Topic 5.10 · teacher page · Higher Level

Running the second derivative

Three panels, one x, and a test that is faster rather than stronger.

The one thing to do with the animation

Move to x = 1 slowly and watch the colour change.

The top curve is coloured by the sign of f″, not f′. At x = 1 the colour flips at exactly the moment the bottom graph crosses zero, and the curve is still rising throughout.

That is the point students miss: concavity is a completely separate question from direction, and a curve can be going up the whole time while changing its mind about how.

The test is faster, not stronger

If f″ is zero at a stationary point the test says nothing at all, and you fall back to a sign check of f′.

y = x⁴ at the origin is the example to have ready: f′ and f″ are both zero and it is still a minimum.

The answers

The functionf(x) = x³ − 3x² + 1, f′ = 3x² − 6x, f″ = 6x − 6.
Stationary pointsx = 0 and x = 2. f″(0) = −6 so (0, 1) is a maximum; f″(2) = 6 so (2, −3) is a minimum.
Point of inflexionf″ = 0 at x = 1, and the concavity genuinely changes, so it is (1, −1).
1. f″(2)6.
2. f″ negative at a stationary pointB, a local maximum.
3. y of the inflexion−1.

Where the marks go

1 markDifferentiating twice correctly.

1 markStating the SIGN of f″ and the conclusion that follows. The sign alone is not an answer.

1 markFor an inflexion, solving f″ = 0 and substituting back into f for the coordinate.

1 markConfirming the concavity actually changes, when the question asks you to justify.

What each wrong answer tells you

They giveWhat it means
0 (Q1)Gave f′(2), which is zero because x = 2 is stationary. They differentiated once and stopped.
−3 (Q1)Gave f(2), the y coordinate. Three similar-looking numbers are in play and the labels matter.
Minimum (Q2)Sign confusion. Concave down is a cap, so the stationary point sits at the top. Draw both shapes rather than restating the rule.
1 (Q3)The x coordinate again. Same habit as Standard Level 5.6, and it does not go away on its own.
0 (Q3)Substituted into f″, which is zero there by definition.

Other things they will say

"Is a point of inflexion always a stationary point?" No, and this example proves it: at (1, −1) the gradient is −3, nowhere near zero. Worth settling early because the two ideas get fused.

"Concave up or convex?" The guide uses concave-up and concave-down, so use those. Convex is correct English and will cost nothing, but consistency with the exam wording is free.

"Why bother with the sign table then?" Because the second derivative test fails when f″ = 0, and because some questions ask for it explicitly. It is a shortcut, not a replacement.

A possible order

 What is happening
1Recall 5.2: what f′ tells you. Then ask what f′ does NOT tell you, and let the gap open.
2The three panels. Sweep slowly, stopping at both stationary points and the inflexion.
3The second derivative test on the worked example, both stationary points.
4Points of inflexion, including the x⁴ counterexample where the test fails.
5Questions 1 to 3.
6Notation: f″(x) and d²y/dx² are the same thing.

Two things not to say

Do not say "f double dash is zero means a point of inflexion". It is the same false shortcut as "f dash is zero means a turning point", one level up, and it is just as wrong.

Do not drop the sign check entirely now that the test exists. The first time f″ comes out zero at a stationary point, a class that has only ever used the test has nothing to fall back 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. Find the inflexionFor y = x³ − 3x² + 4, solve y″ = 0.
    y″ = 6x − 6 = 0 at x = 1, where y = 2.
  2. Name the concavityState where that curve is concave down.
    y″ < 0 for x < 1, so it is concave down on x < 1.

2In context

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

  1. Not every inflexion is stationaryAt that inflexion, find y′ and say what it means.
    y′(1) = 3 − 6 = −3, not zero. The curve is still falling as it changes concavity, so an inflexion need not be a stationary point.
  2. Interpret in contextA company's revenue is rising but its second derivative has just turned negative. Explain what has happened in plain words.
    Revenue is still growing, but the growth is now slowing. The business has passed its steepest moment, which is usually the useful warning and is invisible if only the level is watched.

3Challenge

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

  1. The confusion to pre-emptA student reports that because y″(1) = 0 the curve has a stationary point at x = 1. Correct them.
    y″ = 0 concerns concavity, not gradient. Stationary means y′ = 0. Here y′(1) = −3, so the curve is sloping downwards through a point where it changes from concave down to concave up.
  2. A zero that is not an inflexionFor y = x⁴, y″ = 12x² is zero at x = 0. Decide whether that is an inflexion.
    No. y″ is positive either side, so the concavity never changes. y″ = 0 is necessary for an inflexion but not sufficient; the sign must actually change.

Practicalities

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