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) The maximum is $\frac { 1 } { 2 }$ at $x = 0$; the minimum is $- \frac { 1 } { 2 }$ at $x = - 2$.
C) None
D) The maximum is 2 at $x = 0$; the minimum is $\frac { 1 } { 2 }$ at $x = - 2$.

Question

Quizplus · Accepted Answer

To find the extreme values, we first find the critical points by setting the first derivative equal to 0:
$$y' = \frac{(x^2+2x+2)- (x+1)(2x+2)}{(x^2+2x+2)^2}=0$$
Simplifying gives $x^2+x=0$, so $x = 0$ or $x = -1$. To determine whether these are maximum or minimum, we can use the second derivative test:
$$y'' = \frac{-6x-2}{(x^2+2x+2)^2}$$
At $x=-2$, $y''<0$, so there is a local maximum at $x=-2$. At $x=0$, $y''>0$, so there is a local minimum at $x=0$. Finally, since the denominator of the function is always positive, the maximum value of the function occurs at these critical points. Plugging in $x=-2$ gives $y=\frac{1}{2}$, and plugging in $x=0$ gives $y=-\frac{1}{2}$. Therefore, the maximum value is $\frac{1}{2}$ at $x= -2$, and the minimum value is $-\frac{1}{2}$ at $x= 0$. The correct answer choice is $\boxed{\textbf{(B)}}$.

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 $y = \frac { x + 1 } { x ^ { 2 } + 2 x + 2 }$