Answer
$f(4)\approx2.413$ is the absolute maximum value.
$f(\frac{\sqrt 3}{9})\approx-0.385$ 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.
$$f(t)=t-\sqrt[3]t,\hspace{1.5cm}[-1, 4]$$
$$f(t)=t-t^{1/3}$$
First, $f(t)$ is continuous on the interval $[-1,4]$, so the Closed Interval Method can be applied here.
1) Find critical points of $f$
$$f'(t)=1-\frac{1}{3}t^{-2/3}$$
$$f'(t)=1-\frac{1}{3\sqrt[3]{t^2}}$$
We can easily see that $f'(t)$ does not exist when $t=0$. $t=0$ lies in the domain of $f$, so it is a critical point.
Also, $$f'(t)=0$$
$$\frac{1}{3\sqrt[3]{t^2}}=1$$
$$\sqrt[3]{t^2}=\frac{1}{3}$$
$$t^2=\frac{1}{27}$$
$$t=\pm\frac{1}{3\sqrt 3}=\pm\frac{\sqrt 3}{9}$$
These values also are critical points.
After all, we've found 3 critical points. They all lie in interval $(-1, 4)$ given, so they will all move to the next sections.
2) Calculate the values of $f$ at critical points and endpoints.
- Critical points: $t=0$ and $t=\pm\frac{\sqrt 3}{9}$
$$f(0)=0-\sqrt[3]0=0$$
$$f(\frac{\sqrt 3}{9})=\frac{\sqrt 3}{9}-\sqrt[3]{\frac{\sqrt 3}{9}}\approx-0.385$$
$$f(\frac{-\sqrt 3}{9})=\frac{-\sqrt 3}{9}-\sqrt[3]{\frac{-\sqrt 3}{9}}\approx0.385$$
- End points: $t=-1$ and $t=4$
$$f(-1)=-1-\sqrt[3]{-1}=-1-(-1)=0$$
$$f(4)=4-\sqrt[3]4\approx2.413$$
3) Comparison
Among the results:
- $f(4)\approx2.413$ is the largest, so $f(4)\approx2.413$ is the absolute maximum value.
- $f(\frac{\sqrt 3}{9})\approx-0.385$ is the smallest, so $f(\frac{\sqrt 3}{9})\approx-0.385$ is the absolute minimum value.