Answer
$$
f(x) =2 x+3 x^{2 / 3}
$$
(*)
A relative maximum occurs at $x=-1,$ and the value of the function $f$ at this value is given by:
$$
\begin{aligned}
f(-1) =1.
\end{aligned}
$$
(**)
A relative minimum occurs at $x=0,$ and the value of the function $f$ at this value is given by:
$$
\begin{aligned}
f(0) =0.
\end{aligned}
$$
Work Step by Step
$$
f(x) =2 x+3 x^{2 / 3}
$$
To find the intervals where the function is increasing and decreasing.
First, we find $f^{\prime}(x) $,
$$
\begin{aligned}
f^{\prime}(x) &=2+2 x^{-1 / 3} \\
&=2+\frac{2}{\sqrt[3]{x}}
\end{aligned}
$$
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
$$
\begin{aligned}
f^{\prime}(x) =2+\frac{2}{\sqrt[3]{x}} &=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
\frac{2}{\sqrt[3]{x}} &=-2\\
\Rightarrow\quad\quad\quad\quad\quad\\
\sqrt[3]{x} &=-1
\\
\Rightarrow\quad\quad\quad\quad\quad\\
x &=-1
\end{aligned}
$$
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 $\sqrt[3]{x} $ equal to 0, when $x=0$ So,
The critical numbers are: $x=-1 $ 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
$$
(-\infty, -1 ), \quad (-1 ,0 ) \quad \text {and }\quad ( 0 , \infty) .
$$
(1)
Test a number in the interval $(-\infty, -1) $ say $-2 $:
$$
\begin{aligned}
f^{\prime}(-2) &=2+\frac{2}{\sqrt[3]{(-2)}} \\
&=2-2^{\frac{2}{3}}\\
&\approx 0.41259\\
& \gt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(-\infty , -1) $.
(2)
Test a number in the interval $( -1, 0) $ say $ -0.5$:
$$
\begin{aligned}
f^{\prime}(-0.5) &=2+\frac{2}{\sqrt[3]{(-0.5)}} \\
&\approx -0.51984 \\
&\lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(-1 , 0) $.
(3)
Test a number in the interval $(0, \infty) $ say $1 $:
$$
\begin{aligned}
f^{\prime}(1) &=2+\frac{2}{\sqrt[3]{(1)}} \\
&=4\\
& \gt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(0 , \infty) $.
From (1), (2) and (3) we find that:
(*)
The function $f$ is increasing on interval $ ( -\infty , -1) , $ and is decreasing on interval $ (-1 ,0 ) . $ So, a relative maximum occurs at $x=-1,$ and the value of the function $f$ at this value is given by:
$$
\begin{aligned}
f(-1) &=2 (-1)+3 (-1)^{2 / 3} \\
&=1. \\
\end{aligned}
$$
(**)
The function $f$ is decreasing on interval $ (-1 ,0 ), $ and is increasing on interval $ ( 0 , \infty). $ So, a relative minimum occurs at $x=0,$ and the value of the function $f$ at this value is given by:
$$
\begin{aligned}
f(0) &=2 (0)+3 (0)^{2 / 3} \\
&=0. \\
\end{aligned}
$$
