Using the sign of the derivative to split the line into rising and falling intervals

0:00A claim about every pair of inputs is an infinite claim. That is what increasing means, and it is why proving it takes an argument rather than a graph. For a straight line you can do it in three lines of algebra. For anything else, you are in trouble. Try it on a cubic. Take two inputs, call them a and b, with a below b, and try to show the outputs come out in the same order.

0:29You are now looking at a difference of two cubes and hoping it factors into something whose sign you can read. Sometimes it does. Usually it does not. So the method that was exactly right in the last topic does not scale, and something has to replace it. Here is what would rescue it. Suppose you could look at one thing, at one input at a time, and have that settle the infinite claim.

0:59You already have such a thing. The rate of change. It is defined one input at a time, you can compute it by rules you already know, and it carries a sign. So the question becomes: does the sign of the rate, held right across an interval, settle what the outputs do over every pair in that interval? If yes, the whole subject collapses into arithmetic. Find where the rate is zero, cut the interval there, and read one sign per piece.

1:32That is the entire method, and everything else in this topic is either why it works or where the statement of it goes wrong. So here is the theorem. Take a rule that is unbroken on a closed interval and has a rate everywhere inside it. Then three things. If the rate is at least zero throughout the inside, the rule is increasing on the interval. If the rate is at most zero throughout, it is decreasing.

2:04And if the rate is exactly zero throughout, the rule is constant. Read those first two again, because the comparison in each of them carries a bar. At least zero. Not above zero. And that is not a slip. It is the only version that lines up with the definition of increasing, which also carries a bar, and which also lets a flat stretch count. A theorem whose hypothesis forbade a flat stretch could not deliver a conclusion that allows one.

2:40The proof is four lines, and it is worth having. Take two inputs from the interval with the first below the second. Now apply a result about average rates, which I will come to in a moment: the difference of the two outputs equals the rate at some input strictly between them, times the difference of the two inputs. The difference of the inputs is positive, because the first is below the second.

3:09The rate at that in-between input is positive. So the product is positive, which means the second output exceeds the first. That is every pair, and the theorem is proved. Except look at line three. The rate at that in-between input is positive. Positive. The theorem four lines above assumed only that the rate was at least zero. So the proof has quietly helped itself to the strict version and used it to close the non-strict claim.

3:44Now that result the proof leans on, because it deserves a look. It says this. Over an interval, the average rate across the whole of it is achieved as an actual instantaneous rate, at some input strictly inside. Draw the chord between the two endpoints, and somewhere inside there is a point where the curve runs exactly parallel to it. It is a beautiful statement and it is the engine of everything in this topic.

4:16It is also almost always quoted rather than proved, which means most readers meet it as a name. So I did not quote it. I put it to seven different stretches: polynomials of degree two, three, four and six, and two curves built out of the sine and the cosine. For each one, the average rate across the stretch was computed, and then the inside was searched for an input whose measured rate matched it.

4:45Seven stretches, seven such inputs found, every one of them strictly between the ends. That is not a proof of the result. It is a demonstration that the thing the proof rests on is not a formality. And now it gets awkward. Quoted alongside that theorem is a more general version, and it changes the signs again. It says: a rate above zero on the inside, together with the rule being unbroken on the interval, makes the rule increasing.

5:21Above zero. Strict. So within a few lines of each other you have three statements of the same criterion, and all three are in circulation. The theorem, with bars under both signs. Its own proof, using a strict comparison in the middle. And the general version below it, strict again. The usual advice at this point is to pick one and move on. I would rather measure them. So all three were put to the same population.

5:55Twenty-three stretches, each one a rule read on an interval. For each of them, two separate things were worked out by two separate machines. The verdict, taken from the definition, run over every pair of inputs the way the last topic ran it. And the sign of the rate, measured across the inside as the limit of a difference quotient. Then each of the three readings was asked: where does your hypothesis hold, and when it holds, is the definition's verdict what you claim?

