Answer
The maximum is $f(5)=26$ and the minimum is $f(6) =37/2$
Work Step by Step
Given $$y=\frac{x^{2}+1}{x-4}, \quad[5,6] $$
Since
\begin{align*}
\frac{dy}{dx}&= \frac{\frac{d}{dx}\left(x^2+1\right)\left(x-4\right)-\frac{d}{dx}\left(x-4\right)\left(x^2+1\right)}{\left(x-4\right)^2}\\
&=\frac{x^2-8x-1}{\left(x-4\right)^2}
\end{align*}
Find critical points
\begin{align*}
f'(x)&=0\\
\frac{x^2-8x-1}{\left(x-4\right)^2}&=0\\
x^2-8x-1&=0
\end{align*}
Then $x=4 \pm \sqrt{17}$. Because $4 \pm \sqrt{17}\notin [5,6]$ , $f(x)$ has no critical points on $[5,6] $, since
\begin{align*}
f(5)&= 26\\
f(6)&=37/2
\end{align*}
Then the maximum is $f(5)=26$ and the minimum is $f(6) =37/2$
