Topic 5.6 · teacher page

Running stationary points

The endpoint nobody checks, and why the switchover on this curve is exact.

The one thing to do with the animation

Ask for the greatest value while b is at 3, then drag. With the interval ending at 3 the answer is the local maximum, 16, and the class is right. Keep dragging. At b = 4 the endpoint draws level, and past it the greatest value is somewhere the gradient is not zero at all.

The red ring jumps from the turning point to the end of the interval in front of them. That is the moment: differentiating alone can never find an endpoint, because nothing happens to the gradient there.

Why this curve

f(x) = x³ − 12x has its local maximum at (−2, 16), and f(4) = 64 − 48 = 16 as well. So the switchover is exact and lands on a whole number. On a curve with a messy crossover the moment is a blur and the point is lost.

The answers

Stationary pointsf′(x) = 3x² − 12 = 0 gives x = ±2. Local maximum (−2, 16), local minimum (2, −16).
Sign tablef′(−3) = 15, f′(0) = −12, f′(3) = 15. Up, down, up.
Endpointsf(−3) = 9, f(3) = −9, f(4) = 16, f(5) = 65.
1. y of the local minimum−16.
2. Greatest on [−3, 5]B, 65, at the right-hand end.
3. "The maximum is x = −2"B. That is the location; the value is 16 and the point is (−2, 16).

Where the marks go

1 markDifferentiating and solving f′(x) = 0.

1 markSubstituting back into f. Half the class stops at the x values.

1 markClassifying, with a sign check that actually appears on the page.

1 markOn an interval question, evaluating at the endpoints too and comparing the whole list.

What each wrong answer tells you

They sayWhat it means
2 (Q1)Gave the x coordinate. The habit this page exists to break.
16 (Q1)Right value, wrong turning point. The work was done; the labelling was not.
0 (Q1)Substituted into f′ rather than f. Say it plainly: f′ gives gradients, f gives heights.
16 (Q2)The target. They found the local maximum and stopped. Go back to the slider rather than explaining again.
"Nothing wrong" (Q3)They do not yet distinguish a location from a value. Very common, and it quietly costs marks in every later topic too.

Other things they will say

"Why check the ends? The gradient is not zero there." Exactly the right observation, and the reason endpoints are missed. A maximum on a closed interval does not have to be smooth; it can be a corner of the domain.

"Is it a maximum or a local maximum?" Encourage the word local always. It costs nothing and it is the honest description.

"Can there be no stationary points?" Yes, and it is worth ten seconds: f′(x) = 3x² + 1 is never zero.

A possible order

 What is happening
1Recap 5.2: the sign of f′ and what zero means. Keep it short; this is the application lesson.
2The full method on x³ − 12x, written out with coordinates as pairs. Insist on the substitution line.
3The animation. Greatest value at b = 3, then drag past 4.
4Interval questions properly: list the stationary points and both ends, then compare.
5Questions 1 to 3. Question 3 is about precision of language and is worth discussing aloud.
6Set practice where at least one question gives a domain.

Two things not to say

Do not say "find the maximum" when you mean "find the stationary points". The sloppy phrasing is exactly what produces the x-only answer and the missed endpoint.

Do not introduce the second derivative test. It is Higher Level, it is another thing to get wrong, and the sign check is sufficient and clearer about what is actually happening.

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 themFor y = x³ − 3x² + 4, find the stationary points.
    y′ = 3x² − 6x = 0 at x = 0 and x = 2, giving (0, 4) and (2, 0).
  2. Classify themUse the second derivative to classify each.
    y″ = 6x − 6. At x = 0 it is −6, so a maximum; at x = 2 it is 6, so a minimum.

2In context

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

  1. Greatest on an intervalFor the same curve on 0 ≤ x ≤ 4, find the greatest value.
    y(0) = 4, y(2) = 0, y(4) = 20. The greatest is 20, at the endpoint x = 4, not at the maximum.
  2. Where the endpoint takes overOn 0 ≤ x ≤ b, find the value of b at which the endpoint first matches the local maximum.
    Solve b³ − 3b² + 4 = 4, so b²(b − 3) = 0 and b = 3. Beyond b = 3 the endpoint is the greatest value.

3Challenge

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

  1. Why endpoints must be checkedExplain why finding and classifying stationary points is not enough on a closed interval.
    A maximum found by calculus is only a LOCAL one. On a closed interval the greatest value can sit at an end, where the derivative says nothing. Candidates are the stationary points and both endpoints.
  2. When the second derivative failsFor y = x⁴, y′ and y″ are both zero at x = 0. Decide what kind of point it is and how.
    y′ = 4x³ is negative before 0 and positive after, so it is a minimum. When y″ = 0 the test is inconclusive and the sign of y′ either side must be used.

Practicalities

The interval starts at −3 and only the right end moves, which keeps the demonstration simple. No external library, nothing stored, nothing sent.