Topic 1.16: Working with the Intermediate Value Theorem (IVT)

Why turning works

Let’s name the feet A, B, C and D, going around the table, and pair them across the diagonals, so A goes with C and B goes with D. Watch it rock again. It sits on A and C the whole time and tips back and forth between B and D, like a seesaw balanced on that diagonal. You can see why with two strings stretched across the floor, one under each diagonal. Where they cross, the gold one rides a little higher, so the gold diagonal holds the table up.

I’m going to turn the table exactly a quarter of the way around. Before I do, which pair of feet do you think it’ll be rocking on?

A quarter turn didn’t fix anything, it just swapped which diagonal holds the table up, and that swap turns out to be the key to the whole proof. Let’s keep score with one number, the wobble, which is how far the loose foot hangs off the stone. We give it a plus sign when B and D are rocking, and a minus sign when A and C are.

At zero degrees the wobble was plus six millimeters, and after the quarter turn it’s exactly minus six. In our simple model of the table the floor hasn’t changed at all, because the four feet just traded spots. Now think about the turn in between. Can the wobble get from plus six all the way down to minus six without ever being zero?

The floor has no steps or cracks, so the wobble can’t suddenly jump. It can only change gradually as the feet slide over the stone, and that means somewhere on the way it has to pass through zero. Look at that. At about thirty-seven degrees the two strings meet in the middle, the wobble is zero, and all four feet are down on the stone.

The idea we just used has a name, the Intermediate Value Theorem. Let’s put our wobble on the board, as the table turned from zero to ninety degrees. The theorem names these pieces with letters. The turn runs from a, at zero degrees, to b, at ninety, and the wobbles there, f of a and f of b, are plus six and minus six.

The target, d, is a wobble of zero, and c is the angle where the wobble hits it, about thirty-seven degrees. The theorem says the same thing about any function whose graph has no breaks. Let’s draw a general one, starting at f of a, and ending at f of b.

If f is continuous on the closed interval from a to b, and d is a number between f of a and f of b, then there’s at least one input c between a and b where f of c is exactly d. The curve can wiggle as much as it likes, and even cross the line three times, but it can’t get from one side to the other without crossing at least once.

Keep the letters straight, because this trips people up on the exam. The letters a, b and c are inputs along the bottom, and d is an output, a height on the graph. The theorem won’t tell you where c is, or how many there are. It only promises that at least one exists. Every time you use it, you check two things, that there are no breaks and that the target sits between the two ends.

Let’s use it on a plain equation with no table in sight, p of x equals x cubed plus x minus three, and show it has a zero between one and two. The first thing to check is the condition, and here it’s easy. p is a polynomial, and polynomials are continuous everywhere, so p is continuous on the closed interval from one to two.

Now let’s check the endpoints. p of one is one plus one minus three, which is negative one, and p of two is eight plus two minus three, which is seven. Zero sits between negative one and seven, so the theorem guarantees some c between one and two where p of c is zero.

On the exam, this is the sentence that earns the points. It says why the function is continuous on the interval, it shows the target sitting between the two endpoint outputs, and it ends with at least one c. We didn’t have to find the zero of that cubic at all, because the theorem proves the zero is there without telling us where.

If you do want to know where it is, a graphing calculator can find it. I’ve put p into the calculator for you, so find that zero to three decimal places, the way the exam wants calculator answers.

Can we break it?

Let’s see if we can break the trick. Over by the lemon tree one of the old tiles has lifted, so it sticks up about a centimeter and a half, and one foot is standing right on it. It wobbles here too, and the graph will draw the wobble as the table turns.

This time you’re in charge. Turn the table on the spot and try to find an angle where it stops wobbling. A floor with a step in it gives you a wobble with a jump in it, and a function that can jump can skip right over zero, so the guarantee is gone.

Continuity is the theorem’s first condition, and it has to hold at every angle along the way, because one bad tile anywhere is enough to break it. Last lesson you checked continuity at a single point. A function is continuous on an interval when it’s continuous at every single point of that interval.

Let’s slide an interval along this graph and check it. On the first stretch there’s a sharp corner, which looks dramatic, but the graph doesn’t break, so the function is continuous there, corner and all. Slide it over this hole and it fails, because one missing point is enough to ruin the whole interval, and a jump or a vertical asymptote inside the interval breaks it too.

What happens at the ends of a closed interval? The square root of x on the interval from zero to four simply starts at zero, with nothing at all to its left. At an end of the interval, only the side that’s inside has to behave, so at zero we only need the limit from the right. What happens outside the interval doesn’t count. It matches, because the square root shrinks to zero.

At the left end, the limit from inside the interval has to equal the function’s value there, and at the right end we check that same match from the left. This graph has a jump at two and a hole at five. On which of these intervals is f continuous?

You don’t have to check every point by hand, because the functions you know are already continuous everywhere they’re defined. Polynomials, rational functions, powers and roots, exponentials, logs and the trig functions are all continuous on their domains.

The catch is in that last word, domain. These functions only break at inputs they can’t take, like a zero in a denominator, a negative under a square root, or a zero or negative inside a log.

