Answer
$$
f(x)=\frac{x^{5}-5 x}{5}
$$
(a)
The critical numbers of $f$: $x=\pm1$.
(b)
$f$ is increasing on the intervals :$ (-\infty\lt x\lt -1) , (1\lt x \lt \infty)$
and decreasing on the interval : $ (-1\lt x \lt 1)$
(c)
Relative maximum: $ (-1, \frac{4}{5})$
Relative minimum: $ (1, \frac{4}{5})$
Work Step by Step
$$
f(x)=\frac{x^{5}-5 x}{5}
$$
Note that $f $ is differentiable on the entire real number line and the
derivative of $f$ is
$$
f^{\prime}(x)=x^{4}-1
$$
To determine the critical numbers of $f $ set $f^{\prime}(x)$ equal to zero.
$$
f^{\prime}(x)=x^{4}-1=(x^{2}-1)(x^{2}+1)=0
$$
So,
(a) The critical numbers of $f$: $x=\pm1$.
Because there are no points for which $f^{\prime}(x)$ does not exist, you can conclude that $x=1$and $x=-1$ are the only critical numbers. The table summarizes the testing of the three intervals determined by these two critical numbers.
$$
\begin{array}{|c|c|c|c|}\hline \text { Interval } & {-\infty\lt x\lt -1} & {-1\lt x \lt 1} & {1\lt x \lt \infty} \\ \hline \text { Test Value } & {x=-2} & {x=0} & {x=2} \\ \hline \text { Sign of } f^{\prime}(x) & {f^{\prime}(2) =15\gt 0} & {f^{\prime}\left(0\right)=-1\lt 0} & {f^{\prime}(2)=15 \gt 0} \\ \hline \text { Conclusion } & {\text { Increasing }} & {\text { Decreasing }} & {\text { Increasing }} \\ \hline\end{array}
$$
(b)
$f$ is increasing on the intervals :$ (-\infty\lt x\lt -1) , (1\lt x \lt \infty)$
and decreasing on the interval : $ (-1\lt x \lt 1)$
(c)
Relative maximum: $ (-1, \frac{4}{5})$
Relative minimum: $ (1, \frac{4}{5})$
