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Chapter 6 · Application of Derivatives
A sign change either side of a critical point decides which kind it is
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The idea
Theorem 3 is the previous module's sign argument applied twice at once, and its real content is the third part, not the first two: a critical point where the derivative keeps its sign is neither kind of extreme, and the chapter gives that case a name and a figure rather than leaving it as an exception. Reading the theorem that way makes the classification a single question — does the sign change as the input crosses the point — with three possible answers instead of two, and it removes the commonest error in the module, which is to find a critical point and assume it must be one thing or the other. It also explains why the test is stated for a function that is merely continuous at the point: the two halves of the argument live on either side of it and never need a derivative at it.
What you should be able to do
- State all three parts of Theorem 3 and say which hypothesis each needs
- Explain why the test asks only for continuity at the critical point, not differentiability
- Classify a critical point by testing the sign of the derivative on each side
- Build a table of nearby signs and read the verdict off it
- Recognise the case where the sign does not change, and name it
- Apply the test where the derivative does not exist at the point, by examining the two sides separately
- Say why the first test is the only available one at a point of non-differentiability
- Handle a derivative with a repeated factor, and predict from the factor whether the sign changes there
- Distinguish a local maximum from the local maximum value, in the chapter's own wording
- Read the Summary's version of the test and say what it leaves out
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| First Derivative Test | the rule classifying a critical point by the sign change across it | printed in this chapter (Theorem 3, §6.4, Part I p. 164), and printed earlier on Part I p. 153 for a different theorem |
| critical point | a point where the derivative is zero or does not exist | printed in this chapter (the boxed note, §6.4, Part I p. 164) |
| local maximum value | the value taken at a point of local maxima | printed in this chapter (the boxed note, §6.4, Part I p. 165) |
| local minimum value | the value taken at a point of local minima | printed in this chapter (the boxed note, §6.4, Part I p. 165) |
| point of inflection | the verdict when the sign does not change across a critical point | printed in this chapter (Theorem 3 part (iii) and Fig 6.13, Part I p. 164) |
| point of inflexion | the same verdict, spelt the chapter's other way | printed in this chapter (Example 18, Part I p. 166, and the Summary, Part I p. 186) |
| changes sign | the event the whole test turns on | printed in this chapter (Theorem 3, §6.4, Part I p. 164) |
| not differentiable | possessing no derivative at the point, the second way to be critical | printed in this chapter (Theorem 2 and the boxed note, Part I p. 164) |
| continuous | unbroken at the point, the only hypothesis Theorem 3 places there | printed in this chapter (Theorem 3, §6.4, Part I p. 164) |
| sign change table | the two-column layout recording the derivative's sign on each side | an added name for the layout the chapter prints after Example 17 without a heading |
| test input | a convenient number picked inside each side to settle the sign | an added term; the chapter supplies such numbers in brackets and does not name the technique |
Where people slip up
- "A critical point is either a maximum or a minimum." Part (iii) exists precisely for the third case, and both Example 18 and Miscellaneous Exercise Q10 land on it. Teach the test as a three-way question from the first sentence.
- "The derivative has to exist at the critical point." Theorem 3 asks only that the function be unbroken there. Example 19 runs the test at a point with no derivative, and Fig 6.14 draws two such points. This is the test's main advantage over the next one.
- "I should evaluate the derivative at the critical point to see the sign." It is zero there, or undefined. The sign is read on each side, at a convenient input strictly between this critical point and the next.
- "Any nearby number will do as a test input." Any number strictly between this critical point and the neighbouring one will. A number chosen beyond the next critical point reports the wrong sign, and with three critical points that mistake is easy.
- "A repeated factor still changes sign." A factor raised to an even power never does. Miscellaneous Exercise Q10's squared factor is what makes one of its three critical points a non-extreme, and a student who tests numerically without noticing will get it right by luck and be unable to explain it.
- "The point and the value are the same answer." They are two answers, and the chapter has a boxed note about it. Questions in Exercise 6.3 Q3 ask for both explicitly.
- "The two spellings must mean two different things." They do not. The chapter uses one spelling on Part I p. 164 and the other from Part I p. 166 onward, including in the Summary. Say so once and move on.
