Math Snap

STEP 1

What is this asking?
We're looking for any x-values where the function $f(x)$ has a discontinuity, and then we need to classify each discontinuity as either removable or nonremovable.
Watch out!
Remember, a discontinuity is removable if we can "fill in the hole" by defining the function at that point.
Nonremovable discontinuities are like asymptotes where the function explodes!

STEP 2

1. Simplify the Function
2. Find Potential Discontinuities
3. Classify the Discontinuities

STEP 3

Let's factor the denominator of our function $f(x)$.
We're doing this because factoring helps us see if there are any common factors between the numerator and denominator, which might lead to removable discontinuities.
$$f(x) = \frac{x+2}{x^2 - 2x - 8}$$ $$f(x) = \frac{x+2}{(x-4)(x+2)}$$

STEP 4

Now, we see a common factor of $(x+2)$ in both the numerator and the denominator!
We can simplify $f(x)$ by dividing both by $(x+2)$.
Remember, we can only do this if $x+2 \ne 0$, which means $x \ne -2$.
So, for $x \ne -2$, we have:
$$f(x) = \frac{1}{x-4}$$ This simplified form will help us analyze the discontinuities.

STEP 5

From the simplified form, $f(x) = \frac{1}{x-4}$, we see that the function is undefined when the denominator is zero.
This happens when $x-4=0$, so $x=4$.
This is a potential discontinuity.

STEP 6

We also need to consider the value we excluded earlier, $x = -2$.
Since the original function was undefined at $x = -2$, this is another potential discontinuity.

STEP 7

Let's look at $x = 4$.
As $x$ approaches 4, the function $f(x) = \frac{1}{x-4}$ blows up!
It approaches positive infinity from the right and negative infinity from the left.
This is a nonremovable discontinuity, specifically a vertical asymptote.

STEP 8

Now, let's consider $x = -2$.
In the simplified form, $f(x) = \frac{1}{x-4}$, we can plug in $x = -2$ and get $f(-2) = \frac{1}{-2-4} = \frac{1}{-6} = -\frac{1}{6}$.
Since we can "fill in the hole" at $x = -2$ by defining $f(-2) = -\frac{1}{6}$, this is a removable discontinuity.

Removable discontinuities: $x = -2$
Nonremovable discontinuities: $x = 4$