Graph the equation. Include the coordinates of any local and absolute extreme points and inflection points.
-y$= \frac { 16 x } { x ^ { 2 } + 16 }$

A) local minimum: $ (-4,-1) $ local maximum: $ (4,1) $ inflection point: $ (0,0) $
B) local minimum: $ (4,-2) $ local maximum: $ (-4,2) $ inflection point: $ (0,0) $
C) local minimum: $ (-4,-2) $ local maximum: $ (4,2) $ inflection points: $ (0,0),(-4 \sqrt{3},-4 \sqrt{3}) $, $ (4 \sqrt{3}, 4 \sqrt{3}) $
D) absolute maximum: $ (0,1) $ no inflection point

Question

Graph the equation. Include the coordinates of any local and absolute extreme points and inflection points.
-y$= \frac { 16 x } { x ^ { 2 } + 16 }$

A) local minimum: $ (-4,-1) $ local maximum: $ (4,1) $ inflection point: $ (0,0) $ 
B) local minimum: $ (4,-2) $ local maximum: $ (-4,2) $ inflection point: $ (0,0) $ 
C) local minimum: $ (-4,-2) $ local maximum: $ (4,2) $ inflection points: $ (0,0),(-4 \sqrt{3},-4 \sqrt{3}) $, $ (4 \sqrt{3}, 4 \sqrt{3}) $ 
D) absolute maximum: $ (0,1) $ no inflection point

Quizplus · Accepted Answer

To find the local minimum and maximum points, we need to take the first derivative of the equation and set it to zero. Then we solve for x.

$$\frac{d}{dx}\left(\frac{16x}{x^2+16}\right)=\frac{-16x^2+256}{(x^2+16)^2}=0$$

Solving this, we get $x=\pm 4$. These are the x-coordinates of the local maximum and minimum points. To find the y-coordinates, we need to plug these values into the original equation.

$$y=\frac{16\cdot 4}{4^2+16}=2$$
$$y=\frac{16\cdot (-4)}{(-4)^2+16}=-2$$

Therefore, the local maximum point is $(4,2)$ and the local minimum point is $(-4,-2)$.

To find the inflection points, we need to take the second derivative of the equation and set it to zero. Then we solve for x.

$$\frac{d^2}{dx^2}\left(\frac{16x}{x^2+16}\right)=\frac{-64x(x^2-24)}{(x^2+16)^3}=0$$

Solving this, we get $x=0$ and $x=\pm 4\sqrt{3}$. These are the x-coordinates of the inflection points. To find the y-coordinates, we need to plug them into the original equation.

$$y=\frac{16\cdot 0}{0^2+16}=0$$
$$y=\frac{16\cdot (-4\sqrt{3})}{(-4\sqrt{3})^2+16}=-4\sqrt{3}$$
$$y=\frac{16\cdot (4\sqrt{3})}{(4\sqrt{3})^2+16}=4\sqrt{3}$$

Therefore, the inflection points are $(0,0)$, $(-4\sqrt{3},-4\sqrt{3})$, and $(4\sqrt{3},4\sqrt{3})$.

Choice A is incorrect because it only gives one of the local extreme points and it is incorrect. Choice B is also incorrect as it gives the local extreme points in reverse order. Choice D is incorrect because it does not consider the inflection points. Choice C is the correct choice as it gives all the correct local extreme points and inflection points.

Graph the Equation =16xx2+16= \frac { 16 x } { x ^ { 2 } + 16 }=x2+1616x​

Graph the Equation $= \frac { 16 x } { x ^ { 2 } + 16 }$