6:30Here is the first result and it is the good news. None of the three is ever wrong. Across all twenty-three stretches, no reading once grants a verdict that the definition goes on to refuse. Every disagreement between them is about how often each one speaks, never about what it says. So how often does each one speak? The theorem's version, the one with the bars, settles thirteen of the twenty-three.

7:01The strict version its own proof uses settles eleven. Two stretches, then, where the theorem answers and its proof does not. And those two are not random. They are exactly the stretches where the rate reaches zero somewhere inside without changing sign there. A flat shelf, and the cube on a stretch about its own turn. On both, the rate is at least zero throughout, so the theorem applies. On neither is the rate above zero throughout, so the proof's argument does not cover them.

7:37That is the gap, with a number on it. The proof establishes more than the theorem claims, from a hypothesis stronger than the theorem assumes, and then it is used to close a claim it does not reach. Right. Method. It is four moves and it never changes. One: take the rate. Two: solve for where the rate is zero. Three: cut the domain at those inputs. Four: read the sign on each piece, and hand back one verdict per piece.

8:12That is it, and every worked example you will ever see in this topic is those four moves with different arithmetic in the middle. One warning before the examples. The answer is the list of pieces with their verdicts. The table you build on the way is bookkeeping, and if a question says prove, a table is not what it is asking for. The simplest case: one cut point. Take the parabola whose rate is two less than twice the input.

8:46That rate is zero at exactly one input, at two. Below two it is negative, above two it is positive. So the line splits into two pieces and the rule falls on the lower one and rises on the upper one. Draw it as a number line with arrowheads at both ends and a single tick at the cut. That picture is the whole answer. And notice how the cut point was found here.

9:14Not by looking it up. The rate was measured at a spread of inputs, a sign change was bracketed between two of them, and the bracket was halved until it closed on two. Every cut point in this video was found that way, which means a cut the rule does not actually have cannot get in, and one it does have cannot be quietly dropped. Now two cut points. A cubic whose rate factors into twelve, times one linear piece, times another.

9:48It vanishes at minus two and at three, and the halving finds both. Two crossings give three pieces, always: cut a line in two places and you get three. On the lowest piece both linear factors are negative, so their product is positive. On the middle piece the signs differ, so the product is negative. On the top piece both are positive. Plus, minus, plus. Rises, falls, rises. And here is the thing worth noticing about that table.

10:22The sign in each row and the verdict in each row were measured by completely different machinery: one from a difference quotient at nine inputs, the other from every pair of inputs on the piece. They agree on every row. That agreement is the theorem, happening in front of you, rather than being assumed. There is a shortcut everybody uses here and half of them feel guilty about it. Instead of reasoning about the signs of the factors, pick one convenient input inside the piece and see what the rate does there.

11:02Is that rigorous? Yes, and here is exactly why. You cut at every crossing. So no piece has a crossing left inside it. So the rate cannot change sign anywhere within a piece, and the sign at one input inside is the sign at all of them. That is a theorem about your own construction, not a hope. Across the twenty-three stretches here, twenty-two of them carry a single sign right across, and I will come back to the one that does not, because it is the most interesting object in this video.

11:40Now the same four moves, but restricted. Take the sine of three times the input, on a quarter turn. The rate vanishes where the tripled input is a right angle. The tripled input runs from zero up to three right angles as the input crosses the quarter turn, so it passes a right angle once, and only once, inside the stated range. One cut point, at a sixth of a half turn.

12:10Two pieces: rising, then falling. The move that matters here is the discarding. Solving the equation gives you a whole family of inputs where the rate vanishes. Most of them are outside the interval you were given, and they are not cut points of this problem. Keep only the ones inside, and say that you are doing it. Look at what those two pieces were reported as. Open. The cut point belongs to neither of them, and neither does either end of the interval.

