Answer
$$
f(x) =3 x^{\frac{5}{3}}-15 x^{2 / 3}
$$
(*)
A relative maximum occurs at $x=0,$ and the value of the function $f$ at this value is given by:
$$
f(0) =0,
$$
(**)
A relative minimum occurs at $x=2,$ and the value of the function $f$ at this value is given by:
$$
f(2) =-9\cdot \:2^{\frac{2}{3}} \approx -14.2866
$$
Work Step by Step
$$
f(x) =3 x^{\frac{5}{3}}-15 x^{2 / 3}
$$
To find the intervals where the function is increasing and decreasing.
First, we find $f^{\prime}(x) $,
$$
\begin{aligned}
f^{\prime}(x) &=3(\frac{5}{3}) x^{\frac{2}{3}}-15 (2 / 3) x^{-1 / 3}\\
&=5 x^{\frac{2}{3}}-10 x^{-1 / 3}\\
&=5 x^{\frac{-1}{3}}[x-2 ]\\
&=\frac{5(x-2)}{x^{\frac{1}{3}}}
\end{aligned}
$$
To find the critical numbers, we first find any values of $x$ that make $f^{\prime} =0 $. Now, solve the equation $f^{\prime} =0 $ to get
$$
\begin{aligned}
f^{\prime}(x) =\frac{5(x-2)}{x^{\frac{1}{3}}} &=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
(x-2) &=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
x&=2
\end{aligned}
$$
Next, we find any values of $x$ where $f^{\prime}(x) $ fails to exist. This occurs whenever the denominator of $f^{\prime}(x) $ is $0$, here we observe that the denominator $x^{\frac{1}{3}}$ equal to 0, when $x=0$ So,
The critical numbers : $x=2 $ and $x=0$ .
We can still apply the first derivative test, however, to find where $f$ is increasing and decreasing. Now, check the sign of $f^{\prime}(x)$. in the intervals
$$
(-\infty, 0), \quad (0 ,2 ) \quad \text {and }\quad ( 2 , \infty) .
$$
(1)
Test a number in the interval $(-\infty, 0) $ say $-1 $:
$$
\begin{aligned}
f^{\prime}(-1) &=\frac{5((-1)-2)}{(-1)^{\frac{1}{3}}} \\
&=15\\
& \gt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(-\infty , 0) $.
(2)
Test a number in the interval $( 0, 2) $ say $ 1$:
$$
\begin{aligned}
f^{\prime}(1) &=\frac{5((1)-2)}{(1)^{\frac{1}{3}}} \\
&=-5\\
&\lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(0 , 2) $.
(3)
Test a number in the interval $(2, \infty) $ say $3 $:
$$
\begin{aligned}
f^{\prime}(3) &=\frac{5((3)-2)}{(3)^{\frac{1}{3}}} \\
&=\frac{5}{3^{\frac{1}{3}}}\\
&\approx 3.4668\\
& \gt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(2 , \infty) $.
From (1), (2) and (3) we find that:
(*)
The function $f$ is increasing on interval $ ( -\infty , 0) , $ and is decreasing on interval $ (0 ,2 ) . $ So, a relative maximum occurs at $x=0,$ and the value of the function $f$ at this value is given by:
$$
\begin{aligned}
f(0) &=3 (0)^{\frac{5}{3}}-15 (0)^{2 / 3}\\
&=0. \\
\end{aligned}
$$
(**)
The function $f$ is decreasing on interval $ (0 ,2 ), $ and is increasing on interval $ ( 2 , \infty). $ So, a relative minimum occurs at $x=2,$ and the value of the function $f$ at this value is given by:
$$
\begin{aligned}
f(2) &=3 (2)^{\frac{5}{3}}-15 (2)^{2 / 3} \\
&=-9\cdot \:2^{\frac{2}{3}} \\
&\approx -14.2866
\end{aligned}
$$
