Answer
$$
f(t)=t-\sqrt[3] {t}, \text { on } \,\, [-1,4]
$$
The absolute maximum value is
$$
f(4)=4-\sqrt[3] {4} \approx 2.413,
$$
and the absolute minimum value is
$$
f\left(\frac{1}{3 \sqrt{3}}\right)=\frac{1}{3 \sqrt{3}}-\frac{1}{\sqrt{3}}=-\frac{2 \sqrt{3}}{9}.
$$
Work Step by Step
$$
f(t)=t-\sqrt[3] {t}, \text { on } \,\, [-1,4]
$$
Since $ f$ is continuous on $ [-1,4], $ we can use the Closed Interval Method:
$$
f(t)=t-\sqrt[3] {t},
$$
$$
f^{\prime}(t)=1-\frac{1}{3} t^{-\frac{2}{3}}
$$
$f^{\prime}(t)$ does not exist when $t = 0$. the critical numbers of $f$ occur when $f^{\prime}(t)=0,$ that is,
$$
f^{\prime}(t)=1-\frac{1}{3} t^{-\frac{2}{3}}=0
$$
$ \Leftrightarrow $
$$
1=\frac{1}{3} t^{-\frac{2}{3}} \Leftrightarrow 1=\frac{1}{3t^{\frac{2}{3}}}
$$
$ \Leftrightarrow $
$$
t^{\frac{2}{3}}=\frac{1}{3}
$$
$ \Leftrightarrow $
$$
t=\pm\left(\frac{1}{3}\right)^{3 / 2}=\pm \sqrt{\frac{1}{27}}=\pm \frac{1}{3 \sqrt{3}}=\pm \frac{\sqrt{3}}{9}
$$
Notice that each of these critical numbers lies in the interval $ [-1,4], $
The values of $f $ at these critical numbers are
$$
f\left(\frac{-1}{3 \sqrt{3}}\right)=\frac{-1}{3 \sqrt{3}}-\frac{-1}{\sqrt{3}}=\frac{-1+3}{3 \sqrt{3}}=\frac{2 \sqrt{3}}{9} \approx 0.3849, \\ f\left(\frac{1}{3 \sqrt{3}}\right)=\frac{1}{3 \sqrt{3}}-\frac{1}{\sqrt{3}}=-\frac{2 \sqrt{3}}{9}
$$
The values of $f$ at the endpoints of the interval are
$$
f(-1)=0 ,\quad f(4)=4-\sqrt[3] {4} \approx 2.413.
$$
Comparing these four numbers, we see that the absolute maximum value is
$$
f(4)=4-\sqrt[3] {4} \approx 2.413.
$$
and the absolute minimum value is
$$
f\left(\frac{1}{3 \sqrt{3}}\right)=\frac{1}{3 \sqrt{3}}-\frac{1}{\sqrt{3}}=-\frac{2 \sqrt{3}}{9}
$$
