Question

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

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

identify the coordinates of the local maximum points. select the correct choice below and, if necessary, fill in the answer box to complete your choice.

a. the local maximum 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 local maximum points.

Explanation:

Step1: Find the first derivative

Use the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^2}\). Here \(u = 8x\), \(u^\prime=8\), \(v=x^{2}+9\), \(v^\prime = 2x\).

$$
LATEXBLOCK0
$$

Step2: Find the critical points

Set \(y^\prime = 0\), so \(\frac{72 - 8x^{2}}{(x^{2}+9)^{2}}=0\). Since \((x^{2}+9)^{2}>0\) for all \(x\in R\), then \(72 - 8x^{2}=0\).

$$
LATEXBLOCK1
$$

Step3: Use the second - derivative test

Find the second derivative. First, rewrite \(y^\prime=\frac{72 - 8x^{2}}{(x^{2}+9)^{2}}=(72 - 8x^{2})(x^{2}+9)^{-2}\)
Using the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = 72 - 8x^{2}\), \(u^\prime=-16x\), \(v=(x^{2}+9)^{-2}\), \(v^\prime=-2(x^{2}+9)^{-3}(2x)\)

$$
LATEXBLOCK2
$$

Evaluate \(y^{\prime\prime}\) at \(x = 3\):

$$ y^{\prime\prime}(3)=\frac{16\times3\times(9 - 27)}{(9 + 9)^{3}}=\frac{48\times(-18)}{(18)^{3}}<0 $$

Evaluate \(y^{\prime\prime}\) at \(x=-3\):

$$ y^{\prime\prime}(-3)=\frac{16\times(-3)\times(9 - 27)}{(9 + 9)^{3}}=\frac{-48\times(-18)}{(18)^{3}}>0 $$

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

Answer:

A. The local maximum point(s) is/are \((3,\frac{4}{3})\)