Question

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

$y = \frac{8x}{x^{2}+9}$

a. the absolute minimum point(s) is/are $\left(-3,-\frac{4}{3}\
ight)$
(type an ordered pair. type an exact answer, using radicals as needed. use a comma to separate answers as needed.)
b. there are no absolute minimum points.

find the inflection points. select the correct choice below and, if necessary, fill in the answer box to complete your choice.

a. the inflection point(s) is/are
(type an ordered pair. type an exact answer, using radicals as needed. use a comma to separate answers as needed.)
b. there are no inflection points.

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 \(u = 8x\), \(u^\prime=8\), \(v=x^{2}+9\), \(v^\prime = 2x\).

$$
LATEXBLOCK0
$$

Set \(y^\prime = 0\) for critical points:

$$
LATEXBLOCK1
$$

When \(x = 3\), \(y=\frac{8\times3}{3^{2}+9}=\frac{24}{18}=\frac{4}{3}\); when \(x=-3\), \(y=\frac{8\times(-3)}{(-3)^{2}+9}=-\frac{4}{3}\)

Step2: Find the second - derivative

Use the quotient rule again. Let \(u = 72-8x^{2}\), \(u^\prime=-16x\), \(v=(x^{2}+9)^{2}\), \(v^\prime = 2(x^{2}+9)\times(2x)=4x(x^{2}+9)\)

$$
LATEXBLOCK2
$$

Set \(y^{\prime\prime}=0\)

$$
LATEXBLOCK3
$$

When \(x = 0\), \(y = 0\); when \(x=3\sqrt{3}\), \(y=\frac{8\times3\sqrt{3}}{(3\sqrt{3})^{2}+9}=\frac{24\sqrt{3}}{27 + 9}=\frac{\sqrt{3}}{2}\); when \(x=-3\sqrt{3}\), \(y=\frac{8\times(-3\sqrt{3})}{(-3\sqrt{3})^{2}+9}=-\frac{\sqrt{3}}{2}\)

Answer:

The inflection points are \((-3\sqrt{3},-\frac{\sqrt{3}}{2})\), \((0,0)\), \((3\sqrt{3},\frac{\sqrt{3}}{2})\)