## Calculus with Applications (10th Edition)

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