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

1 Answer
Narad T.
Nov 10, 2017

The domain is x in (-oo,-2) uu (1,oo)
The range is y in (-oo,-sqrt80/3) uu(sqrt80/3, oo)

Explanation:

The denominator must be !=0

Therefore,

sqrt(x^2+x-2)!=0, =>, x^2+x-2>0

(x+2)(x-1)>0

As the coefficient of x^2 is >0, so

x in (-oo,-2) uu (1,oo)

The domain is x in (-oo,-2) uu (1,oo)

To find the range, proceed as follows :

Let y=(3x+1)/sqrt(x^2+x-2)

Rearranging this equation

ysqrt(x^2+x-2)=(3x+1)

Squaring both sides

(ysqrt(x^2+x-2))^2=(3x+1)^2

y^2(x^2+x-2)=9x^2+6x+1

Rearranging

x^2(y^2-9)+x(y^2-6)-(2y^2+1)=0

This is a quadratic equation in x, in order to have solutions, the discriminant must be >=0

The discriminant is

Delta=b^2-4ac=(y^2-6)^2+4(y^2-9)(2y^2+1)>=0

y^4-12y^2+36+8y^4-68y^2-36>=0

9y^4-80y^2>=0

y^2(9y^2-80)>=0

y=0, S=O/

y=+-sqrt80/3

The range is y in (-oo,-sqrt80/3) uu(sqrt80/3, oo)

graph{(3x+1)/sqrt(x^2+x-2) [-28.87, 28.88, -14.43, 14.43]}

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