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

Jan 29, 2017

Domain is $x \in \left[0 , \infty\right)$, sans x = 1.
Range is $f \in \left(- \infty , \infty\right)$.
Asymptotes : $\uparrow x = 1 \downarrow \mathmr{and} y = 0 \rightarrow$.

#### Explanation:

To make f real, $x \ge 0$.

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

As $x \to {0}_{\pm} , f \to \pm \infty$.

As $x \to \infty , f \to 0$.

The asymptotes keep the two branches of the graph, in #Q_1 and

Q_4, respectively.

So, domain is $x \in \left[0 , \infty\right)$, sans x = 1.

Range is $f \in \left(- \infty , \infty\right)$.

See the illustrative Socratic graph.

graph{(sqrtx/(x^2+x-2)-y)(x-.99+.01y) =0 [-10, 10, -5, 5]}