Answer
The number for which the function $ f\left( x \right)=\left\{ \begin{align}
& x+6\text{ if }x\le 0 \\
& \text{6 if }02
\end{align} \right.$ is discontinuous at $ x=2$.
Work Step by Step
Consider the function $ f\left( x \right)=\left\{ \begin{align}
& x+6\text{ if }x\le 0 \\
& \text{6 if }02
\end{align} \right.$,
First check the discontinuity of the function at $ x=0$
Find the value of $ f\left( x \right)$ at $ x=0$,
From the definition of the function, for $ x=0$, $ f\left( x \right)=x+6$
Then the value of $ f\left( x \right)$ at $ x=0$ is,
$\begin{align}
& f\left( 0 \right)=0+6 \\
& =6
\end{align}$
The function is defined at the point $ x=0$.
Now find the value of $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)$,
First find the left-hand limit of $\,f\left( x \right)$,
That is,
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=0+6=6$
Now find the right-hand limit of $\,f\left( x \right)$,
That is,
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=6$
Since the left-hand limit and right-hand limit are equal, that is $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=6=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)$
Thus $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=6$.
From the above steps, $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=6=f\left( 0 \right)$
Thus, the function satisfies all the properties of being continuous.
Hence, the function $ f\left( x \right)=\left\{ \begin{align}
& x+6\text{ if }x\le 0 \\
& \text{6 if }02
\end{align} \right.$ is continuous at $ x=0$.
Now check the discontinuity of the function at $ x=2$
Find the value of $ f\left( x \right)$ at $ x=2$,
From the definition of the function,
$ f\left( 2 \right)=6$
The function is defined at the point $ x=2$.
Now find the value of $\,\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)$,
First find the left-hand limit of $\,f\left( x \right)$,
That is,
$\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)=6$
Now find the right-hand limit of $\,f\left( x \right)$,
That is,
$\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)={{2}^{2}}+1=5$
Since the left-hand limit and right-hand limit are not equal, that is $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f\left( x \right)\ne \underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f\left( x \right)$
Thus, $\,\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)$ do not exist.
Thus, the function does not satisfy the second property of being continuous. Hence, the function $ f\left( x \right)=\left\{ \begin{align}
& x+6\text{ if }x\le 0 \\
& \text{6 if }02
\end{align} \right.$ is discontinuous at $ x=2$.
Thus, the number for which the function $ f\left( x \right)=\left\{ \begin{align}
& x+6\text{ if }x\le 0 \\
& \text{6 if }02
\end{align} \right.$ is discontinuous at $ x=2$.
