Teacher page: the second derivative and the three graphs (AA 5.7)

The one thing to do with the figure

Press Hide the labels and ask which is which.

The three panels stay, the titles go. Students then have to use the structure: where f turns, f′ crosses zero; where f′ turns, f″ crosses zero; and the degree drops by one each time.

That is exactly the question that gets set, and it is much harder than reading three labelled graphs.

The inflexion here is not stationary

At x = 2 the gradient is −3, not zero. f″ crosses zero so the bend changes, but the curve is falling throughout.

Students who have only seen stationary inflexions assume the two always coincide, and then cannot find a non-stationary one at all.

The answers

The function	f = x³ − 6x² + 9x, f′ = 3x² − 12x + 9, f″ = 6x − 12.
Key points	Maximum (1, 4), minimum (3, 0), inflexion (2, 2).
1. f″(1)	−6.
2. f′ has a minimum at x = 2	B, f has a point of inflexion.
3. y of the inflexion	2.

Where the marks go

1 markGetting the zeros in the right places when sketching one graph from another.

1 markNaming which feature of which graph justifies the conclusion.

1 markSubstituting back into f for a coordinate, not stopping at the x value.

What each wrong answer tells you

They give	What it means
0 (Q1)	Gave f′(1), which is zero because x = 1 is stationary.
4 (Q1)	Gave f(1), the y value at the maximum. Three similar numbers are in play.
"f has a minimum" (Q2)	Confusing a turning point OF f′ with f′ crossing zero. One level out.
Sketching f′ as a cubic	Degree did not drop. A structural error visible before any detail.
−3 (Q3)	Gave the gradient at the inflexion rather than the y coordinate. Right number, wrong question.

Other things they will say

"Is an inflexion always a stationary point?" No, and this example is the counterexample. Settle it early or the two ideas fuse permanently.

"Concave up or convex?" The guide says concave-up and concave-down. Both are correct English; match the exam wording for free.

"Can f and f′ cross?" Yes, and it means nothing. They are different quantities on the same axes, so an intersection has no interpretation.

A possible order

	What is happening
1	Recall 5.2. Then ask what f′ does NOT tell you.
2	The three panels labelled, sweeping slowly through x = 1, 2, 3.
3	Hide the labels. Let them argue it out.
4	Reading f from a given f′, which is the standard question.
5	Questions 1 to 3.
6	The non-stationary inflexion, stated plainly.

Two things not to say

Do not always draw the three graphs in the same order with the same labels. Students memorise the picture rather than the relationship.

Do not say "f double dash zero means inflexion". It is the same false shortcut as "f dash zero means turning point", and 5.8 will need it to be false.

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.

Find the stationary pointsFor y = x⁴ − 4x³, solve y′ = 0.
y′ = 4x²(x − 3) = 0 at x = 0 and x = 3.
Classify x = 3Use the second derivative at x = 3.
y″ = 12x² − 24x, so y″(3) = 36, positive, a minimum. y(3) = −27.

2In context

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

The inconclusive oneAt x = 0 both y′ and y″ are zero. Decide what kind of point it is.
y′ = 4x²(x − 3) is negative either side of 0, since 4x² > 0 and x − 3 < 0. The curve falls through x = 0, so it is a stationary point of inflexion, neither a maximum nor a minimum.
Find the inflexionsSolve y″ = 0 for the same curve.
12x(x − 2) = 0 at x = 0 and x = 2, where y = 0 and y = −16.

3Challenge

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

Two kinds of inflexionCompare the inflexions at x = 0 and x = 2 on that curve.
Both change concavity, but at x = 0 the tangent is horizontal, since y′(0) = 0, and at x = 2 it is not, since y′(2) = −16. Stationary inflexions are a special case, not the definition.
Why the test can failState the limitation of the second derivative test and the reliable alternative.
When y″ = 0 the test gives no answer, as at x = 0 here. The reliable method is the sign of y′ either side, which always decides, and the second derivative is only a shortcut for the common case.

Practicalities

Works on a phone. Nothing is loaded from any other site, so it runs behind a school firewall, and nothing a student does is saved or sent anywhere.