Question

What is the domain and range of `f(x) = (x-1)/(x^2+1)`?

Answer 1

`"Domain":x inRR`
`"Range":f(x)in[-(sqrt(2)+1)/2,(sqrt(2)-1)/2]`

Explanation:

Considering that all real values of `x` will give a non-zero value for `x^2+1`, we can say that for `f(x)`, domain = `x inRR`

For range, we need the maximum and minimum.
`f(x)=(x-1)/(x^2+1)`

`f'(x)=((x^2+1)-2x(x-1))/(x^2+1)^2=(x^2+1-2x^2+2x)/(x^2+1)=(-x^2+2x+1)/(x^2+1)`

The maximum and minimum values occur when `f'(x)=0`

`x^2-2x-1=0`

`x=(2+-sqrt((-2)^2-4(-1)))/2`

`x=(2+-sqrt8)/2=(2+-2sqrt(2))/2=1+-sqrt2`

Now, we input our `x` values into `f(x)`:
`(1+sqrt(2)-1)/((1+sqrt(2))^2+1)=(sqrt(2)-1)/2`
`(1-sqrt(2)-1)/((1-sqrt(2))^2+1)=-(sqrt(2)+1)/2`

`f(x)in[-(sqrt(2)+1)/2,(sqrt(2)-1)/2]`