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

1 Answer
May 7, 2017

Answer:

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]