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

Nov 10, 2017

The domain is $x \in \left(- \infty , - 2\right) \cup \left(1 , \infty\right)$
The range is $y \in \left(- \infty , - \frac{\sqrt{80}}{3}\right) \cup \left(\frac{\sqrt{80}}{3} , \infty\right)$

#### Explanation:

The denominator must be $\ne 0$

Therefore,

$\sqrt{{x}^{2} + x - 2} \ne 0$, $\implies$, ${x}^{2} + x - 2 > 0$

$\left(x + 2\right) \left(x - 1\right) > 0$

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

$x \in \left(- \infty , - 2\right) \cup \left(1 , \infty\right)$

The domain is $x \in \left(- \infty , - 2\right) \cup \left(1 , \infty\right)$

To find the range, proceed as follows :

Let $y = \frac{3 x + 1}{\sqrt{{x}^{2} + x - 2}}$

Rearranging this equation

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

Squaring both sides

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

${y}^{2} \left({x}^{2} + x - 2\right) = 9 {x}^{2} + 6 x + 1$

Rearranging

${x}^{2} \left({y}^{2} - 9\right) + x \left({y}^{2} - 6\right) - \left(2 {y}^{2} + 1\right) = 0$

This is a quadratic equation in $x$, in order to have solutions, the discriminant must be $\ge 0$

The discriminant is

$\Delta = {b}^{2} - 4 a c = {\left({y}^{2} - 6\right)}^{2} + 4 \left({y}^{2} - 9\right) \left(2 {y}^{2} + 1\right) \ge 0$

${y}^{4} - 12 {y}^{2} + 36 + 8 {y}^{4} - 68 {y}^{2} - 36 \ge 0$

$9 {y}^{4} - 80 {y}^{2} \ge 0$

${y}^{2} \left(9 {y}^{2} - 80\right) \ge 0$

$y = 0$, $S = \emptyset$

$y = \pm \frac{\sqrt{80}}{3}$

The range is $y \in \left(- \infty , - \frac{\sqrt{80}}{3}\right) \cup \left(\frac{\sqrt{80}}{3} , \infty\right)$

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