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

Aug 16, 2017

Domain: $x \ge 0 | x \ne 1 \mathmr{and} \left[0 , 1\right) \cup \left(1 , \infty\right)$
Range: $f \left(x\right) \in \mathbb{R} \mathmr{and} \left(- \infty , \infty\right)$

#### Explanation:

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

For domain under root should be $\ge 0 \therefore x \ge 0$ ,

denominator should not be zero , i.e $\left(x - 1\right) \ne 0$

$\therefore x \ne 1 \mathmr{and} \left(x + 2\right) \ne 0 \mathmr{and} x \ne - 2$ . So restriction is

$x \ge 0 , x \ne 1$

Domain: $x \ge 0 | x \ne 1 \mathmr{and} \left[0 , 1\right) \cup \left(1 , \infty\right)$

Range: Any real number , i.e $f \left(x\right) \in \mathbb{R} \mathmr{and} \left(- \infty , \infty\right)$

graph{sqrt(x)/(x^2+x-2) [-10, 10, -5, 5]} [Ans]