# What are the removable discontinuities of f(x) = (x^3 - 3x^2 - x + 3) / (x+1)?

Jun 26, 2015

The only discontinuity is at $x = - 1$ and it is removable.

#### Explanation:

Rational functions are continuous on their domains. (They are continuous except where the denominator is zero.)

$f \left(x\right) = \frac{{x}^{3} - 3 {x}^{2} - x + 3}{x + 1}$ Is continuous except at $x = - 1$

Not the when $x = - 1$, the numerator is $0$, so we know that $x - \left(- 1\right) = x + 1$ is a factor of the numerator.

We'll need the factoring in a minute, so either do the division or factor by grouping:

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

$= {x}^{2} \left(x - 3\right) - 1 \left(x - 3\right)$

$= \left({x}^{2} - 1\right) \left(x - 3\right) = \left(x + 1\right) \left(x - 1\right) \left(x - 3\right)$

${\lim}_{x \rightarrow - 1} \frac{{x}^{3} - 3 {x}^{2} - x + 3}{x + 1} = {\lim}_{x \rightarrow - 1} \frac{\left(x + 1\right) \left(x - 1\right) \left(x - 3\right)}{x + 1}$

$= {\lim}_{x \rightarrow - 1} \left(\left(x - 1\right) \left(x - 3\right)\right)$

$= \left(- 2\right) \left(- 4\right) = 8$

Because this limit exists, the discontinuity is removable.

To remove the discontinuity, define $g$ by:
$g \left(x\right) = \left\{\begin{matrix}f \left(x\right) & \text{if" & x != -1 \\ 8 & "if} & x = - 1\end{matrix}\right.$
Note that $g \left(x\right) = f \left(x\right)$ for all $x \ne - 1$ but $g$ is continuous at $- 1$.