$\text { Find the } x \text {-values (if any) at which the function } f ( x ) = \frac { x + 2 } { x ^ { 2 } + 6 x + 8 } \text { is not continuous. }$ Which of the discontinuities are removable?
A) no points of discontinuity
B) $x = - 2$ (not removable), $x = - 4$ (removable)
C) $x = - 2$ (removable), $x = - 4$ (not removable)
D) no points of continuity
E) $x = - 2$ (not removable), $x = - 4$ (not removable)

Question

$\text { Find the } x \text {-values (if any) at which the function } f ( x ) = \frac { x + 2 } { x ^ { 2 } + 6 x + 8 } \text { is not continuous. }$ Which of the discontinuities are removable? 
A) no points of discontinuity
B) $x = - 2$ (not removable), $x = - 4$ (removable)
C) $x = - 2$ (removable), $x = - 4$ (not removable)
D) no points of continuity
E) $x = - 2$ (not removable), $x = - 4$ (not removable)

Quizplus · Accepted Answer

The function is not continuous at any $x$ where the denominator is equal to 0, since division by zero is undefined. Therefore, we need to find the roots of $x^2+6x+8$. Factoring, we have $(x+2)(x+4)=0$, so the roots are $x=-2$ and $x=-4$.

To determine if these points are removable discontinuities, we need to check if the limit of the function exists at those points.

At $x=-2$, we have:

$$\lim_{x	o -2} \frac{x+2}{x^2+6x+8}=\frac{0}{0+0+8}=0$$

Notice that if we simplify the function by canceling the factor of $x+2$ in both the numerator and denominator, we get:

$$f(x)=\frac{x+2}{x^2+6x+8}=\frac{1}{x+4}$$

This function is defined and continuous everywhere except for $x=-4$, where there is a vertical asymptote. Therefore, the discontinuity at $x=-2$ is removable.

At $x=-4$, we have:

$$\lim_{x	o -4} \frac{x+2}{x^2+6x+8}=\frac{-2}{0+0+8}=-\frac{1}{4}$$

Since the limit exists and is finite, the discontinuity at $x=-4$ is not removable.

Therefore, the points of discontinuity are $x=-2$ (removable) and $x=-4$ (not removable), so the answer is $\boxed{	extbf{(C)}}$.