Answer
Function $y$, at $x=-1$, has the absolute and local minimum value of $-1/2$. It also has the absolute and local maximum value of $1/2$ at $x=1$.
Work Step by Step
$$y=\frac{x}{x^2+1}$$
- Domain: $x\in R$
1) Find all the critical points:
Find the derivative of the function: $$y'=\frac{(x)'(x^2+1)-x(x^2+1)'}{(x^2+1)^2}=\frac{(x^2+1)-x(2x)}{(x^2+1)^2}$$ $$y'=\frac{x^2+1-2x^2}{(x^2+1)^2}=\frac{1-x^2}{(x^2+1)^2}$$
- For $y'=0$, $(1-x^2)=0$ or $x=\pm1$.
- There is no value of $x$ for which $y'$ is undefined.
All these points $x=\{-1,1\}$ are critical points of function $y$ here.
2) Evaluate function $y$ at the critical points:
- For $x=-1$: $$y=\frac{-1}{(-1)^2+1}=-\frac{1}{1+1}=-\frac{1}{2}$$
- For $x=1$: $$y=\frac{1}{1^2+1}=\frac{1}{2}$$
Now check the graph of function $y$.
We see that function $y$, at $x=-1$, has the absolute and local minimum value of $-1/2$. It also has the absolute and local maximum value of $1/2$ at $x=1$.
