Answer
Yes, the function $ f\left( x \right)={{x}^{2}}-3x+7$ is continuous at $4$.
Work Step by Step
Consider the function $ f\left( x \right)={{x}^{2}}-3x+7$,
First check whether the function is defined at the point $ a $ or not.
Find the value of $ f\left( x \right)$ at $ a=4$,
$\begin{align}
& f\left( 4 \right)={{\left( 4 \right)}^{2}}-3\left( 4 \right)+7 \\
& =16-12+7 \\
& =11
\end{align}$
The function is defined at the point $ a=4$.
Now find the value of $\,\underset{x\to 4}{\mathop{\lim }}\,{{x}^{2}}-3x+7$,
$\begin{align}
& \,\underset{x\to 4}{\mathop{\lim }}\,f\left( x \right)=\,\underset{x\to 4}{\mathop{\lim }}\,\left( {{x}^{2}}-3x+7 \right) \\
& ={{\left( 4 \right)}^{2}}-3\left( 4 \right)+7 \\
& =16-12+7 \\
& =11
\end{align}$
Thus, $\,\underset{x\to 4}{\mathop{\lim }}\,\left( {{x}^{2}}-3x+7 \right)=11$
Now check whether $\,\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ or not.
From the above, $\,\underset{x\to 4}{\mathop{\lim }}\,\left( {{x}^{2}}-3x+7 \right)=11\text{ and }f\left( 4 \right)=11$
Therefore, $\,\underset{x\to 4}{\mathop{\lim }}\,f\left( x \right)=f\left( 4 \right)$
Thus, the function satisfies all the properties of being continuous.
Hence, the function $ f\left( x \right)={{x}^{2}}-3x+7$ is continuous at $4$.
