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