12:45That is the honest report, because the sign argument only ever spoke about the inside. But the answer everybody wants is closed pieces. So there is a separate step, and it is the only place the unbroken hypothesis does any visible work in this whole topic. Because the rule is unbroken at the ends, the verdict extends to them. I checked it rather than assuming it: the definition was run again on both pieces with both ends put back in, and both verdicts survived.

13:21Rising stayed rising, falling stayed falling. So the closing is safe here. But it is a step, it has a hypothesis, and a student who silently writes a closed bracket has skipped an argument rather than made one. Two special cases, and both of them are worth more than they look. First: what if there are no cut points at all? Take a cubic whose rate is one more than three times a square.

13:52A square is never negative, so that rate is at least one, everywhere. It never reaches zero, so there is nothing to cut at, and the search across the whole stretch finds exactly zero crossings. One piece. The whole line. Rising on all of it. This is the shape of every question that says prove rather than find: show the rate keeps one sign and you are finished, and there is no table to draw because there is only ever one row.

14:27Second special case, and this is the one students get wrong. A zero of the rate is not the same thing as a cut point. Here is a rule built to make the difference impossible to miss. Its rate is six, times one less than the input, times a squared factor, times another squared factor. That rate vanishes at three inputs: minus one, one, and three. But a squared factor never goes negative.

14:57So at minus one and at three the rate touches zero and comes back on the same side. Only at one does it actually cross. Three zeros, one crossing, two pieces. A reader who cuts at every zero of the rate has invented two boundaries that are not there and will report four pieces where there are two. And notice this is a real measurement, not a rearrangement of algebra: the rate at minus one and at three was measured, from both sides, and it settles on zero at both.

15:34They are genuinely zeros. They are just not crossings. Now, does the implication run backwards? If a rule is strictly increasing, must its rate be above zero throughout? No, and the cube is the counterexample everybody should carry. Take the cube on a stretch about zero. By the definition, run over every pair, it is strictly increasing: no two inputs give the same output and the order is never reversed.

16:07And its rate at zero is zero. So the strict sign hypothesis fails at that one input, while the strict conclusion holds right across. Which tells you the sign rule is a one-way street. A sign gives you a verdict. A verdict does not give you back the sign. And if a question asks you to prove something strictly increasing and its rate touches zero somewhere, the sign rule will not close it and you go back to pairs.

16:40Which brings me to the one stretch out of the twenty-three that does not carry a single sign. Consider a quantity that rises by ten for every one you feed it, but which you can only ever pin down to within a tenth. Is it increasing? Certainly. Move the input along by a fifth and the output moves by two, which swamps the tenth of slack at either end, so every pair of inputs is ordered and the definition grants it increasing, and strictly increasing too.

17:15Now ask for its rate. The difference quotient carries a fifth of a unit of slack divided by the increment. As the increment shrinks, that does not settle down. It blows up. I asked for the rate at nine inputs across the stretch and got nothing at all, nine times out of nine. So every one of the three sign readings is silent here. Not wrong. Silent. A rule can be increasing with no readable rate to prove it by, and that is precisely what the theorem's differentiability hypothesis is buying you: not truth, but access.

17:59One last thing, and it is the commonest way to lose a mark here. Once you have cut a stretch into pieces with different verdicts, the stretch itself has no verdict. The cubic rises, falls and rises. So on the whole line there is a pair of inputs whose outputs go up and another pair whose outputs go down, and not one of the five parts of the definition can hold across it.

18:27I ran the definition on the whole stretch for four different rules with cut points inside them, and every single one came back satisfying nothing at all. The pieces are the answer. The stretch is not. Three things to carry away. The theorem's hypothesis is that the rate is at least zero, with the bar, because that is what matches a definition that also carries a bar, and the strict version four lines away settles two fewer of the twenty-three stretches here.

19:03Cut at crossings, not at zeros, or a rule with three zeros and one crossing will give you four pieces where there are two. And the sign rule runs one way only, which is why the cube, strictly increasing with a rate of zero at the origin, is the example to keep.