# Let f(x)=(x^3 - 4x)/(x^3 +x^2-6x), how do you find all points of discontinuity of f(x)?

Aug 29, 2015

Factor the numerator and denominator to identify their zeros to determine the points of discontinuity and their types.

#### Explanation:

$f \left(x\right) = \frac{{x}^{3} - 4 x}{{x}^{3} + {x}^{2} - 6 x} = \frac{x \left(x - 2\right) \left(x + 2\right)}{x \left(x - 2\right) \left(x + 3\right)} = \frac{x + 2}{x + 3}$

with exclusions $x \ne 0$ and $x \ne 2$.

There are removable discontinuities at $x = 0$ and $x = 2$ where both the numerator and the denominator are zero, so $f \left(x\right)$ is undefined at those points, but the left and right limits agree.

There is a simple pole at $x = - 3$, where the denominator is zero and the numerator is non-zero.

${\lim}_{x \to - 3 +} f \left(x\right) = - \infty$
${\lim}_{x \to - 3 -} f \left(x\right) = + \infty$

Apart from these discontinuities, $f \left(x\right)$ is well defined and continuous.

graph{(x+2)/(x+3) [-10, 10, -5, 5]}

Aug 29, 2015

That function has discontinuities at 0, 2, and -3.

#### Explanation:

A rational function is continuous on its domain.

So the points of discontinuity for a rational function are the point outside the domain.

${x}^{3} + {x}^{2} - 6 x = 0$

$x \left({x}^{2} + x - 6\right) = x \left(x - 2\right) \left(x + 3\right) = 0$

The points outside the domain are: $0 , \text{ "2, " and } - 3$

Note

Because this was posted in the topic "Classifying Discontinuities, I should probably add that

$f \left(x\right) = \frac{x \left(x + 2\right) \left(x - 2\right)}{x \left(x - 2\right) \left(x + 3\right)} = \frac{x + 2}{x + 3}$ for $x \ne 0 , 2$

So the only infinite limit occurs at $x = - 3$. There is a non-removable (infinite) discontinuity at $- 3$

The discontinuities at $0$ and $2$ are removable.