Topic 1.13: Removing Discontinuities

Find every break

Now let’s do it the way the exam asks. Take f of x equals x squared minus four, over x squared minus two x. We want every place it’s discontinuous, and what kind of break each one is.

A rational function is continuous everywhere it’s defined, so the only suspects are where the denominator is zero. The bottom, x squared minus two x, is x times x minus two, which is zero at x equals zero and x equals two. Factor the top too, x minus two times x plus two, and now you can see the shared factor. For every x other than zero and two, f is the same as x plus two, over x.

As x approaches two, that’s four over two, so the limit is two. Since f of two doesn’t exist, x equals two is a removable discontinuity, a hole at the point two, two. At x equals zero it’s a different story. In x plus two over x, the top heads to two while the bottom heads to zero.

Just left of zero the bottom is negative, so f dives down without bound, and just right of zero it shoots up. That makes x equals zero an infinite discontinuity, a vertical asymptote. The mistake I see most often is assuming every zero of the denominator gives a vertical asymptote. This function proves it wrong, because its denominator is zero at both zero and two, but zero gives an asymptote and two gives a hole.

To repair the hole, we define f of two to be two, the limit, and then this new function is continuous at two. Nothing we define at zero can rescue that asymptote. On the exam you’d write it like this. The original f isn’t continuous at two because f of two doesn’t exist, and since its limit is two, defining f of two as two removes the discontinuity.

Now you try one. For h of x equals x squared plus two x minus fifteen, over x minus three, what value should h of three be so that h is continuous at three? Now see if you can spot which break in this next function can be repaired. For g of x equals x plus one, over x squared minus one, which statement is true?

Make the pieces meet

We sometimes use different rules for different inputs, so we need to check what happens where we switch rules. Take p of x equals k x plus one when x is less than two, and x squared minus k when x is two or more. The number k changes both pieces at once, so watch the gap at x equals two open and close as I slide k back and forth.

From the left we use the first piece, so the limit is two k plus one. From the right, and at two itself, we use the second piece, so both the limit and the value are four minus k.

For the pieces to meet, those have to be equal. Two k plus one equals four minus k, three k equals three, and k equals one. Slide k to one and look at that, the two pieces lock together at the point two, three.

This is the sentence that earns full credit, so I’ll write every piece. The limit from the left is three and the limit from the right is three, so the limit as x approaches two is three. P of two is also three, so the limit equals p of two, and p is continuous at x equals two.

On a real AP exam, answers that just plugged the boundary number into both pieces, with no limits and no function value, lost a point. Even when it feels obvious, write both limits and the value.

Now I want you to find k yourself. For u of x equals x squared plus one when x is less than two, and k x minus one when x is two or more, what value of k makes u continuous at two? One more thing before we leave this function. Which of these answers would earn the justification point on the exam?

Your taxes are a piecewise function too, and this one you’ll actually live with. In 2026 a single person pays ten percent federal income tax on taxable income up to twelve thousand four hundred dollars. Above that, they pay twelve percent on each dollar, up to fifty thousand four hundred. Let’s write it as a piecewise function, where the second piece starts with some number, k, before the twelve percent part, and x minus twelve thousand four hundred is how far over the line you are.

What does k have to be so that the tax is continuous at twelve thousand four hundred dollars? That number is there to keep the tax continuous, and it settles an argument you’ll hear someday. People say a raise can push you into a higher bracket and cost you money. In this tax schedule, though, the tax has no jump at the line.

Above the line, each extra dollar costs you twelve cents of tax, so you keep eighty-eight cents of it, and below the line you keep ninety, so a raise always leaves you with more. Only the rate jumps, from ten percent to twelve, so the tax graph bends at the line and pulls away from the ten-percent line, but a bend like that, a corner, is still continuous.

Fix the jets

Now we have everything we need to fix the jets. Their longitude is a piecewise function too, one piece before the line, and one piece after it. The minus eight t is the jets heading west, losing eight degrees of longitude every hour.

We already know one value can’t fix a jump. Nothing says we have to keep the second piece the way it is, though, so let’s add the same number k to every reading after the line. The whole question is which k makes the two pieces meet, and you’re going to find it on the jets themselves.

We’ve paused the jets right at the line over the Pacific. Last time you tried moving one dot, and it never worked, so this time you get to move the whole second piece. Drag k until the two pieces of the longitude graph join into one unbroken line.

Let’s fly them across one more time with the patch. Negative one seventy-nine point nine nine, negative one eighty, negative one eighty point zero one, the count just keeps going, and every screen stays on. Even a map program that just connects the dots gets it right now, one short hop across the line instead of a streak across the entire planet.

The reports don’t tell us exactly how engineers fixed the real jets, but what you just used is a standard way to handle this jump, because the count keeps going instead of wrapping around. You can now check continuity at any point by asking whether f of c exists, whether the limit exists, and whether the two are equal.

When one test fails, you can name the break, removable, jump or infinite, the three common kinds. A single value fixes only the removable kind, and a jump can need a whole piece to move. That’s exactly what you just did for a formation of fighter jets over the Pacific.