- "If the second test is available I should always use it." The chapter's own note after Example 23 says otherwise: it chose the first there because it was shorter. Choose by the algebra in front of you.
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Worked answers: Exercise 6.1 · Exercise 6.2 · Exercise 6.3 · Miscellaneous Exercise · this video explains Exercise 6.3 Q3, Miscellaneous Exercise Q10
Transcript3,004 words
You have a critical point. A place where the rate is nought, or where there is no rate at all. What you do not have is an answer. Here are three curves. Each one has a critical point at the same place, marked the same way, and from that mark alone they are indistinguishable. On the first, the curve rises to the mark and falls away after it. A top.
On the second, it falls to the mark and rises away. A bottom. On the third, it rises to the mark, goes flat for an instant, and carries straight on rising. Neither. So finding the candidate was never the answer. It was the shortlist. The question now is how to tell which of the three you are looking at, and the whole of the answer is one word: sign. Here is the rule that settles it.
Take a critical point. Look at the rate just to the left of it, and at the rate just to the right of it. If the rate is positive on the left and negative on the right, the curve was climbing and then was falling. The point is a local maximum. If the rate is negative on the left and positive on the right, the curve was falling and then was climbing. The point is a local minimum.
And if the sign is the same on both sides, the point is neither. That is it. One question, asked twice, once on each side. Notice what the rule never asks for. It never asks for the rate at the point. It asks for the rate near the point, on each side, and that is a different thing entirely. That distinction is not a technicality. It is the reason this test works where nothing else does.
At a critical point the rate is nought, or it does not exist. Either way there is nothing there to read a sign off. So the rule steps to one side, reads a sign, steps to the other side, reads a sign, and never once evaluates anything at the point itself. Which means the point is allowed to be a corner. Take three plus the size of the input. To the left of nought the rule is three less the input, and its rate is minus one. To the right it is three plus the input, and its rate is plus one.
At nought the two sides settle on different numbers, so there is no rate at nought at all. And the test does not care. Negative on the left, positive on the right: a local minimum, with value three. This checker measured every rate in this video the same way, as a limit of a difference quotient taken separately from each side. Nothing was differentiated symbolically anywhere. At the corner, the left side settles on minus one, the right on plus one, and the two together settle on nothing. Measured, not asserted.
Now the third part, the one that gets read as an afterthought. It is not an afterthought. It is the part that stops you from guessing. Without it, the sentence in your head is: I have found a critical point, so it is a maximum or a minimum, and I only have to work out which. That sentence is false, and it is the commonest mistake in this whole subject.
Here is how common the third case actually is. Take rates built as a product of three factors, each raised to a power of one, two or three: twenty-seven different rates, eighty-one critical points between them. Twenty-seven of those eighty-one turn out to be neither a maximum nor a minimum. Exactly a third. Not an exception. Not a special case. A third of everything the family produced. This case has a name. It is called a point of inflection, and you will meet the word spelt two ways. They mean the same thing.
Whether the name is deserved is a question this video comes back to, because it turns out not always to be. Take a worked example and run the test properly. A cubic whose rate factors into three, times the input less one, times the input plus one. That rate is nought at one and at minus one, and nowhere else. Two critical points, and no opinion yet about either. Take minus one first. Just to its left the two factors are both negative, so their product is positive, and the rate is positive.
Just to its right the first factor is still negative and the second has turned positive, so the product is negative. Positive then negative: a local maximum. And the value there is five. Now take one. Just to its left the product is negative. Just to its right both factors are positive, so the product is positive. Negative then positive: a local minimum. And the value there is one. Two answers per point, and both of them are asked for. Where, and how much.
The point of local maxima is minus one. The local maximum value is five. Those are different sentences and you are expected to write both. That worked because of something nobody said out loud. Look at which numbers were used to read those signs. A number just below the critical point, and a number just above it. Not any old number on that side. A number close in. Here is why it matters, and it is worth a measurement rather than a warning.
Take every critical point in this video, and instead of reading the sign right beside it, read the sign at every eighth of a unit out to four units on each side. That is nine hundred and forty-seven readings. A hundred and forty-eight of them disagree with the reading taken right beside the point. They fall on nine of the thirty-two sides walked, and the nearest disagreement sits only a unit and three eighths from its own point.
