Question

Let `f(x) = x / (x^2+1)`, how do you use the first derivative test to determine which critical numbers, if any give relative extrema?

Answer 1

The critical numbers are when `x=-1` and `x=1`
there is a minimum at `(-1,-1/2)`
and a maximum at `(1,1/2)`

Explanation:

The derivative of a quotient is `(u/v)'=(u'v-uv')/v^2`

`f(x)=x/(x^2+1)`
so `f'(x)=(1(x^2+1)-(x*2x))/(x^2+1)^2=(x^2+1-2x^2)/(x^2+1)^2`
`=(1-x^2)/(x^2+1)^2=((1+x)(1-x))/(x^2+1)^2`
The denominator is always `>0`
so
`f'(x)=0` when `x=-1` and when `x=1`
Let's do a sign chart

`x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-1` `color(white)(aaaa)` `1` `color(white)(aaaa)` `+oo`
`f'(x)` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `-`
`f(x)` `color(white)(aaaaaa)` `darr` `color(white)(aaaa)` `uarr` `color(white)(aaaa)` `darr`

so, there is a minimum at `x=-1` and a maximum at `x=1`
Also, the limit of `f(x)` as `x->+-oo` is
limit `f(x)=0^-`
`x->-oo`

limit `f(x)=0^+`
`x->+oo`

Also, `f(-x)=-f(x)`, there is a symmetry about the Origin

graph{x/(x^2+1) [-2.222, 2.222, -1.111, 1.11]}