Answer
$$
f(x)=\frac{-2}{x+1}
$$
$f(x)$ concave upward on
$$(-\infty,-1 ) $$.
$f(x)$ concave downward on
$$(-1 , \infty).$$
No Inflection point .
Work Step by Step
$$
f(x)=\frac{-2}{x+1}
$$
By the quotient rule,we can find the first derivative as follows
$$
\begin{aligned}
f^{\prime}(x) &=\frac{0+2(1)}{(x+1)^{2}}\\
&=\frac{2}{(x+1)^{2}},
\end{aligned}
$$
and the second derivative is
$$
\begin{aligned}
f^{\prime\prime}(x) &=\frac{0-2(2)(x+1)(1)}{(x+1)^{4}}\\
&=\frac{-4}{(x+1)^{3}}.
\end{aligned}
$$
A function $f$ is concave upward on an interval if $f^{\prime\prime}(x) \gt 0$ for all $x$ in the interval, so
$$
\begin{aligned}
f^{\prime\prime}(x) & =\frac{-4}{(x+1)^{3}} \gt 0 .
\end{aligned}
$$
when
$$
\begin{aligned}
(x+1)^{3} &\lt 0 \\
(x+1) & \lt 0 \\
x & \lt -1
\end{aligned}
$$
so $f(x)$ is concave upward on $(-\infty,-1 )$.
Also function $f$ is concave upward on an interval if $f^{\prime\prime}(x) \lt 0$ for all $x$ in the interval, so
$$
\begin{aligned}
f^{\prime\prime}(x) & =\frac{-4}{(x+1)^{3}} \lt 0 .
\end{aligned}
$$
when
$$
\begin{aligned}
(x+1)^{3} & \gt 0 \\
(x+1) & \gt 0 \\
x &\gt -1
\end{aligned}
$$
so $f(x)$ is concave downward on $(-1 , \infty)$.
Since $f^{\prime\prime}(x)\ne 0 $ for all $x$, $f^{\prime\prime}(x) $ does not exist at $x=-1$ and the sign of $f^{\prime\prime}(x)$ changes from positive to negative, the graph changes from concave upward to concave downward at that point, but $ f(-1))$ does not exist. .
So , no inflection point
