# How do you find the discontinuities for y=(4x)/(x^3-9x)?

Jun 17, 2015

Factor the denominator to get:

$y = \frac{4 x}{x \left(x - 3\right) \left(x + 3\right)} = \frac{4}{\left(x - 3\right) \left(x + 3\right)}$

with exclusion $x \ne 0$

This has a removable discontinuity at $x = 0$ and simple poles at $x = \pm 3$

#### Explanation:

$y = f \left(x\right) = \frac{4 x}{x \left(x - 3\right) \left(x + 3\right)} = \frac{4}{\left(x - 3\right) \left(x + 3\right)} \cdot \frac{x}{x}$

$= \frac{4}{\left(x - 3\right) \left(x + 3\right)}$ with exclusion $x \ne 0$

$f \left(0\right)$ is undefined since $\frac{0}{0}$ is undefined,

So $f \left(x\right)$ has a discontinuity at $x = 0$.

This is a removable discontinuity as we can simply define:

$g \left(0\right) = {\lim}_{x \to 0} f \left(x\right) = - \frac{4}{9}$

$g \left(x\right) = f \left(x\right)$ when $x \ne 0$

$f \left(- 3\right)$ and $f \left(3\right)$ are simple pole discontinuities, where the denominator is zero and the numerator is non-zero.

graph{4x/(x(x-3)(x+3)) [-10, 10, -5, 5]}