Answer
$$
f(x)=x^{\frac{2}{3}}
$$
(a)
The critical numbers : $x=0$.
(b)
The function is increasing on interval $ (0, \infty). $
(c)
The function is decreasing on interval $(-\infty, 0 ). $
Work Step by Step
$$
f(x)=x^{\frac{2}{3}}
$$
We find $f^{\prime}(x) $ first, using the power rule and the chain rule.
$$
\begin{aligned}
f^{\prime}(x) &=\frac{2}{3} x^{\frac{-1}{3}}\\
&=\frac{2}{3x^{\frac{1}{3}}}
\end{aligned}
$$
To find the critical numbers, we first find any values of $x$ that make $f^{\prime}(x)=0 $ , but here $f^{\prime}(x) $ is never $0$. Next, we find any values of $x$ where $f^{\prime}(x) $ fails to exist. This occurs whenever the denominator of $f^{\prime}(x) $ is $0$, so it fails to exist at $x=0$. Therefore, $x=0$ is the only critical number.
(a)
The critical numbers : $x=0$.
We can still apply the first derivative test, however, to find where $f$ is increasing and decreasing. The number $0$ divides the number line into two intervals: $(-\infty, 0)$ and $(0, \infty)$.
use a test point in each of these intervals as follows:
(1) Test a number in the interval $(-\infty, 0)$ say $-1$:
$$
\begin{aligned}
f^{\prime}(-1) &=\frac{2}{3(-1)^{\frac{1}{3}}} \\
&=-\frac{2}{3}\\
& \lt 0
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(-\infty, 0).$
(2) Test a number in the interval $(0, \infty)$ say $1$:
$$
\begin{aligned}
f^{\prime}(1) &=\frac{2}{3(1)^{\frac{1}{3}}} \\
&=\frac{2}{3}\\
&\gt 0
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(0, \infty ).$
So,
(b) The function is increasing on interval $ (0, \infty), $
(c) The function is decreasing on interval $(-\infty, 0 ). $
