Topic 5.2 · teacher page

Running increasing, decreasing and stationary

The toggle that ends a misconception most classes carry into the exam, and how to make the sign test a habit rather than a ritual.

The one thing to do with the animation

Get the wrong answer on the board first. On f(x) = x³ − 3x, ask what f′ = 0 means. The class will say "maximum or minimum", and on that curve they are right, twice. Write it up as a rule.

Then switch the toggle to x³ and slide to zero. Gradient zero, increasing on both sides, no turn. Their rule is now visibly broken, and they wrote it, so the correction lands. Replace it with "solve f′ = 0, then check the sign either side".

If you only have five minutes on this page, that is the five minutes.

What the colours are doing

The upper curve is drawn green where f′ is positive and red where it is negative, segment by segment, and the lower graph is tinted the same way above and below its axis. So the link between the sign of the derivative and the direction of the curve is something students see rather than something you assert. Point at the colour change and the zero crossing in the same breath; they are the same event.

The answers

Sign tablef′(x) = 3(x² − 1) is zero at x = ±1. f′(−2) = 9, f′(0) = −3, f′(2) = 9.
Turning pointsLocal maximum at (−1, 2), local minimum at (1, −2).
1. Where increasingB. x < −1 and x > 1.
2. Positive both sidesC. A stationary point that is neither a maximum nor a minimum.
3. y of the minimum−2. From f(1) = 1 − 3.

Where the marks go

1 markSolving f′(x) = 0. Routine.

1 markSubstituting back into f for the y value. Half the class stops at the x value and loses this every time.

1 markStating the intervals with correct inequalities, in terms of x. "Where f′ is positive" is a restatement, not an answer.

1 markJustifying max against min. A sign check either side is sufficient at this level; the second derivative test is Higher Level and not needed here.

What each wrong answer tells you

They sayWhat it means
x > 1 only (Q1)They solved the inequality on one side and stopped. Ask them to test x = −2.
−1 < x < 1 (Q1)Sign flipped: that is where f′ is negative. Very common, and usually a hurry rather than a misunderstanding.
"Everywhere" (Q1)They see x² and conclude positive, forgetting the minus 3. Substitute x = 0 in front of them.
Maximum or minimum (Q2)The target of the lesson. If this is still appearing at the end, go back to the toggle rather than explaining again.
1 (Q3)Gave the x coordinate. The single most expensive habit in this sub-topic.
0 (Q3)Substituted into f′ rather than f. Worth naming out loud: f′ gives gradients, f gives heights.
2 (Q3)Right value, wrong turning point. They did the work and mixed up which was which.

Other things they will say

"It is the maximum, so it is the biggest value." f(−1) = 2 but f(3) = 18. The word local is doing real work. If an interval is given, the ends have to be checked too, and that is a frequent exam structure.

"Can f′ be zero over a whole stretch?" Yes, for a constant function. A good question and it costs ten seconds.

"Is a stationary point of inflexion examinable?" Recognising that f′ = 0 does not guarantee a turning point is fair game at Standard Level. The vocabulary and the concavity treatment belong to Higher Level, so teach the idea and go light on the name.

A possible order

 What is happening
1Sketch a hill on the board. Where is the gradient positive, negative, zero? Get the three statements agreed in words.
2The animation on x³ − 3x. Let them state the "f′ = 0 means a turning point" rule and write it up.
3Switch to x³. Break the rule. Replace it with the sign check.
4The sign table, properly laid out, and substituting back into f. Make them write the coordinates as pairs.
5Questions 1 to 3. Take question 2 to the whole class and question 3 individually.
6Local against global, with f(3) = 18 as the counterexample.

Two things not to say

Do not say "f′ = 0 gives the turning points". It is the sentence this whole page exists to prevent, and it is very easy to say while pointing at a curve where it happens to be true.

Do not reach for the second derivative to settle max against min. It is not needed at Standard Level, it is an extra thing to get wrong, and the sign check is both sufficient and more honest about what is going on.

Questions to set

Three tiers, ramping the way practice should: the method on its own, then the method inside something real, then a challenge. Set the tier the class in front of you needs rather than one undifferentiated sheet. Answers are given so these can go straight onto a board.

1Fluency

The method on its own, with friendly numbers. Set these first and move on quickly once they are secure.

  1. Sign of the gradientA graph of f rises, levels, then falls. State the sign of f′ in each part.
    Positive, then zero, then negative.
  2. Match the zeroWhat is true of the graph of f at a point where f′ = 0?
    The tangent is horizontal. It is a stationary point, which may be a maximum, a minimum or neither.

2In context

The same skill inside a real situation, where the first job is working out what is being asked.

  1. Sketch from the derivativef′(x) > 0 for x < 2 and f′(x) < 0 for x > 2. Describe f.
    It rises to a maximum at x = 2 then falls. The value of the maximum cannot be found from this information alone.
  2. Never negativeFor y = x³, find y′ and explain what that means about where the curve falls.
    y′ = 3x², which is never negative. The curve never falls anywhere, so it has a stationary point at x = 0 that is not a maximum or a minimum.

3Challenge

Reasoning, working backwards, or spotting an error. These are where the top grades are decided.

  1. The common mix-upA student says that because f′(3) = 0, f has a maximum at x = 3. Explain why that does not follow.
    f′ = 0 only says the tangent is horizontal. It could be a minimum, or a point of inflexion as x³ has at 0. The second derivative or the sign of f′ either side decides which.
  2. Read the second derivative toof′ is positive everywhere but decreasing. Describe the shape of f.
    f rises everywhere, but less and less steeply, so it is concave down. Growth that is slowing but still growth, which is the shape most students draw as a plateau.

Practicalities

Two synced panels fit a phone screen in portrait. The toggle rebuilds both graphs, so a student can flip back and forth freely. No external library, nothing stored, nothing sent.