Answer
$$
f(x)=x^{\frac{7}{3}}+x^{\frac{4}{3}}
$$
critical numbers are: $\frac{-4}{7}$ and $0$.
By Part 2 of the second derivative test, $\frac{-4}{7}$ leads to a relative maximum.
By using the first derivative test, $0$ leads to a relative minimum.
Work Step by Step
$$
f(x)=x^{\frac{7}{3}}+x^{\frac{4}{3}}
$$
First, find the points where the derivative is $0$
Here
$$
\begin{aligned}
f^{\prime}(x) &=(\frac{7}{3})x^{\frac{4}{3}}+ (\frac{4}{3})x^{\frac{1}{3}}\\
&=\frac{x^{\frac{1}{3}}}{3}(7x+4)
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x)& =\frac{x^{\frac{1}{3}}}{3}(7x+4)=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
7x+4 =0 \quad &\text {or} \quad x =0 \\
\\
\Rightarrow\quad\quad\quad\quad\quad\\
7x =-4 \quad &\text {or} \quad x =0 \\
\Rightarrow\quad\quad\quad\quad\quad\\
x =\frac{-4}{7} \quad &\text {or} \quad x =0 \\
\end{aligned}
$$
critical numbers are: $\frac{-4}{7}$ and $0$.
Now use the second derivative test. The second derivative is
$$
\begin{aligned}
f^{\prime\prime}(x)&=\frac{7}{3} (\frac{4}{3})x^{\frac{1}{3}}+ \frac{4}{3} (\frac{1}{3})x^{\frac{-2}{3}} \\
&=\frac{28}{9} x^{\frac{1}{3}}+ \frac{4}{9}x^{\frac{-2}{3}}.
\end{aligned}
$$
Evaluate $f^{\prime\prime}(x)$ first at $x=\frac{-4}{7}$, getting
$$
\begin{aligned}
f^{\prime\prime}(\frac{-4}{7})&=\frac{28}{9} (\frac{-4}{7})^{\frac{1}{3}}+ \frac{4}{9}(\frac{-4}{7})^{\frac{-2}{3}}\\
&=-\frac{4^{\frac{1}{3}}\cdot \:7^{\frac{2}{3}}}{3}\\
& \approx -1.93626 \lt 0
\end{aligned}
$$
so that by Part 2 of the second derivative test, $\frac{-4}{7}$ leads to a relative maximum.
Now, evaluate $f^{\prime\prime}(x)$ at $x=0$, getting
$$
\begin{aligned}
f^{\prime\prime}(0)&=\frac{28}{9} (0)^{\frac{1}{3}}+ \frac{4}{9}(0)^{\frac{-2}{3}}\\
&=0
\end{aligned}
$$
then the test gives no information about extrema, so we can use the first derivative test.
Check the sign of $f^{\prime}(x)$ in the intervals
$$
(\frac{-4}{7}, 0 ), \quad (0,\infty).
$$
(1)
Test a number in the interval $(\frac{-4}{7}, 0 )$ say $\frac{-1}{2}$:
$$
\begin{aligned}
f^{\prime}(\frac{-1}{2}) &=(\frac{7}{3})(\frac{-1}{2})^{\frac{4}{3}}+ (\frac{4}{3})(\frac{-1}{2})^{\frac{1}{3}}\\
&=-\frac{1}{6\cdot \:2^{\frac{1}{3}}} \\
&\approx -0.13228
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(\frac{-4}{7}, 0 )$
(2)
Test a number in the interval $(0, \infty )$ say $1$:
$$
\begin{aligned}
f^{\prime}(1) &=(\frac{7}{3})(1)^{\frac{4}{3}}+ (\frac{4}{3})(1)^{\frac{1}{3}}\\
&=\frac{11}{3} \\
&\approx 3.6666
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is positive in that interval, so $f(x)$ is
increasing on $(0, \infty ).$
So that by the first derivative test, $0$ leads to a relative minimum.
