Stationary points, local and global

Here is f(x) = x³ − 12x. It has a local maximum and a local minimum, both marked. Drag the right-hand end of the interval and watch which point is actually the highest one in range.

b = 3.00

16greatest value, at the local maximum

−16least value, at the local minimum

Push b past 4 and the greatest value stops being the turning point.

The method

Differentiate to get f′(x).
Solve f′(x) = 0. This gives the x values only.
Substitute each x back into f to get the y values.
Decide which is which by checking the sign of f′ either side.

Worked example

Find and classify the stationary points of f(x) = x³ − 12x.

f′(x)= 3x² − 12 3x² − 12= 0, so x² = 4 and x = ±2 two values, so two points f(−2)= −8 + 24 = 16 f(2)= 8 − 24 = −16

x	−3	−2	0	2	3
f′(x)	15	0	−12	0	15
f is	up	flat	down	flat	up

Up then down, so (−2, 16) is a local maximum. Down then up, so (2, −16) is a local minimum.

Write points as coordinates. "The maximum is at x = −2" and "the maximum is 16" are different statements, and a question usually wants the pair. Giving only the x value is the most expensive habit on this sub-topic.

Local is not greatest

This is the bit the animation is for, and it is the bit that gets missed.

On the interval from −3 to 3, the greatest value really is the local maximum, 16.

On the interval from −3 to 5, it is not. f(5) = 125 − 60 = 65, far higher, and it happens at the end of the interval where the gradient is nowhere near zero.

So when a question asks for the greatest or least value on an interval, the stationary points are only part of the answer. Work out f at the stationary points and at both ends, then compare the list. Differentiating alone will never find an endpoint, because nothing special happens to the gradient there.

Your turn

1. For f(x) = x³ − 12x, give the y coordinate of the local minimum.

2. On the interval from −3 to 5, what is the greatest value of f(x) = x³ − 12x?

16, at the local maximum. 65, at the right-hand end of the interval. −16, at the local minimum. There is no greatest value.

3. A student writes "the maximum is x = −2". What is wrong with that?

Nothing, it is correct. x = −2 is where it happens; the maximum value is 16, and the point is (−2, 16). The maximum is at x = 2, not x = −2. It should say minimum.

Where the marks go

Solving f′(x) = 0 is routine. Substituting back for the y values is where half the available marks sit, and it is the step most often skipped.

Justify your classification. A sign check either side is enough at this level, and it has to actually appear on the page: an unsupported "this is a maximum" usually earns nothing.

If an interval is given, check the endpoints. The word "greatest" or "least" with a domain attached is the signal.

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