Answer
$$
f(x)=\frac{x^{2}+1}{x^{2}-1}
$$
the graph is concave upward:
$$
(-\infty \lt x\lt-1), (1 \lt x\lt \infty)
$$
the graph is concave downward :
$$
(-1 \lt x \lt 1) .
$$
Work Step by Step
$$
f(x)=\frac{x^{2}+1}{x^{2}-1}
$$
Differentiating twice produces the following.
$$
f^{\prime }(x)=\frac{-4x}{(x^{2}-1)^{2}}
$$
$$
f^{\prime \prime }(x)=\frac{4(3x^{2}+1)}{(x^{2}-1)^{3}}
$$
There are no points at which $f^{\prime \prime }(x)=0$, but at $x=\pm 1$ $f$ is not continuous. So, test for concavity in the intervals $(-\infty , -1), (-1 , 1)$ and $(1 , \infty)$. The results are shown in the following table:
$$
\begin{array}{|c|c|c|c|c|}\hline \text { Intervals: } & {-\infty0} & {f^{\prime \prime}<0} & {f^{\prime \prime}>0} \\ \hline \text { Conclusion: } & {\text { Concave upward }} & {\text { Concave downward }} & {\text { Concave upward }} \\ \hline\end{array}
$$
So, Concave upward:
$$
(-\infty \lt x\lt-1), (1 \lt x\lt \infty).
$$
Concave downward:
$$
(-1 \lt x \lt 1) .
$$
