text{ if }x>6
\end{align} \right.$ is discontinuous at $ x=6$
Work Step by Step
Consider the function $ f\left( x \right)=\left\{ \begin{align}
& 7x\text{ if }x<6 \\
& \text{41 if }x=6 \\
& {{x}^{2}}+6\text{ if }x>6
\end{align} \right.$,
Check the discontinuity of the function at $ x=6$
Find the value of $ f\left( x \right)$ at $ x=6$,
From the definition of the function,
$ f\left( 6 \right)=41$
The function is defined at the point $ x=6$.
Now find the value of $\,\underset{x\to 6}{\mathop{\lim }}\,f\left( x \right)$,
First find the left-hand limit of $\,f\left( x \right)$,
That is,
$\underset{x\to {{6}^{-}}}{\mathop{\lim }}\,f\left( x \right)=7\left( 6 \right)=42$
Now find the right-hand limit of $\,f\left( x \right)$,
That is,
$\underset{x\to {{6}^{+}}}{\mathop{\lim }}\,f\left( x \right)={{6}^{2}}+6=42$
Since the left-hand limit and right-hand limit are equal, that is $\underset{x\to {{6}^{-}}}{\mathop{\lim }}\,f\left( x \right)=42=\underset{x\to {{6}^{+}}}{\mathop{\lim }}\,f\left( x \right)$
Thus $\,\underset{x\to 6}{\mathop{\lim }}\,f\left( x \right)=42$
From the above steps, $\,\underset{x\to 6}{\mathop{\lim }}\,f\left( x \right)\ne f\left( 6 \right)$
Thus, the function does not satisfy the third property of being continuous.
Hence, the function $ f\left( x \right)=\left\{ \begin{align}
& 7x\text{ if }x<6 \\
& \text{41 if }x=6 \\
& {{x}^{2}}+6\text{ if }x>6
\end{align} \right.$ is discontinuous at $ x=6$.
Thus, the number for which the function $ f\left( x \right)=\left\{ \begin{align}
& 7x\text{ if }x<6 \\
& \text{41 if }x=6 \\
& {{x}^{2}}+6\text{ if }x>6
\end{align} \right.$ is discontinuous at $ x=6$.
