# What is the domain and range of f(x) = sqrt x /( x^2 + x - 2)?

Jan 6, 2018

The domain is $x \in \left[0 , 1\right) \cup \left(1 , + \infty\right)$.
The range is $y \in \mathbb{R}$

#### Explanation:

The function is

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

Factorise the denominator

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

From the sqrt sign , $x \ge 0$

Therefore,

The domain is

$x \in \left[0 , 1\right) \cup \left(1 , + \infty\right)$

Let,

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

When $x = 0$, $f \left(0\right) = 0$

${\lim}_{x \to {1}^{-}} \frac{\sqrt{x}}{\left(x + 2\right) \left(x - 1\right)} = \frac{1}{0} ^ \equiv - \infty$

${\lim}_{x \to {1}^{+}} \frac{\sqrt{x}}{\left(x + 2\right) \left(x - 1\right)} = \frac{1}{0} ^ + = + \infty$

The range is $y \in \mathbb{R}$

graph{sqrt(x)/(x^2+x-2) [-8.89, 8.89, -4.444, 4.445]}