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

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Answer 1 · 2017-05-07T12:03:13

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

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

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]

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