So somewhere on this side and just beside the point are different instructions, and the difference is not small. The rule to carry is this. Pick your test number strictly between this critical point and the next one along. Between two consecutive critical points, the rate cannot change sign, because to change sign it would have to pass through nought, and that would make another critical point in between. Cross one, and you are reading somebody else's answer.
Now a curve that lands on the third case. A cubic whose rate is six times the square of the input less one. That rate is nought at one. So one is a critical point, and the shortlist has one name on it. But a square is never negative. Just to the left of one the rate is positive. Just to the right of one the rate is positive. The sign never changes. The curve pauses and carries on.
So one is neither a maximum nor a minimum, and a walk across the whole stretch confirms it is the only critical point there is, with nowhere the rate fails to exist. The comparison agrees. Look at the values around the point rather than at any rate, and no neighbourhood makes one a highest value or a lowest one. There is a wider fact hiding in that. A rate that never changes sign anywhere leaves a curve with no turning points at all. It climbs, or it falls, the whole way.
There is a second test for this job, which uses the rate of the rate. It is often quicker, and it is not always available. It is not available at a corner, because there is no first rate there to take the rate of. So at three plus the size of the input, at a peak with a sharp point, at any join where two pieces meet at an angle, the sign test is not the shorter road. It is the only road.
That is the honest reason to learn it first. But it is also sometimes just quicker, and it is worth seeing that chosen deliberately. Take the shortest distance from a point sitting up the vertical axis to a curve that squares its input. Write the distance in terms of how high up the curve you are. There is one place where its rate crosses nought. Measured at three different starting heights, that place lands half a unit below the height you started at, every time.
The rate runs negative below it and positive above it, so it is a lowest value, and the distance there is the square root of one less than four times the height, halved. One sign on each side, and the question was finished. No second rate needed. Here is the one place in this subject where you can call the answer before you substitute anything. Suppose a rate arrives already factored. Something like the input less two, cubed, times the input plus one, squared, times seven times the input, less two.
Three critical points: two, minus one, and two sevenths. Now think about what happens to each factor as you walk across its own root. A factor raised to an odd power changes sign there, because the thing inside it does. A factor raised to an even power does not, because a square, or a fourth power, is never negative. So the sign of the whole product changes at a root exactly when that root's exponent is odd.
At minus one the exponent is two. Even. No sign change. Neither kind of extreme. At two sevenths the exponent is one, and the product runs positive then negative: a local maximum. At two the exponent is three, and the product runs negative then positive: a local minimum. All three verdicts, from parity alone. That is a prediction, so it deserves to be checked rather than believed. Take the twenty-seven rates from earlier, with their eighty-one critical points.
For each one, read whether the sign actually changes, from the rate's own values on either side. Then, separately, predict whether it ought to change, from the exponent and nothing else. The two agree at all eighty-one. And both answers get used, so the agreement means something: fifty-four of the roots carry an odd exponent, and twenty-seven carry an even one. Those twenty-seven are exactly the ones that come back as neither kind of extreme. The remaining fifty-four split into thirty-four minima and twenty maxima.
One more thing about how that family was built, because it matters for whether any of this is evidence. The rates were multiplied out into coefficients, and each rule was built from its rate by arithmetic on that list of coefficients. Then the rule's rate was measured back, by difference quotient, at nine inputs apiece. It matched at two hundred and forty-two of the two hundred and forty-three. The one miss is the steepest rule of the family read farthest out, where a rate above seventeen hundred is being asked to close on a millionth. The tolerance is absolute, so it bites hardest where the numbers are largest.
Back to that third verdict, and to its name. A point of inflection is supposed to be a place where the curve changes the way it bends. Concave one side, convex the other. That is a claim about the curve, not a label for the cases that did not fit. So it can be checked. Bending can be measured without any rate at all. Take the value at a point, take the values a step either side, average those two, and see which is higher.
If the average of the neighbours is above the middle, the curve is cupping upward there. If below, it is capping over. If they are equal, it is straight. Now run that on every point in this video that the sign test called neither a maximum nor a minimum. There are four of them. At three, the bending genuinely does change across the point. The name is earned. At the fourth, it does not.
