Answer
Function $y$, at $x=-2$, has the absolute and local minimum value of $-1/2$. It also has the absolute and local maximum value of $1/2$ at $x=0$.
Work Step by Step
$$y=\frac{x+1}{x^2+2x+2}$$
We have $$x^2+2x+2=(x^2+2x+1)+1=(x+1)^2+1\gt0$$ for all $x\in R$
Therefore, function $y$ is defined on $R$.
1) Find all the critical points:
Find the derivative of the function: $$y'=\frac{(x+1)'(x^2+2x+2)-(x+1)(x^2+2x+2)'}{(x^2+2x+2)^2}$$ $$y'=\frac{(x^2+2x+2)-(x+1)(2x+2)}{(x^2+2x+2)^2}$$ $$y'=\frac{x^2+2x+2-(2x^2+4x+2)}{(x^2+2x+2)^2}$$ $$y'=\frac{-x^2-2x}{(x^2+2x+2)^2}=-\frac{x(x+2)}{(x^2+2x+2)^2}$$
- For $y'=0$, either $x=0$ or $x=-2$.
- There is no value of $x$ for which $y'$ is undefined, since it has been shown that $(x^2+2x+2)\ne0$ for all $x\in R$.
So $x=-2$ and $x=0$ are critical points of function $y$ here.
2) Evaluate function $y$ at the critical points:
- For $x=-2$: $$y=\frac{-2+1}{(-2)^2+2(-2)+2}=\frac{-1}{4-4+2}=-\frac{1}{2}$$
- For $x=0$: $$y=\frac{0+1}{0^2+2\times0+2}=\frac{1}{2}$$
Now check the graph of function $y$.
We see that function $y$, at $x=-2$, has the absolute and local minimum value of $-1/2$. It also has the absolute and local maximum value of $1/2$ at $x=0$.
