Answer
$$
h(x)=\frac{x^{2}-1}{2x-1}
$$
the graph is concave upward:
$$
(-\infty \lt x\lt \frac{1}{2})
$$
the graph is concave downward :
$$
(\frac{1}{2} \lt x \lt \infty) .
$$
Work Step by Step
$$
h(x)=\frac{x^{2}-1}{2x-1}
$$
Differentiating twice produces the following.
$$
h^{\prime }(x)=\frac{2(x^{2}-x+1)}{(2x-1)^{2}}
$$
$$
h^{\prime \prime }(x)=\frac{-6}{(2x-1)^{3}}
$$
There are no points at which $f^{\prime \prime }(x)=0$, but at $x=\frac{1}{2}$ $f$ is not continuous. So, test for concavity in the intervals $(-\infty , \frac{1}{2}) , (\frac{1}{2} , \infty)$ . The results are shown in the following table:
$$
\begin{array}{|c|c|c|c|c|}\hline \text { Intervals: } & {-\infty0} & {h^{\prime \prime}<0} \\ \hline \text { Conclusion: } & { \text { Concave upward }} & {\text { Concave downward }} \\ \hline\end{array}
$$
So, Concave upward:
$$
(-\infty \lt x\lt \frac{1}{2})
$$
Concave downward:
$$
(\frac{1}{2} \lt x \lt \infty) .
$$
