sketch the graph of the following function. indicate where the function is increasing or decreasing, where…

sketch the graph of the following function. indicate where the function is increasing or decreasing, where any relative extrema occur, where asymptotes occur, where the graph is concave up or concave down, where any points of inflection occur, and where any intercepts occur. f(x) = (8x - 3)/x (type an equation. use a comma to separate answers as needed.) b. the function has no slant asymptotes. on what interval(s) is f concave up and on what interval(s) is f concave down? select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. a. the function is concave down on and is never concave up. (simplify your answer. type your answer in interval notation. use a comma to separate answers as needed.) b. the function is concave up on and is never concave down. (simplify your answer. type your answer in interval notation. use a comma to separate answers as needed.) c. the function is concave up on and concave down on. (simplify your answers. type your answers in interval notation. use a comma to separate answers as needed.)
Answer
Explanation:
Step1: Rewrite the function
Rewrite $f(x)=\frac{8x - 3}{x}=8-\frac{3}{x}$.
Step2: Find the first - derivative
Using the power rule, if $y = 8-3x^{-1}$, then $f^\prime(x)=3x^{-2}=\frac{3}{x^{2}}$. Since $f^\prime(x)=\frac{3}{x^{2}}>0$ for all $x\neq0$, the function is increasing on $(-\infty,0)\cup(0,\infty)$. There are no relative extrema as $f^\prime(x)$ never equals 0.
Step3: Find the second - derivative
Differentiate $f^\prime(x)=\frac{3}{x^{2}} = 3x^{-2}$ using the power rule. $f^{\prime\prime}(x)=-6x^{-3}=-\frac{6}{x^{3}}$.
Step4: Determine concavity
Set $f^{\prime\prime}(x)=0$, but $-\frac{6}{x^{3}} = 0$ has no solution. - When $x<0$, $f^{\prime\prime}(x)>0$, so the function is concave up on $(-\infty,0)$. - When $x>0$, $f^{\prime\prime}(x)<0$, so the function is concave down on $(0,\infty)$.
Answer:
C. The function is concave up on $(-\infty,0)$ and concave down on $(0,\infty)$