Answer
$$
g(x)=\frac{x^{2}+4}{4-x^{2}}
$$
the graph is concave upward:
$$
(-2 \lt x\lt 2)
$$
the graph is concave downward :
$$
(-\infty \lt x \lt -2), (2 \lt x \lt \infty) .
$$
Work Step by Step
$$
g(x)=\frac{x^{2}+4}{4-x^{2}}
$$
Differentiating twice produces the following.
$$
g^{\prime }(x)=\frac{16x}{(4-x^{2})^{2}}
$$
and
$$
g^{\prime \prime}(x)=\frac{16\left(3 x^{2}+4\right)}{\left(4-x^{2}\right)^{3}}=\frac{16\left(3 x^{2}+4\right)}{(2-x)^{3}(2+x)^{3}}
$$
There are no points at which $f^{\prime \prime }(x)=0$, but at $x=\pm 2$ $f$ is not continuous. So, test for concavity in the intervals $(-\infty , -2), (-2 , 2)$ and $(2 , \infty)$. The results are shown in the following table:
$$
\begin{array}{|c|c|c|c|c|c|}\hline {-\infty\lt x\lt -2} & {-2\lt x\lt 2} & {2\lt x\lt \infty} \\ \hline {g^{\prime \prime} \lt 0} & {g^{\prime \prime}\gt 0} & {g^{\prime \prime} \lt 0} \\ \hline {\text { Concave downward }} & {\text { Concave upward }} & {\text { Concave downward }} \\ \hline\end{array}
$$
So, Concave upward:
$$
(-2 \lt x\lt 2)
$$
Concave downward:
$$
(-\infty \lt x \lt -2), (2 \lt x \lt \infty) .
$$
