Answer
$f(3)=125$ is the absolute maximum value.
$f(0)=-64$ is the absolute minimum value
Work Step by Step
The Closed Interval Method
1) Function $f$ must be continuous on the closed interval $[a, b].$
2) Find the values of $f$ at the critical points in $(a,b)$ and the values of $f$ at the endpoints.
3) Compare the values.
- The largest is the absolute maximum value.
- The smallest is the absolute minimum value.
$$f(t)=(t^2-4)^3,\hspace{1.5cm}[-2, 3]$$
First, $f(t)$ is continuous on interval $[-2,3]$, so the Closed Interval Method can be applied here.
1) Find critical points of $f$
$$f'(t)=3(t^2-4)^2(t^2-4)'$$
$$f'(t)=3(t^2-4)^22t$$
$$f'(t)=6t(t^2-4)^2.$$
Therefore, $$f'(t)=0$$
$$6t(t^2-4)^2=0$$
$$t=0\hspace{0.5cm}or\hspace{0.5cm}t^2=4$$
$$t=0\hspace{0.5cm}or\hspace{0.5cm}t=\pm2$$
These values all belong to the domain of $f$, so they are all critical points. However, only $t=0$ and $t=2$ lie in the interval $(-2, 3)$, so only these two will move to the next sections.
2) Calculate the values of $f$ at critical points and endpoints.
- Critical points: $t=0$ and $t=2$
$$f(0)=(0^2-4)^3=(-4)^3=-64$$
$$f(2)=(2^2-4)^3=0^3=0$$
- End points: $t=-2$ and $t=3$
$$f(-2)=[(-2)^2-4]^3=0^3=0$$
$$f(3)=(3^2-4)^3=5^3=125$$
3) Comparison
Among the results:
- $f(3)=125$ is the largest, so $f(3)=125$ is the absolute maximum value.
- $f(0)=-64$ is the smallest, so $f(0)=-64$ is the absolute minimum value.
