Answer
$$
f(x)=\frac{1-x}{3+x}; \quad\quad[0,3]
$$
The graph has no critical numbers.
The absolute maximum, $\frac{1}{3} \approx 0.3333 $ occurs when $x=0, $
and the absolute minimum, $\frac{-1}{3} \approx -0.3333 $ occurs when $x=3.$
Work Step by Step
$$
f(x)=\frac{1-x}{3+x}; \quad\quad[0,3]
$$
First look for critical numbers in the interval $(0,3)$.
find the points where the derivative is $0$
$$
\begin{aligned}
f^{\prime}(x) &=\frac{1-x}{3+x};\\
&=\frac{(3+x) (-1)-(1-x)(1)}{(3+x)^{2}}\\
&=\frac{-4}{(3+x)^{2}}
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x) &=\frac{-4}{(3+x)^{2}}=0
\end{aligned}
$$
Since this equation has no solution,
So,the graph has no critical numbers.
Now, evaluate the function at the endpoints:
$$
\text{when } x=0: f(0)=\frac{1-(0)}{3+(0)}=\frac{1}{3} \approx 0.3333\\
\text{when } x=3: f(3)=\frac{1-(3)}{3+(3)}=\frac{-2}{6}\approx -0.3333\\
$$
We conclude that, the absolute maximum, $\frac{1}{3} \approx 0.3333 $ occurs when $x=0 $ and the absolute minimum, $\frac{-1}{3} \approx -0.3333 $ occurs when $x=3.$
