Calculus: turning points, maxima, minima and motion

International GCSE Higher uses differentiation to find turning points, to solve 'largest' and 'smallest' problems, and to describe motion.

Practise this lesson free (6 questions)

Turning points

At a turning point the gradient is 0, so solve dydx = 0.
y = x3 โˆ’ 12x: dydx = 3x2 โˆ’ 12 = 0, so x2 = 4 and x = โˆ’2 or 2.
y-values: x = โˆ’2 gives y = 16; x = 2 gives y = โˆ’16.

Maximum or minimum?

Use the shape of the graph. A positive cubic rises, falls, then rises again, so the left turning point is a maximum and the right one a minimum.
You can also check the gradient either side: + then โˆ’ is a maximum; โˆ’ then + is a minimum.
For y = x3 โˆ’ 12x: (โˆ’2, 16) is a maximum and (2, โˆ’16) a minimum.

Motion and practical problems

Displacement s, velocity v = dsdt, acceleration a = dvdt.
s = t3 โˆ’ 4t2 + 5t: v = 3t2 โˆ’ 8t + 5 and a = 6t โˆ’ 8.
For a practical problem, write the quantity in one variable, differentiate and set equal to 0.
A = x(40 โˆ’ 2x): dAdx = 40 โˆ’ 4x = 0, so x = 10 and A = 200.
Worked example

s = t3 โˆ’ 4t2 + 5t. Find the velocity when t = 2.

  1. v = dsdt = 3t2 โˆ’ 8t + 5
  2. t = 2: 12 โˆ’ 16 + 5 = 1

Answer: v = 1

Key idea

Solve dy/dx = 0 for turning points; decide maximum or minimum from the shape or the gradient either side. Velocity = ds/dt and acceleration = dv/dt.

Key words
Gradient
The steepness of a line: change in y รท change in x. It is m in y = mx + c.
Turning point
The point where a curve changes from going down to going up or the other way: a minimum or maximum.

All GCSE and International GCSE Mathematics key terms and definitions โ†’

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Extra practice

2 more questions

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  1. Find the y-coordinate of the maximum point of y = x3 โˆ’ 27x.

    Show the answer

    Answer: 54

    3x2 โˆ’ 27 = 0 gives x = ยฑ3. The maximum is at x = โˆ’3: (โˆ’3)3 โˆ’ 27(โˆ’3) = โˆ’27 + 81 = 54.

  2. y = x3 โˆ’ 12x. What kind of turning point is at x = โˆ’2?

    1. It is not a turning point
    2. Neither
    3. A minimum
    4. A maximum
    Show the answer

    Answer: A maximum

    dydx = 3x2 โˆ’ 12 = 0 at x = โˆ’2. d2ydx2 = 6x = โˆ’12, negative, so a maximum.

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Sources and references

Checked by the GCSE-ready editorial team, 17 October 2026. Last updated 2 October 2026. Spotted a mistake? Tell us and we will fix it.

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