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

May 7, 2017

Domain : $x : \left(- \infty , - 2\right) \cup \left(1 , \infty\right)$
Range : $f \left(x\right) : \left(- 3 , - \infty\right) \cup \left(3 , \infty\right)$

#### Explanation:

$f \left(x\right) = \frac{3 x - 1}{\sqrt{{x}^{2} + x - 2}} = \frac{3 x - 1}{\sqrt{\left(x + 2\right) \left(x - 1\right)}}$

Domain: denominator must not be $0$ and under root must not be $< 0$ So $x + 2 \ne 0 \therefore x \ne - 2$ and $x - 1 \ne 0 \therefore x \ne 1$

For under root calculation: critical points are $x = - 2 \mathmr{and} x = 1$
when $x < - 2 , \left(x + 2\right) \cdot \left(x - 1\right) = \left(- \cdot -\right) = +$

when $- 2 < x < 1 , \left(x + 2\right) \cdot \left(x - 1\right) = \left(+ \cdot -\right) = -$

when $x > 1 , \left(x + 2\right) \cdot \left(x - 1\right) = \left(= \cdot +\right) = +$

So in domain : $x < - 2 \mathmr{and} x > 1 \mathmr{and} \left(- \infty , - 2\right) \cup \left(1 , \infty\right)$

Horizontal asymptote is at $y = \frac{3}{\pm 1} = \pm 3$

So range : $f \left(x\right) : \left(- 3 , - \infty\right) \cup \left(3 , \infty\right)$ graph{(3x-1)/(x^2+x-2)^0.5 [-20, 20, -10, 10]} [Ans]