We just learned how to determine the continuity of a function at a given point by using a graph. But often, you won't actually be given the luxury of a graph. And instead of being given a specific point to test the continuity at, you'll be asked to specifically find where a function is discontinuous by just being given that function. Remember that when we were working with graphs, we saw that discontinuities occurred whenever a function had a jump, a hole, or an asymptote, which happens often for rational functions and for piecewise functions. So that means that we need to be able to determine where these discontinuities are when working with both of these types of functions.
Now here, I'm going to walk you through how to do that step by step. So let's go ahead and get started. Let's first just jump right into our example here, and we want to determine the values of x here, if any, for which our function is discontinuous. Now the first function that we are given here is \( f(x) = \frac{x^2 - x - 6}{x+2} \). Now polynomials by themselves are continuous everywhere, but because these polynomials are being divided and we're dealing with a rational function here, we know that this rational function may have holes or asymptotes.
So it would be discontinuous there. But how do we find where these discontinuities are? Well, for rational functions, it's actually rather easy to determine where these discontinuities are because it's just going to be where our denominator is equal to 0. So here our denominator is \( x + 2 \). So if we take that denominator and set it equal to 0 and then just solve for x here by subtracting 2 from both sides, I end up with \( x = -2 \).
And this is where my function is discontinuous. Now one of two things could be happening here. This could be an asymptote, or if we were able to factor this numerator and cancel out one of the factors with the denominator, this would end up being a hole or a removable discontinuity. Now it's not likely that you'll be asked often to figure out what type of discontinuity you're dealing with. But if you're curious here, for this function, you actually would end up with a hole or a movable discontinuity.
So on a graph, you would have to stop, pick up your pen, and continue on the other side of that hole. Now that's for a rational function, but let's go ahead and move on to our piecewise function. Now here we want to do the same thing. We want to determine the values of x, if any, for which our function is discontinuous. Now here our function is equal to 4 for x less than negative one, and it's equal to \( x + 1 \) for x greater than or equal to negative one.
We don't need to worry about discontinuities within each piece, but because this is a piecewise function, there may be a discontinuity, like a hole or a jump, in between my two pieces. Now, in order to determine if there is a discontinuity there, that means that we want to compare the limit and the function value for the point or points in between each piece. Now here, since I only have two pieces, I only have to worry about one point here, and that's \( x = -1 \). And for this point, I want to look at my limit and my function value. So I want to find the limit of \( f(x) \) as \( x \) approaches negative 1, and I also want to find my function value \( f(-1) \).
Because if these values are equal to each other, I know that my function is continuous there. But if they are not equal to each other, I know that I have a discontinuity there. So let's go ahead and find these values starting first with our limit. Now because this is a piecewise function, I need to be careful here and look at my one-sided limits. So I'm going to start with my left sided limit, the limit of \( f(x) \) as \( x \) approaches negative one from that left side.
Now looking at my piecewise function, I know that coming in from the left side, my function is going to be equal to 4 for these values of x less than negative one. So I know that that left sided limit is simply 4. Now my limit from the other side, my right sided limit, the limit of \( f(x) \) as \( x \) approaches negative one from that right side. Looking back at my piecewise function, I know that coming in from that right side, my function is going to be equal to \( x + 1 \) because these values are greater than negative 1. So that means that this limit is going to be equal to negative 1 plus 1, which is simply 0, so that right sided limit is 0.
Now looking at my one-sided limits, one of them is 4 and the other one is 0. Because these are not the same, that tells me that my regular limit, the limit of \( f(x) \) as \( x \) approaches negative one, simply does not exist. Now because this limit does not exist, I sort of already know that it's not going to be the same as my function value. But let's go ahead and continue on here. Now my function value \( f(-1) \), we actually already found that when finding our right sided limit.
I know that that function value is just negative one plus 1, which is equal to 0. So with that function value of 0 here, I see that comparing my limits to my function value, these are definitely not the same thing. So I am dealing with a discontinuity here. So for this piecewise function, it is discontinuous for \( x = -1 \). Now if you're having trouble visualizing what this may look like, because we're dealing with a piecewise function here and my first piece is just a constant whereas my second one is just a linear function, I am dealing with a jump discontinuity here, meaning that I definitely could not trace this function without picking up my pen.
Now that we know how to determine where the discontinuities of a function are, let's continue practicing with this. Thanks for watching, and I'll see you in the next one.