Answer
Concave upward on
$$
(-8, \infty ) ,
$$
concave downward on
$$
(-\infty, -8 )
$$
inflection point at
$$(-8, 768).$$
Work Step by Step
$$
f(x)=x^{\frac{7}{3}}+56 x^{\frac{4}{3}}
$$
The first derivative is
$$
\begin{aligned}
f^{\prime}(x) &=(\frac{7}{3})x^{\frac{4}{3}}+56(\frac{4}{3}) x^{\frac{1}{3}}\\
&=\frac{7}{3}x^{\frac{4}{3}}+\frac{224}{3} x^{\frac{1}{3}} ,
\end{aligned}
$$
and the second derivative is
$$
\begin{aligned}
f^{\prime\prime}(x) &=\frac{7}{3} (\frac{4}{3})x^{\frac{1}{3}}+\frac{224}{3}(\frac{1}{3}) x^{\frac{-2}{3}} \\
&=\frac{28}{9} x^{\frac{1}{3}}+\frac{224}{9}x^{\frac{-2}{3}} \\
\end{aligned}
$$
We factor $f^{\prime\prime}(x)$ as
$$
\begin{aligned}
f^{\prime\prime}(x) &=\frac{28}{9} x^{\frac{1}{3}}+\frac{224}{9}x^{\frac{-2}{3}}\\
&=\frac{28(x+8)}{9 x^{\frac{2}{3}}}
\end{aligned}
$$
look for an $x$-value that makes $f^{\prime\prime}(x)=0$.
Set $f^{\prime\prime}(x)$ equal to $0$ and solve for $x$.
$$
\begin{aligned}
f^{\prime\prime}(x) =\frac{28(x+8)}{9 x^{\frac{2}{3}}}=0 \quad \text {when } \quad x=-8
\end{aligned}
$$
but also where $f^{\prime\prime}(x)$ does not exist this happens
at $x=0$. So Check the sign of $f^{\prime\prime}(x)$ in the three intervals
$$
(-\infty, -8 ), \quad (-8,0), \quad (0,\infty).
$$
(1)
Test a number in the interval $(-\infty, -8 )$ say $ -27$
$$
\begin{aligned}
f^{\prime\prime}(-27) &=\frac{28((-27)+8)}{9 (-27)^{\frac{2}{3}}} \\
&=\frac{28((-19)}{9 (9)} \\
&=\frac{-532}{81} \\
& \lt 0
\end{aligned}
$$
to see that $ f^{\prime\prime}(x)$ is negative in that interval, so $f(x)$ is concave downward on $(-\infty, -8 )$
(2)
Test a number in the interval $( -8,0) $ say $ -1$
$$
\begin{aligned}
f^{\prime\prime}(-1) &=\frac{28((-1)+8)}{9 (-1)^{\frac{2}{3}}} \\
&=\frac{28((7)}{9 (1)} \\
&=\frac{196}{9} \\
& \gt 0
\end{aligned}
$$
to see that $ f^{\prime\prime}(x)$ is positive there, so $f(x)$ is concave upward on $ (-8, 0 )$.
(3)
Test a number in the interval $ (0, \infty )$ say $ 1$
$$
\begin{aligned}
f^{\prime\prime}(1) &=\frac{28((1)+8)}{9 (1)^{\frac{2}{3}}} \\
&=\frac{28((9)}{9 (1)} \\
&=28 \\
& \gt 0
\end{aligned}
$$
to see that $ f^{\prime\prime}(x)$ is positive there, so $f(x)$ is concave upward on $ (0, \infty )$.
Therefore, the function $f$ is
concave upward on
$$
(-8, \infty )
$$
and concave downward on
$$
( -\infty,-8 ).
$$
Since the sign of $f^{\prime\prime}(x)$ changes from negative
to positive at $ x=-8 $, the graph changes from concave downward to concave upward at that point, so $(-8 , f(-8))$ is inflection point.
Now ,we find $f( -8)$as follows:
$$
f(-8)=(-8)^{\frac{7}{3}}+56 (-8)^{\frac{4}{3}}=-128+896=768
$$
Finally, we have inflection point at $(-8, 768).$
