Answer
$f(t)$ has the absolute maximum value of $2$ at $x=0$ and the absolute minimum value of $-1$ at $x=3$ in the defined domain.
Work Step by Step
$$f(t)=2-|t|\hspace{1cm}-1\le t\le3$$
1) For $t\in[-1,0)$, we have $|t|=-t$. The function $f(t)$ would then be $$f(t)=2+t$$
While for $t\in[0,3]$, we have $|t|=t$. The function $f(t)$ would then be $$f(t)=2-t$$
So here we have 2 different functions for$f(t)$ in 2 separate intervals. Therefore, we need to consider $t=0$ as one of the endpoints besides $t=-1$ and $t=3$.
2) We can skip the lengthy work of deciding whether $t=0$ is a critical point or not by checking the differentiability of $f'(t)$ at $t=0$, since we will eventually check $f(t)$ for $t=0$ as an endpoint.
Find the derivative $f'(t)$ for $t\ne0$:
- For $t\in[-1,0)$: $f'(t)=1$
- For $t\in[0,3]$: $f'(t)=-1$
None of the values of $t$ could get $f'(t)=0$ or undefined. So $f(t)$ has no critical points.
2) Evaluate $f(t)$ at the endpoints:
- For $t=-1$, we would use $f(t)=2+t$: $$f(-1)=2+(-1)=2-1=1$$
- For $x=0$, we would use $f(t)=2-t$: $$f(0)=2-0=2$$
- For $x=3$, we would use $f(t)=2-t$: $$f(3)=2-3=-1$$
This means $f(t)$ has the absolute maximum value of $2$ at $x=0$ and the absolute minimum value of $-1$ at $x=3$ in the defined domain.