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

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

Answer

Explanation:

Step1: Find the first - derivative

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

Step2: Use the second - derivative test

Find the second - derivative. Use the quotient rule again, (u = 72 - 8x^{2}), (u^\prime=-16x), (v=(x^{2}+9)^{2}), (v^\prime = 2(x^{2}+9)\times(2x)=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}}=\frac{48\times(-18)}{18^{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}}=\frac{-48\times(-18)}{18^{3}}>0), (y(-3)=\frac{8\times(-3)}{9 + 9}=-\frac{4}{3})

Answer:

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