Answer
$$
f(x)=\frac{8+x}{8-x}; \quad\quad[4,6]
$$
The graph has no critical numbers.
The absolute maximum, $7 $ occurs when $x=6 $
and the absolute minimum, $3 $ occurs when $x=4.$
Work Step by Step
$$
f(x)=\frac{8+x}{8-x}; \quad\quad[4,6]
$$
First look for critical numbers in the interval $(4,6)$.
find the points where the derivative is $0$
$$
\begin{aligned}
f^{\prime}(x) &=\frac{8+x}{8-x};\\
&=\frac{(8-x) (1)-(8+x)(-1)}{(8-x)^{2}}\\
&=\frac{16}{(8-x)^{2}}
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x) &=\frac{16}{(8-x)^{2}}=0
\end{aligned}
$$
Since this equation has no solution, also $ f^{\prime}(x) $is not exist when $x=8 $ but $8 \notin [4,6] $, so the graph has no critical numbers.
Now, evaluate the function at the endpoints:
$$
\text{when } x=4: f(4)=\frac{8+(4)}{8-(4)} =3\\
\text{when } x=6: f(6)=\frac{8+(6)}{8-(6)} =7\\
$$
We conclude that, the absolute maximum, $7 $ occurs when $x=6 $ and the absolute minimum, $3 $ occurs when $x=4.$
