Question

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

Answer 1

Domain is `RR` and range is`[1/3,3]`

Explanation:

As `x^2+x+1=(x^2+2xx x xx1/2+1/4)+3/4`

= `(x+1/2)^2+3/4`

the denominator `x^2+x+1>0,AAx inRR`

Hence, domain is `RR`

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

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

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

Hence at `x=0`, `f(x)=1` and also as `x->+-oo`, `f(x)->1`

Further `(df)/(dx)=0`, when

`(x^2+x+1)d/(dx)(x^2-x+1)-(x^2-x+1)d/(dx)(x^2+x+1)=0`

or `(x^2+x+1)(2x-1)-(x^2-x+1)(2x+1)=0`

or `2x^2-2=0` i.e. `x=+-1`

annd at `x=1`, `f(x)=1/3` and at `x=-1`, `f(x)=3`

Hence range is `[1/3,3]`

graph{(x^2+x+1-2x)/(x^2+x+1) [-5, 5, -0.5, 4.5]}

Answer 2

The domain is `x in RR`
The range is `y in [1/3,3]`

Explanation:

The function is `f(x)=(x^2-x+1)/(x^2+x+1)`

The discriminant of the denominator is

`Delta=1^2-4*1*1=-3`

As the discriminant is `>0`, `x^2+x+1>0`

Therefore,

The domain is `x in RR`

Proceed as follows to determine the range

Let `y=(x^2-x+1)/(x^2+x+1)`

`y(x^2+x+1)=x^2-x+1`

`x^2(y-1)-x(y+1)+(y-1)=0`

This is a quadratic equation in `x^2`, and in order for this equation to have solutions, the discriminant must be `>=0`

`Delta=(-(y+1))^2-4(y-1)(y-1)>=0`

`y^2+2y+1-4y^2+8y-4>=0`

`-3y^2+10y-3>=0`

`y_1=(-10+sqrt(100-4*(-3)(-3)))/(-6)=(-10+8)/(-6)=1/3`

`y_2=(-10-sqrt(100-4*(-3)(-3)))/(-6)=(-10-8)/(-6)=3`

As the coefficient of `y^2` is `>0`, the range is

`y in [1/3,3]`

graph{(x^2-x+1)/(x^2+x+1) [-6.474, 6.013, -2.12, 4.12]}