How do you find the domain and range of 1 /( x^3-9x)?

Jun 6, 2017

$x \in \mathbb{R} , x \ne 0 , x \ne \pm 3$
$y \in \mathbb{R} , y \ne 0$

Explanation:

$\text{the denominator of " y=1/(x^3-9x)" cannot be zero}$

$\text{as this would make y undefined}$

$\text{equating the denominator to zero and solving gives the}$
$\text{values that x cannot be}$

$\text{solve } {x}^{3} - 9 x = 0 \Rightarrow x \left(x - 3\right) \left(x + 3\right) = 0$

$\Rightarrow x = 0 , x = \pm 3 \leftarrow \textcolor{red}{\text{excluded values}}$

$\text{domain is } x \in \mathbb{R} , x \ne 0 , x \ne \pm 3$

$\text{to find any excluded values in the range}$
$\text{consider the horizontal asymptote of the function}$

$\text{divide terms on numerator/denominator by the highest}$
$\text{power of x, that is } {x}^{3}$

$y = \frac{\frac{1}{x} ^ 3}{{x}^{3} / {x}^{3} - \frac{9 x}{x} ^ 3} = \frac{\frac{1}{x} ^ 3}{1 - \frac{9}{x} ^ 2}$

as $x \to \pm \infty , y \to \frac{0}{1 - 0} = 0 \leftarrow \textcolor{red}{\text{ excluded value}}$

$\text{range is } y \in \mathbb{R} , y \ne 0$