#### Answer

$$
f(x)=\frac{3}{x-5}
$$
$f(x)$ concave upward on
$$(5,\infty )$$.
$f(x)$ concave downward on
$$(-\infty , 5).$$
No Inflection point .

#### Work Step by Step

$$
f(x)=\frac{3}{x-5}
$$
By the quotient rule,we can find the first derivative as follows
$$
\begin{aligned}
f^{\prime}(x) &=\frac{0-3(1)}{(x-5)^{2}}\\
&=\frac{-3}{(x-5)^{2}},
\end{aligned}
$$
and the second derivative is
$$
\begin{aligned}
f^{\prime\prime}(x) &=\frac{0-3(-2)(x-5)(1)}{(x-5)^{4}}\\
&=\frac{6}{(x-5)^{3}}.
\end{aligned}
$$
A function $f$ is concave upward on an interval if $f^{\prime\prime}(x) \gt 0$ for all $x$ in the interval, so
$$
\begin{aligned}
f^{\prime\prime}(x) & =\frac{6}{(x-5)^{3}} \gt 0 .
\end{aligned}
$$
when
$$
\begin{aligned}
(x-5)^{3} & \gt 0 \\
(x-5) & \gt 0 \\
x & \gt 5
\end{aligned}
$$
so $f(x)$ is concave upward on $(5,\infty )$.
Also a function $f$ is concave downward on an interval if
$f^{\prime\prime}(x) \lt 0$ for all $x$ in the interval, so
$$
\begin{aligned}
f^{\prime\prime}(x) & =\frac{6}{(x-5)^{3}} \lt 0 .
\end{aligned}
$$
when
$$
\begin{aligned}
(x-5)^{3} &\lt 0 \\
(x-5) &\lt 0 \\
x &\lt 5
\end{aligned}
$$
so $f(x)$ is concave downward on $(-\infty , 5)$.
Since $f^{\prime\prime}(x)\ne 0 $ for all $x$ and the sign of $f^{\prime\prime}(x)$ changes from positive to negative at $5$, the graph changes from concave upward to concave downward at that point, but $ f(5))$ does not exist. .
So , no inflection point