Question

How do you determine the intervals for which the function is increasing or decreasing given `f(x)=x/(x^2+1)`?

Answer 1

The limits when `f(x)` is increasing is `x in [-1, 1]`
The limits when `f(x)` is decreasing is `x in ] -oo,-1 [ uu ] 1, oo[`

Explanation:

We must calculate the derivative `f'(x)=(u'v-uv')/v^2`

`u=x` `=>` `u'=1`
`v=x^2+1` `=>` `v'=2x`

`:. 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`

`f'(x)=0` when `x=1` and `x=-1`

We do a sign chart for `f'(x)`

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

The limits when `f(x)` is increasing is `x in [-1, 1]`
The limits when `f(x)` is decreasing is `x in ] -oo,-1 [ uu ] 1, oo[`
graph{x/(x^2+1) [-3.077, 3.08, -1.538, 1.54]}