Question

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

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

a. the local 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 local minimum points.

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

a. the absolute 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 absolute maximum points

Explanation:

Step1: Find the first derivative

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

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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 \(u = 72 - 8x^{2}\), \(u^\prime=-16x\), \(v=(x^{2}+9)^{2}\), \(v^\prime = 2(x^{2}+9)\times(2x)=4x(x^{2}+9)\)

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When \(x = 3\), \(y^{\prime\prime}=\frac{16\times3\times(9 - 27)}{(9 + 9)^{3}}=\frac{48\times(-18)}{18^{3}}<0\), so \(x = 3\) is a local maximum point. When \(x=-3\), \(y^{\prime\prime}=\frac{16\times(-3)\times(9 - 27)}{(9 + 9)^{3}}=\frac{-48\times(-18)}{18^{3}}>0\), so \(x=-3\) is a local minimum point.

Step3: Analyze the behavior as \(x\to\pm\infty\)

\(\lim_{x\to\pm\infty}y=\lim_{x\to\pm\infty}\frac{8x}{x^{2}+9}=\lim_{x\to\pm\infty}\frac{\frac{8x}{x^{2}}}{\frac{x^{2}}{x^{2}}+\frac{9}{x^{2}}}=0\)

Since \(\lim_{x\to\pm\infty}y = 0\) and \(y(3)=\frac{8\times3}{9 + 9}=\frac{4}{3}\), \(y(-3)=\frac{8\times(-3)}{9 + 9}=-\frac{4}{3}\)

Answer:

The absolute maximum point is \((3,\frac{4}{3})\) and the absolute minimum point is \((-3,-\frac{4}{3})\). So for the absolute maximum points, the answer is A. The absolute maximum point(s) is/are \((3,\frac{4}{3})\)