Find the extreme values of the function and where they occur.
-$$y = \frac { x + 1 } { x ^ { 2 } + 2 x + 2 }$$
A) The maximum is $- \frac { 1 } { 2 }$ at $x = 0$; the minimum is $\frac { 1 } { 2 }$ at $x = - 2$.
B) None
C) The maximum is 2 at $x = 0$; the minimum is $\frac { 1 } { 2 }$ at $x = - 2$.
D) The maximum is $\frac { 1 } { 2 }$ at $x = 0$; the minimum is $- \frac { 1 } { 2 }$ at $x = - 2$.

Question

Quizplus · Accepted Answer

To find the extreme values, we first take the derivative of the function and set it equal to zero:
\begin{align*}
y' &= \frac{(x^2+2x+2)-(x+1)(2x+2)}{(x^2+2x+2)^2} \
&= \frac{-x^2}{(x^2+2x+2)^2} \
\end{align*}
Setting $y'=0$, we get $x=0$. To determine whether this is a maximum or minimum, we can use the second derivative test:
\begin{align*}
y'' &= \frac{(-x^2)'(x^2+2x+2)^2 - (-x^2)(2(x^2+2x+2)(2x+2))}{(x^2+2x+2)^4} \
&= \frac{-2x(x^2+1)}{(x^2+2x+2)^3}
\end{align*}
$y''(0) = 0$, which means the second derivative test is inconclusive. Instead, we can look at the behavior of the function as $x$ approaches $\pm\infty$:
\begin{align*}
\lim_{x	o\pm\infty}y &= \lim_{x	o\pm\infty}\frac{x+1}{x^2+2x+2} \
&= \lim_{x	o\pm\infty}\frac{1/x+1/x^2}{1+2/x+2/x^2} \
&= 0
\end{align*}
This means that there are no vertical asymptotes and the function approaches zero as $x$ goes to $\pm\infty$. Therefore, the function has a global maximum at $x=0$ and a global minimum at $x=-2$. Plugging these values back into the function, we get $y(0) = \frac{1}{2}$ and $y(-2) = -\frac{1}{2}$. So the answer is (D).

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

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