# What is the domain and range of f(x)={ x^2 - 81 }/ {x^2 - 4x}?

Dec 30, 2017

${D}_{f} = \mathbb{R} - \left\{0 , 4\right\} = \left(- \infty , 0\right) \cup \left(0 , 4\right) \cup \left(4 , + \infty\right)$ , Range = $f \left({D}_{f}\right) = \left(- \infty , \frac{81 - 9 \sqrt{65}}{8}\right] \cup \left[\frac{81 + 9 \sqrt{65}}{8} , + \infty\right)$

#### Explanation:

$f \left(x\right) = \frac{{x}^{2} - 81}{{x}^{2} - 4 x}$

In order for this function to be defined we need ${x}^{2} - 4 x \ne 0$

We have ${x}^{2} - 4 x = 0$ $\iff$ $x \left(x - 4\right) = 0$ $\iff$ $\left(x = 0 , x = 4\right)$

So ${D}_{f} = \mathbb{R} - \left\{0 , 4\right\} = \left(- \infty , 0\right) \cup \left(0 , 4\right) \cup \left(4 , + \infty\right)$

For $x$$\in {D}_{f}$ ,

$f \left(x\right) = \frac{{x}^{2} - 81}{{x}^{2} - 4 x}$ $=$ $\frac{\left(x - 9\right) \left(x + 9\right)}{{x}^{2} - 4 x}$

$f \left(x\right) = 0 \iff \left(x = 9 , x = - 9\right)$

• $\frac{{x}^{2} - 81}{{x}^{2} - 4 x} = y$ $\iff$ ${x}^{2} - 81 = y \left({x}^{2} - 4 x\right)$

${x}^{2} - 81 = y {x}^{2} - 4 x y$

• Adding $\textcolor{g r e e n}{4 y x}$ in both sides,

${x}^{2} - 81 + 4 y x = y {x}^{2}$

• Substracting $\textcolor{red}{y {x}^{2}}$ from both sides

${x}^{2} - 81 + 4 y x - y {x}^{2} = 0$ $\iff$

${x}^{2} \left(1 - y\right) + 4 x y - 81 = 0$

This is quadratic equation for $x$ so
$a = 1 - y$
$b = 4 y$
$c = - 81$

We need $D = {b}^{2} - 4 \cdot a \cdot c \ge 0$ $\iff$
$16 {y}^{2} - 4 \left(1 - y\right) \cdot \left(- 81\right) \ge 0$ $\iff$
$16 {y}^{2} + 324 \left(1 - y\right) \ge 0$ $\iff$
$16 {y}^{2} - 324 y + 324 \ge 0$ $\iff$
$4 {y}^{2} - 81 y + 81 \ge 0$

${y}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$=$ $\frac{81 \pm \sqrt{6561 - 1296}}{8}$

$=$ $\frac{81 \pm \sqrt{5265}}{8}$

$=$ $\frac{81 \pm 9 \sqrt{65}}{8}$

$4 {y}^{2} - 81 y + 81 \ge 0$ $\iff$ (y<=(81-9sqrt65)/8 or y>=(81+9sqrt65)/8)

so, $f \left(x\right) \le \frac{81 - 9 \sqrt{65}}{8}$ or $f \left(x\right) \ge \frac{81 + 9 \sqrt{65}}{8}$

Which means, $f \left({D}_{f}\right) = \left(- \infty , \frac{81 - 9 \sqrt{65}}{8}\right] \cup \left[\frac{81 + 9 \sqrt{65}}{8} , + \infty\right)$