That fourth one is a corner where a gentle straight climb meets a steeper straight climb. Positive rate on the left, positive rate on the right, so the sign test says neither, and calls it a point of inflection. But both pieces are straight. The bending is nought on the left and nought on the right. Nothing changes. So the third verdict is sound as a verdict and unreliable as a name. What it really says is: not an extreme. It does not always say: the bending turned over.
Now the thing that is usually left out, and it is the largest thing in this topic. Read the rule once more, slowly. If the rate is positive throughout some stretch to the left, and negative throughout some stretch to the right. Throughout. That is a hypothesis, and hypotheses can fail. What if there is no stretch on the left where the rate keeps one sign? Here is a rule where that happens. Take the square of the input, and lift it by a wave whose crowding grows without bound as the input approaches nought.
Away from nought it is an ordinary smooth curve. At nought its value is nought. And every value it takes anywhere else is above nought, because the square is positive and the lift never brings it down to zero. So nought is a local minimum. Not by any rate argument. By looking at the values, which is the definition. Now ask the sign test what it thinks. Walk out from nought, a two hundred and fifty-sixth at a time, as far as a half, on each side.
The rate reads on a hundred and twenty-six of the hundred and twenty-eight inputs on each side. And it does not keep one sign. It changes sign ten times on the right and nine on the left. Worse than that for the test: on each side there is a pair of inputs carrying opposite signs already within a sixty-fourth of the point. So there is no neighbourhood you could hand the test that has a single sign in it. Not a small one, not any one.
Put that pair of readings to the three parts. Positive then negative? No. Negative then positive? No. The same sign on both sides? No. None of the three applies. There is a fourth answer, and the fourth answer is: this test has nothing to say here. And it is not saying nothing about nothing. There is a genuine local minimum sitting at that point, which the test has missed entirely.
Take the wave away and the same square is covered at once: negative on the left, positive on the right, a local minimum. It was never the square that broke it. So hold the rule at the right strength. It is not: every critical point is a maximum, a minimum, or a point of inflection. It is: if a critical point has a definite sign on each side, then those two signs decide which of the three it is.
The if is doing real work. Drop it and you have a rule that is false. In practice, for every polynomial you will meet, and every reasonable combination of the usual functions, the hypothesis holds and you will never see the fourth case. That is a fact about the functions you have been handed, not a fact about the theorem. Knowing which of the two you are relying on is most of what it means to understand a theorem at all.
Now the standard exercise, run properly, because every one of these wants two answers and not one. The square: lowest at nought, and the lowest value is nought. A cubic less three times the input: highest at minus one with value two, and lowest at one with value minus two. The sine plus the cosine, on a quarter turn: the rate crosses nought at half a quarter turn, the signs run positive then negative, and the highest value there squares to two.
The sine less the cosine, over a whole turn: it turns twice, once each way, at three eighths of a turn and at seven eighths. Both values square to two, and they sit on opposite sides of nought. A cubic with two turns: highest at one with value nineteen, lowest at three with value fifteen. Half the input plus two over it, on the positive numbers: lowest at two, with value two.
The reciprocal of two more than the square: highest at nought, with value one half. And the input times the square root of one less it, which lives only between nought and one: its rate crosses nought at two thirds, and the highest value is twice the square root of three, over nine. That last one is the one that punishes carelessness. Walk your test inputs past one and you are asking the rule about numbers it was never defined at.
So the whole of this topic is one question asked on two sides of a point. Does the sign change, and if so, which way round? Positive to negative is a top. Negative to positive is a bottom. No change is neither, and that third answer is a third of what a whole family of rates produced, not a curiosity. The rate at the point is never used, which is exactly why the test reaches corners that the other test cannot.
Read your signs close in, strictly between one critical point and the next, or you will read somebody else's answer. And when a rate arrives factored, the parity of an exponent tells you the verdict before you substitute anything: odd changes sign, even does not, checked here at eighty-one roots out of eighty-one. Two answers per question, always. The point, and the value at it. And keep the if. The test speaks when each side has a sign. When neither side has one, the test is silent, and something can still be there.
Where this fits
Either side of this one
- Local maxima and minima, and why critical points include the non-differentiable onesClass 12 · Ch 6, Application of Derivatives
- The quicker second-derivative test, and the case where it tells you nothingClass 12 · Ch 6, Application of Derivatives