Find the x-values (if any) at which f is not continuous. Which of the discontinuties are removable?

f(x)=(x)/(x^2-4)

I'm not sure how to do this problem but I know that it is not continnous at x=2?

f is discontinuous at x = ±2

since the numerator (x) is not zero there, f(x) has vertical asymptotes at x=±2.

These are not removable.

Removable discontinuities occur where f(x) = 0/0 and the common factor can be removed, such as

f(x) = (x-2)/(x^2-4)
= (x-2)/(x-2)(x+2)

At x=2, f(x) = 0/0

but everywhere else, f(x) = 1/(x+2) because you can remove the common factor of (x-2). The "hole" is removable by defining

f(2) = 1/4 which is 1/(x+2)

Got it, thanks for the help!

To determine the x-values at which the function f(x)=(x)/(x^2-4) is not continuous, we need to check for potential discontinuities in the function.

We can start by checking for any values of x that make the denominator equal to zero, as this would result in an undefined value for f(x). In this case, the denominator (x^2 - 4) would be zero if x = 2 or x = -2.

So, the possible values for which f(x) might be discontinuous are x = 2 and x = -2.

Now, let's examine each value to see if it is actually a discontinuity and if it is removable.

At x = 2:
We can evaluate f(2) by substituting it into the function:
f(2) = 2 / ((2)^2 - 4) = 2 / (4 - 4) = 2 / 0 (which is undefined)

Therefore, f(x) is not continuous at x = 2.

At x = -2:
We can evaluate f(-2) by substituting it into the function:
f(-2) = (-2) / ((-2)^2 - 4) = (-2) / (4 - 4) = (-2) / 0 (which is undefined)

Therefore, f(x) is not continuous at x = -2 as well.

Both of these discontinuities are non-removable, meaning they cannot be eliminated or canceled by simplifying or modifying the function.

To summarize:
- The function f(x) = (x) / (x^2 - 4) is not continuous at x = 2 and x = -2.
- Neither of these discontinuities is removable.

To find the x-values at which the function is not continuous, we need to look for points where there might be a discontinuity. The potential sources of discontinuity are:

1. Vertical asymptotes: A vertical asymptote occurs at x = a if the limit of the function as x approaches a from the left or right is positive or negative infinity.

2. Holes: A hole occurs at x = a if there is a common factor between the numerator and denominator that cancels out, creating a removable discontinuity.

3. Jump discontinuities: A jump discontinuity occurs if there is a jump in the function when approaching a particular x-value from the left and the right.

We can start by looking for vertical asymptotes. In this case, the denominator x^2 - 4 can be factored as (x - 2)(x + 2). So the function has vertical asymptotes at x = 2 and x = -2.

Next, we need to check if there are any holes in the function. To do this, factor out the common factor (x - 2) from the numerator x(x - 2) and denominator (x - 2)(x + 2):

f(x) = (x)/(x^2 - 4) = x/(x - 2)(x + 2)

We can now see that there is a common factor of (x - 2) that cancels out. This means that the function has a hole at x = 2. So, the discontinuity at x = 2 is removable.

To summarize, the function f(x) = (x)/(x^2 - 4) is not continuous at x = 2 because it has a removable discontinuity (hole) at that point.