Answer
$$
f(x)=\frac{1}{270}(-3x^{5}+40x^{3}+135x)
$$
the graph is concave upward:
$$
(-\infty \lt x\lt-2), (0 \lt x\lt 2)
$$
the graph is concave downward :
$$
(-2 \lt x \lt 0), (2 \lt x \lt \infty) .
$$
Work Step by Step
$$
f(x)=\frac{1}{270}(-3x^{5}+40x^{3}+135x)
$$
Begin by observing that $f$ is continuous on the entire real number line.
Next, find the second derivative of $f$
$$
f^{\prime }(x)=\frac{1}{270}(-15x^{4}+120x^{2}+135)
$$
$$
f^{\prime \prime }(x)=\frac{1}{270}(-60x^{3}+240x)=\frac{-2x}{9}(x-2)(x+2).
$$
Because $f^{\prime \prime }(x)=0$ when $x=0 ,\pm 2$ and $f^{\prime \prime }(x)$ is defined on the entire real number line, we should test $f^{\prime \prime }(x)$ in the intervals $(-\infty , -2), (-2, 0), (0, 2)$and $(2, \infty )$The results are shown in the following table:
$$
\begin{array}{|c|c|c|c|c|c|}\hline {-\infty\lt x\lt -2} & {-2\lt x\lt 0} & {0\lt x\lt 2} & {2\lt x\lt \infty} \\ \hline {y^{\prime \prime}\gt 0} & {y^{\prime \prime}\lt 0} & {y^{\prime \prime}\gt 0} & {y^{\prime \prime}\lt 0} \\ \hline {\text { Concave upward }} & {\text { Concave downward }} & {\text { Concave upward }} & {\text { Concave downward }} \\ \hline\end{array}
$$
So, Concave upward:
$$
(-\infty \lt x\lt-2), (0 \lt x\lt 2)
$$
Concave downward:
$$
(-2 \lt x \lt 0), (2 \lt x \lt \infty) .
$$
