Answer
$$
f(x)=x^{4}-32 x+4
$$
(a)
The critical numbers of $f$: $x=2$.
(b)
$f$ is increasing on the intervals :$ (2\lt x \lt \infty)$
and decreasing on the interval : $ (-\infty \lt x \lt 2)$
(c)
Relative minimum: $ (2, -44)$
Work Step by Step
$$
f(x)=x^{4}-32 x+4
$$
Note that $f $ is differentiable on the entire real number line and the
derivative of $f$ is
$$
f^{\prime}(x)=4x^{3}-32
$$
To determine the critical numbers of $f $ set $f^{\prime}(x)$ equal to zero.
$$
f^{\prime}(x)=4x^{3}-32=4(x^{3}-8)=0
$$
So,
(a) The critical numbers of $f$: $x=2$.
Because there are no points for which $f^{\prime}(x)$ does not exist, you can conclude that $x=2$ is the only critical number. The table summarizes the testing of the two intervals determined by critical number.
$$
\begin{array}{|c|c|c|c|}\hline \text { Interval } & {-\infty\lt x\lt 2} & {2\lt x \lt \infty} \\ \hline \text { Test Value } & {x=0} & {x=3} \\ \hline \text { Sign of } f^{\prime}(x) & {f^{\prime}(0) =-32\lt 0} & {f^{\prime}\left(3 \right)=76 \gt 0} \\ \hline \text { Conclusion } & {\text { Decreasing }} & {\text { Increasing }} \\ \hline\end{array}
$$
(b)
$f$ is increasing on the intervals :$ (2\lt x \lt \infty)$
and decreasing on the interval : $ (-\infty \lt x \lt 2)$
(c)
Relative minimum: $ (2, -44)$
