identify the coordinates of any local and absolute extreme points and inflection points graph the…

identify the coordinates of any local and absolute extreme points and inflection points graph the function.\n\n$y = \\frac{8x}{x^{2}+9}$\n\na. the absolute maximum point(s) is/are $\\left(3,\\frac{4}{3}\\right)$\n\n(type an ordered pair. type an exact answer, using radicals as needed. use a comma to separate answers as needed.)\n\nb. there are no absolute maximum points\n\nidentify the coordinates of the absolute minimum points. select the correct choice below and, if necessary, fill in the answer box to complete your choice.\n\na. the absolute minimum point(s) is/are\n\n(type an ordered pair. type an exact answer, using radicals as needed. use a comma to separate answers as needed.)\n\nb. there are no absolute minimum points.

identify the coordinates of any local and absolute extreme points and inflection points graph the function.\n\n$y = \\frac{8x}{x^{2}+9}$\n\na. the absolute maximum point(s) is/are $\\left(3,\\frac{4}{3}\\right)$\n\n(type an ordered pair. type an exact answer, using radicals as needed. use a comma to separate answers as needed.)\n\nb. there are no absolute maximum points\n\nidentify the coordinates of the absolute minimum points. select the correct choice below and, if necessary, fill in the answer box to complete your choice.\n\na. the absolute minimum point(s) is/are\n\n(type an ordered pair. type an exact answer, using radicals as needed. use a comma to separate answers as needed.)\n\nb. there are no absolute minimum points.

Answer

Explanation:

Step1: Find the first derivative

Use the quotient rule ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}), where (u = 8x), (u^\prime=8), (v=x^{2}+9), (v^\prime = 2x). [ \begin{align*} y^\prime&=\frac{8(x^{2}+9)-8x\times(2x)}{(x^{2}+9)^{2}}\ &=\frac{8x^{2}+72 - 16x^{2}}{(x^{2}+9)^{2}}\ &=\frac{72 - 8x^{2}}{(x^{2}+9)^{2}} \end{align*} ] Set (y^\prime = 0), then (72-8x^{2}=0), (x^{2}=9), (x=\pm3).

Step2: Find the second derivative

Use the quotient rule again. Let (u = 72 - 8x^{2}), (u^\prime=-16x), (v=(x^{2}+9)^{2}), (v^\prime = 2(x^{2}+9)\times2x=4x(x^{2}+9)) [ \begin{align*} y^{\prime\prime}&=\frac{-16x(x^{2}+9)^{2}-(72 - 8x^{2})\times4x(x^{2}+9)}{(x^{2}+9)^{4}}\ &=\frac{-16x(x^{2}+9)-4x(72 - 8x^{2})}{(x^{2}+9)^{3}}\ &=\frac{-16x^{3}-144x-288x + 32x^{3}}{(x^{2}+9)^{3}}\ &=\frac{16x^{3}-432x}{(x^{2}+9)^{3}}\ &=\frac{16x(x^{2}-27)}{(x^{2}+9)^{3}} \end{align*} ] When (x = 3), (y^{\prime\prime}(3)=\frac{16\times3\times(9 - 27)}{(9 + 9)^{3}}<0), so (y(3)=\frac{8\times3}{9 + 9}=\frac{4}{3}) is a local maximum. When (x=-3), (y^{\prime\prime}(-3)=\frac{16\times(-3)\times(9 - 27)}{(9 + 9)^{3}}>0), so (y(-3)=\frac{8\times(-3)}{9 + 9}=-\frac{4}{3}) is a local minimum.

Since (\lim_{x\rightarrow\pm\infty}y=\lim_{x\rightarrow\pm\infty}\frac{8x}{x^{2}+9}=0), the local maximum (y(3)=\frac{4}{3}) is the absolute maximum and the local minimum (y(-3)=-\frac{4}{3}) is the absolute minimum.

Answer:

A. The absolute minimum point(s) is/are (\left(-3,-\frac{4}{3}\right))