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

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Answer 1 · 2017-11-10T18:00:59

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

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]}