Answer
$$
f(x)=x^{1 / 3}+1
$$
(a) the critical number of $f$ is $x=0 $
(b) $f$ is increasing on :$ (-\infty\lt x\lt \infty) $
(c) No relative extrema.
Work Step by Step
$$
f(x)=x^{1 / 3}+1
$$
the derivative of $f$ is
$$
f^{\prime}(x)=\frac{1}{3} x^{-2 / 3}=\frac{1}{3 x^{2 / 3}}
$$
$f^{\prime}(x) $ does not exist when $x=0 $.
So,
(a) the critical number of $f$ is $x=0 $
Because there are no points for which $f^{\prime}(x)$ does not exist, you can conclude that $x=0$ is the only critical number. The table summarizes the testing of the two intervals determined by these critical number.
$$
\begin{array}{|c|c|c|c|}\hline \text { Interval } & {-\infty\lt x\lt 0} & {0 \lt x \lt \infty} \\ \hline \text { Test Value } & {x=-1} & {x=1} \\ \hline \text { Sign of } f^{\prime}(x) & {f^{\prime}(-1) =\frac{1}{3} \gt 0} & {f^{\prime}\left(1\right)= \frac{1}{3} \lt 0} \\ \hline \text { Conclusion } & {\text { Increasing }} & {\text { Increasing }} \\ \hline\end{array}
$$
(b)
$f$ is increasing on :$ (-\infty\lt x\lt \infty) $
(c)
No relative extrema.
