Find the extreme values of the function and where they occur.
-$$y = \frac { 9 x } { x ^ { 2 } + 1 }$$
A) The minimum value is 0 at x = 0.
B) The minimum value is 0 at x = 1. The maximum value is 0 at x = -1.
C) The maximum value is 0 at x = 0.
D) The minimum value is - 92 at x = -1. The maximum value is 92 at x = 1.

Question

Quizplus · Accepted Answer

To find the extreme values of the function, we take the derivative:
$$y' = \frac{9(1-x^2)}{(x^2+1)^2}$$
Setting this equal to 0, we get $x = \pm 1$. We also need to check the endpoints of the domain, which are $x = \pm \infty$. As $x$ approaches infinity, $y$ approaches 0 from above, and as $x$ approaches negative infinity, $y$ approaches 0 from below.
Evaluating $y$ at the critical points and endpoints, we get:
\begin{align*}
&x = -\infty: y = 0 \
&x = -1: y = -\frac{9}{2} \
&x = 1: y = \frac{9}{2} \
&x = \infty: y = 0
\end{align*}
Therefore, the minimum value is $-\frac{9}{2}$ at $x=-1$, and the maximum value is $\frac{9}{2}$ at $x=1$. Hence, the best choice is $\boxed{	ext{D}}$.

Find the Extreme Values of the Function and Where They y=9xx2+1y = \frac { 9 x } { x ^ { 2 } + 1 }y=x2+19x​

Find the Extreme Values of the Function and Where They $y = \frac { 9 x } { x ^ { 2 } + 1 }$