Math / Algebra

QuestionFind the $x$ -values (if any) at which $f$ is not continuous. State whether the discontinuities are removable or nonremovable. (Enter your answers as comma-separated lists. If an answer does not exist, enter DNE.) $f(x)=\frac{x+2}{x^{2}-2 x-8}$ removable discontinuities $x=$ $\square$ nonremovable discontinuities $\quad x=$ $\square$

Studdy Solution

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.

STEP 9

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

Was this helpful?

Studdy Solution

STEP 1

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

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.

STEP 9

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

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

STEP 2

1. Simplify the Function2. Find Potential Discontinuities3. Classify the Discontinuities

STEP 3

STEP 4

STEP 5

From the simplified form, f(x)=1x−4f(x) = \frac{1}{x-4}f(x)=x−41​, we see that the function is undefined when the denominator is zero.This happens when x−4=0x-4=0x−4=0, so x=4x=4x=4.This is a **potential discontinuity**.

STEP 6

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

STEP 7

Let's look at x=4x = 4x=4.As xxx approaches **4**, the function f(x)=1x−4f(x) = \frac{1}{x-4}f(x)=x−41​ 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

STEP 9

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

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

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

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.

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.

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.

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