Topic 2.4: Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist

The zipper

Back on the court, I’ve put a probe at zero point three seconds, with one flash a tenth of a second before it and one a tenth after. The slope from the left joins the flash before to the ball, and the slope from the right joins the ball to the flash after. The slope from the left is about negative two and a half and the slope from the right is about negative three and a half, so right now they disagree by almost a whole meter per second.

Watch what happens when I make the strobe faster and faster, so the flashes crowd in around zero point three. The two slopes close in on each other like a zipper, and they lock together at negative two point nine four. That’s within a few hundredths of the negative two point nine our table gave us, because there we used a flash on each side too.

The number the two sides zip together on is the derivative, h prime of zero point three, and that’s what differentiable really means, that from the left and from the right you get one finite slope. Now let’s put the probe right on the bounce, at the instant the ball touches the floor, with flashes a tenth of a second on either side.

Make a guess first, whether the two slopes will meet. Then drag the strobe slider so the flashes squeeze in around the bounce, and watch what the slopes do.

Look at what the calculator’s trick does here. It connects the flash before to the flash after and reports negative zero point seven. That isn’t the velocity at that instant, because in our model there isn’t one. It’s just splitting the difference between negative seven and five point six.

Now you set the bounciness. Drag it until the calculator’s centered slope reads exactly zero, then look at the slopes from the left and from the right.

There’s another way to see the difference. If we zoom way in on the top of a hop, the curve straightens out into a flat line with one slope, zero, so a smooth turnaround has a perfectly good derivative. Zoom in on the bounce, though, ten times, a hundred, a thousand, and it never straightens out. It’s a V at every zoom, and that’s what we call a corner.

On the board, the simplest corner there is belongs to the absolute value function at zero, the very example the AP course uses. Coming in from the left, with a step delta x below zero, the difference quotient is negative one every single time, and from the right it’s positive one, no matter how small the step gets. The left-hand limit of the difference quotient is negative one and the right-hand limit is one. Since they’re different, the two-sided limit doesn’t exist, and that means f prime of zero doesn’t exist.

Ask a TI-84 the same question, and nDeriv subtracts two equal heights, a thousandth on either side, so it confidently reports a derivative of zero. It never looks at the point itself, so it never sees the corner. The calculator in this lesson checks both sides and refuses to answer here, which is exactly right.

Where derivatives fail

Corners aren’t the only way a derivative can fail. This is x to the two-thirds, and at zero it comes to a sharp point called a cusp, where the slopes shoot off toward negative infinity on the left and positive infinity on the right. This one is the cube root of x, the other example the AP course uses. It passes through zero smoothly, but standing straight up, so its tangent line there is vertical.

A vertical line has no slope, since its run is zero. From both sides the slopes agree, but they agree on infinity, and infinity isn’t a slope, which is why we said one finite slope. Your TI-84 will tell you the derivative there is exactly one hundred. The cube root of a thousandth is a tenth, so the calculator divides two tenths by two thousandths and gets a hundred.

A smaller step gives a bigger number, and none of them is the derivative. The last way is the easiest to spot, a break in the graph, like a jump or a hole, and the next board shows why no slope survives there. Which one of these has a derivative at x equals zero?

To see why a break kills the derivative, write the change in height from a to a plus delta x as the slope of the little chord between them, times the run, delta x. If f is differentiable at a, that slope closes in on a finite number, f prime of a, while delta x goes to zero.

The change in height goes to f prime of a times zero, which is zero, so the height has no room to jump, and that’s exactly what continuous means. That’s a theorem worth knowing by heart. If a function is differentiable at a point, it has to be continuous there.

If the exam’s function is called R, you write it just like this, R is continuous because R is differentiable. On the 2025 exam, that one sentence was the point students missed most on its whole question. The arrow only goes one way, though. The absolute value function is continuous at zero, with no break at all, and it still has no derivative there, so continuous does not mean differentiable.

One last case is a missing point. If a isn’t in the domain of f, like the hole in this graph at x equals one, the difference quotient has no f of a to start from. That means a isn’t in the domain of f prime either. Suppose g is not continuous at x equals four. What can you say about g prime of four?

Prove it like a grader

Let’s finish the bounce the way an AP grader would want it. First we restart the clock, so zero is the instant the ball hits the floor. Here’s its height on either side of that instant, the end of the fall and the start of the rebound.

The negative seven t says the ball hits the floor falling at seven meters per second, and the five point six t says it leaves rising at five point six. Step one is continuity. From the left the height goes to zero, from the right it goes to zero, and h of zero is zero, so h is continuous at the bounce. That’s good news, since the ball doesn’t teleport.

Step two is the slopes. From the left, the difference quotient is negative seven delta t minus four point nine delta t squared, all over delta t. Dividing through by delta t leaves negative seven minus four point nine delta t, and that heads to negative seven as delta t goes to zero. From the right, the same steps give five point six minus four point nine delta t, which heads to five point six.

Step three is the sentence that earns the point, and it names exactly which limits disagree. H is continuous at t equals zero, but the left-hand and right-hand limits of its difference quotient are negative seven and five point six, and since they aren’t equal, h prime of zero does not exist.

Both numbers make sense, too, because the ball falls into the floor at seven meters per second and bounces up at eighty percent of that, five point six. These two formulas meet at x equals two, so decide whether their slopes meet too. Is f differentiable at x equals two?

Every corner in the picture

Look at this photo again with everything you know now. Between the flashes on either side of almost any instant, you can read off how fast the ball is rising or falling, the same way you’d read a table on the exam. The slow-motion video on your phone works the same way, because every frame is just another flash.

Wherever the left slope and the right slope zip together on one finite number, the derivative exists, and wherever they don’t, at a corner, a cusp, a vertical tangent or a break, it doesn’t, however confident the calculator sounds. Let’s end with you on the probe. Drag it onto every instant in this photo where h prime does not exist.

Let’s zoom in on the short time a real ball spends pressed against the floor. It squashes for a split second, and for this zoom we’ll follow the ball’s center. The ball’s center traces a tiny smooth valley instead of a corner. Now you round the corner. Give the ball a contact time with the dial, then squeeze the flashes until the two slopes meet.

Look how close the flashes had to be, though. With this squash, they had to be hundred-thousandths of a second apart before the slopes agreed, so to any ordinary camera, a bounce still looks like a corner.

Our model makes the bounce instant, and in that model the corner is real, and now you can prove it. You came in with a photo of a bouncing ball. Now you can read how fast it’s rising or falling at almost any instant, and you can point to exactly where a derivative can’t exist, even when your calculator swears it can.