Question

How do you find the domain and range of `f(x)= (3x-1)/(sqrt(x^2+x-2))`?

Answer 1

Domain : `x : (-oo,-2) uu (1,oo)`
Range : `f(x) : (-3, -oo) uu (3,oo)`

Explanation:

`f(x)= (3x-1)/sqrt(x^2+x-2) = (3x-1)/sqrt((x+2)(x-1))`

Domain: denominator must not be `0` and under root must not be `<0` So `x+2!=0 :. x != -2` and `x-1!=0 :. x != 1`

For under root calculation: critical points are `x=-2 and x= 1`
when `x <-2 , (x+2)* (x-1) = (-*- )= +`

when `-2 < x < 1 , (x+2)* (x-1) = (+*- ) = -`

when `x >1 , (x+2)* (x-1) = (=*+ )= +`

So in domain : `x < -2 and x> 1 or (-oo,-2) uu (1,oo)`

Horizontal asymptote is at `y= 3/(+-1)=+-3`

So range : `f(x) : (-3, -oo) uu (3,oo)` graph{(3x-1)/(x^2+x-2)^0.5 [-20, 20, -10, 10]} [Ans]