Let’s find where this one is continuous, q of x equals x squared minus four, over x squared minus x minus two. It’s rational, so it’s continuous everywhere except where the bottom is zero, and the top factors into x minus two times x plus two, and the bottom into x minus two times x plus one, so two and negative one are out.

That leaves three intervals of continuity. The x minus two does cancel, but canceling doesn’t put two back in the domain, it only tells you the break at two is a hole. For the square root, x plus three has to be at least zero, so x can start at negative three, and that endpoint is included, because zero is fine inside a square root. For the log, x minus one has to be positive, so x has to be greater than one, and that endpoint stays out.

Now you try one. Which choice gives all the intervals where h of x, the square root of four minus x, over x plus one, is continuous?

This is why the theorem checks continuity first. One over x starts at negative one and ends at one on this interval, so you might expect it to cross zero. The theorem doesn’t apply here, because one over x isn’t even defined at zero. Sure enough, it never equals zero, and it jumps right over, like the table on that raised tile. When a condition fails, the theorem stays silent, so you can’t conclude anything from it either way, and that’s going to matter in a minute, about thirty-nine kilometers up.

Mach 1 in the missing data

On the fourteenth of October, 2012, Felix Baumgartner stood on a tiny step outside a capsule, thirty-nine kilometers above New Mexico, with a black sky over his head. Then he leaned forward and let go. His Mach number compares his speed with the speed of sound in the air around him, so Mach one means exactly the speed of sound, and anything above one is supersonic.

When he stepped off he was standing still, so his Mach number was zero, and fifty seconds later it was one point two five, faster than sound. Now suppose we lost every single reading in between, so all we have left are those two numbers. Can we still prove there was an instant when he was going exactly Mach one, right on the line?

First we need the condition. M is his speed divided by the speed of sound. His speed can’t jump, because that would take an infinite force, and the speed of sound changes gradually with the air temperature, with no jumps either. Both speeds change continuously, and the speed of sound never drops to zero, so M, their quotient, is continuous too. It’s the rational-function rule again, since a quotient only breaks where its bottom is zero.

M is continuous on the interval from zero to fifty seconds, and one sits between zero and one point two five. That means the Intermediate Value Theorem guarantees a moment c when M of c is exactly one.

Let’s bring the data back. His team’s records put that moment at thirty-four seconds after he jumped, a time strictly between zero and fifty, just like the theorem promised. The theorem could never have told us thirty-four, because it only proves the moment exists, and the measurements found it. He kept speeding up until about fifty seconds, and then the thicker air started slowing him down, so his Mach number came back down through one about sixty-four seconds in.

Let’s black out the data again. Put one pin on each side of the moment he slowed back through Mach one, then squeeze them together until your window is less than two seconds wide.

This one’s trickier. With the pins at zero and a hundred seconds, M is zero at the start and about point four eight at the end. Does the theorem guarantee he hit Mach one in that time?

How the exam asks it

The AP exam asks you to justify claims like Felix’s, and the people grading look for very specific pieces. One student wrote this answer, and at first glance it looks pretty good. A grader would still take a point away from this answer. What’s missing? Later in the year, when an exam question says a function is differentiable, write that it’s differentiable and so continuous, and next unit you’ll see why that works.

Exam questions love tables. This one gives values of a function g that’s continuous from zero to seven, and asks for the least number of zeros g could have. Let’s go through it together. From zero to one, g goes from three to negative one, so it crosses zero, and from one to two it climbs back to two, which is another crossing.

From two to four both readings are positive, so the theorem says nothing there. From four to five and from five to seven the sign flips again, so that’s four separate intervals and at least four zeros. We say at least four, because the graph could wiggle across more times between the readings, and the table can’t rule that out.

Let’s keep the same table and change the target to four. What’s the least number of solutions to g of x equals four on the interval from zero to seven?

Back to the terrace

Let’s finish what we started, back on the terrace, at a new spot by the railing with a different patch of stone under the feet. At zero degrees the wobble is plus five millimeters, and after a quarter turn it’s minus five, the same size with the opposite sign, just like before.

Let’s write the proof the way you would on the exam. The floor has no steps, so the wobble is continuous from zero to ninety degrees, and zero sits between plus five and minus five.

The Intermediate Value Theorem guarantees at least one angle between zero and ninety degrees where the wobble is zero, and at that angle all four feet are on the ground. A real table tilts a little as it turns, which our model ignores, but in 2005 a physicist at CERN named Andre Martin proved the trick works for real square tables too.

You proved it’ll work, so now do it yourself. Find one angle where the wobble is positive and one where it’s negative, then turn the table between them until it settles.

A function is continuous on an interval when it has no breaks at any point in it. When that interval is closed, it hits every value between its two endpoint outputs at least once.

That’s the Intermediate Value Theorem. It proves a moment exists without telling you where, and when there’s a break, like that raised tile, it doesn’t tell you anything at all. The next time a table at a cafe wobbles, skip the napkin. Turn it, nice and slow, and if anyone asks, you can tell them exactly why it works.