# What is the range of f(x)=1/{(x-3)(x+4)} ?

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Jim G. Share
Feb 9, 2018

$y \in \mathbb{R} , y \ne 0$

#### Explanation:

$f \left(x\right) = \frac{1}{{x}^{2} + x - 12} \text{ expanding factors in denominator}$

$\text{divide terms on numerator/denominator by the highest}$
$\text{power of x that is } {x}^{2}$

$f \left(x\right) = \frac{\frac{1}{x} ^ 2}{{x}^{2} / {x}^{2} + \frac{x}{x} ^ 2 - \frac{12}{x} ^ 2} = \frac{\frac{1}{x} ^ 2}{1 + \frac{1}{x} - \frac{12}{x} ^ 2}$

$\text{as } x \to \pm \infty , f \left(x\right) \to \frac{0}{1 + 0 - 0}$

$\Rightarrow y = 0 \leftarrow \textcolor{red}{\text{is an excluded value}}$

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

$\left(- \infty , 0\right) \cup \left(0 , + \infty\right) \leftarrow \textcolor{b l u e}{\text{in interval notation}}$
graph{1/(x^2+x-12) [-10, 10, -5, 5]}

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