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

Dec 26, 2017

Domation: $0 \le x < 1 \mathmr{and} x > 1$
Range: $y \in \mathbb{R}$

#### Explanation:

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

Domation:
$x \ge 0 \mathmr{and} x \ne - 2 , 1 \iff 0 \le x < 1 \mathmr{and} x > 1$

Range:
$f ' \left(x\right) = \frac{\frac{1}{2 \sqrt{x}} \left({x}^{2} + x - 2\right) - \sqrt{x} \left(2 x + 1\right)}{{\left[\left(x + 2\right) \left(x - 1\right)\right]}^{2}}$
(note!! $f ' \left(0\right) = N a N$)

let $f ' \left(x\right) = 0 \implies \frac{{x}^{2} + x - 2}{2 \sqrt{x}} - \sqrt{x} \left(2 x + 1\right) = 0$
$\iff \frac{{x}^{2} + x - 2}{2 \sqrt{x}} = \sqrt{x} \left(2 x + 1\right)$
$\iff {x}^{2} + x - 2 = 2 x \left(2 x + 1\right)$
$\iff {x}^{2} + x - 2 = 4 {x}^{2} + 2 x$
$\iff 3 {x}^{2} + x + 2 = 0$
$\implies$
there is no solution for this (I take $x \in \mathbb{R}$)
$\implies$
There aren't any max or min
$\implies$
We will check what happens in each range ($0 \le x < 1 \mathmr{and} x > 1$)
$\implies$
$f ' \left(\frac{1}{2}\right) = \frac{\frac{1}{2 \sqrt{\frac{1}{2}}} \left({\left(\frac{1}{2}\right)}^{2} + \frac{1}{2} - 2\right) - \sqrt{\frac{1}{2}} \left(2 \left(\frac{1}{2}\right) + 1\right)}{{\left[\left(\frac{1}{2} + 2\right) \left(\frac{1}{2} - 1\right)\right]}^{2}} =$
$= \frac{\frac{\sqrt{2}}{4} \left(- \frac{5}{4}\right) - \sqrt{\frac{1}{2}} \left(2\right)}{\left(+\right)} =$
$= - \frac{5 \sqrt{2}}{16} - \frac{16 \sqrt{2}}{16} =$
$= \frac{- 21 \sqrt{2}}{16} = \left(-\right) < 0$

$f ' \left(4\right) = \frac{\frac{1}{2 \sqrt{4}} \left({\left(4\right)}^{2} + 4 - 2\right) - \sqrt{4} \left(2 \left(4\right) + 1\right)}{{\left[\left(4 + 2\right) \left(4 - 1\right)\right]}^{2}} =$
$= \frac{\frac{1}{4} \left(18\right) - \left(2\right) \left(9\right)}{\left(+\right)} =$
$2 \cdot 9 = 18 , \frac{18}{4} < 18 \implies \frac{1}{4} \left(18\right) - \left(2\right) \left(9\right) < 0$
$= \left(-\right) < 0$

Well, it was a long algebra but in the end we know now that $f \left(x\right)$ goes down for all the range $0 \le x < 1 \mathmr{and} x > 1$

Now let's do simpler algebra:
$f \left(0\right) = 0$
${\lim}_{x \rightarrow {1}^{-}} f = - \infty$
${\lim}_{x \rightarrow {1}^{+}} f = + \infty$
${\lim}_{x \rightarrow \infty} f = {0}^{+}$

SO:
$0 \le y < - \infty \mathmr{and} 0 > y \iff y \in \mathbb{R}$