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

Apr 30, 2017

Domain of $x$ in interval notation is $\left[0 , 1\right) \cup \left(1 , \infty\right)$ and range is $\mathbb{R}$.

Explanation:

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

As we have $\sqrt{x}$ in numerator, $x$ cannot take negative values, hence we should have $x \ge 0$

Further denominator can be factorized to $\left(x + 2\right) \left(x - 1\right)$, hence $x$ cannot take values $- 2$ and $1$,

and hence domain of $x$ in interval notation is $\left[0 , 1\right) \cup \left(1 , \infty\right)$

However, as $f \left(x\right)$ can take any value from $- \infty$ (when $x \to 1$ from left) to $\infty$ (when $x \to 1$ from right) as degree of denominator is higher than that of denominator. Further, $f \left(x\right) = 0$ when $x = 0$..

hence range is $f \left(x\right) \in \mathbb{R}$

graph{sqrtx/(x^2+x-2) [-7.58, 12.42, -4.96, 5.04]}