Answer
$h(x)$ has the absolute maximum value of $\ln4$ at $x=3$ and the absolute minimum value of $0$ at $x=0$.
Work Step by Step
$$h(x)=\ln(x+1)\hspace{1cm}0\le x\le3$$
1) Find the critical points of $h(x)$
Find $h'(x)$: $$h'(x)=\frac{(x+1)'}{x+1}=\frac{1}{x+1}$$
There are no values of $x$ for which $h'(x)=0$; but for $x=-1$, $h'(x)$ is not defined. So $x=-1$ is a critical point of $h(x)$.
However, since $-1\notin[0,3]$, we do not consider $x=-1$ here.
2) Evaluate $h(x)$ at the endpoints:
- For $x=0$: $$h(x)=\ln(0+1)=\ln1=0$$
- For $x=3$: $$h(x)=\ln(3+1)=\ln4$$
From the values of $h(x)$ above, we conclude that $h(x)$ has the absolute maximum value of $\ln4$ at $x=3$ and the absolute minimum value of $0$ at $x=0$.
