Stationary Points and Turning Points

Finding turning points, maximum or minimum from the second derivative, the biggest volume or area, and a stationary point with a missing number (Higher).

In the exam for Cambridge IGCSE Maths 0580, Edexcel IGCSE Maths 4MA1, Edexcel Further Pure Maths 4PM1 · 44 real past-paper questions below

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What a Stationary Point Is (Higher)

At the top of a hill or the bottom of a valley the curve is flat for a moment. Its gradient is 0 there.

Rule

A stationary point is where $\frac{d y}{d x} = 0$ . A hill top is a maximum; a valley bottom is a minimum.

Key idea

“Turning point” means the same thing. The questions use both names.

Key idea

Puzzle: you have 40 m of fence for a rectangular garden. Which shape gives the most area? Sides $x$ and $20 - x$ give $A = 20 x - x^{2}$ . Set $\frac{d A}{d x} = 20 - 2 x = 0$ : a 10 m square, area 100 m².

Finding the Points

Differentiate, set it equal to 0, solve, then find y.

Worked example

1
$y = x^{3} - 3 x^{2} - 9 x + 5$ — The curve.
2
$\frac{d y}{d x} = 3 x^{2} - 6 x - 9$ — Differentiate each term with the power rule.
3
$3 x^{2} - 6 x - 9 = 0$ — Flat means gradient 0, so set dy/dx equal to 0.
4
$x^{2} - 2 x - 3 = 0$ — Divide every term by 3 to make the numbers smaller.
5
$(x - 3) (x + 1) = 0$ — Factorise. Two numbers that multiply to −3 and add to −2 are −3 and +1.
6
$x = 3 or x = - 1$ — Each bracket equal to 0 gives an x value.
7
$y = (- 1)^{3} - 3 (- 1)^{2} - 9 (- 1) + 5$ — Now find y. Put x = −1 into the curve (not dy/dx), with brackets.
8
$y = - 1 - 3 + 9 + 5 = 10$ — (−1)³ = −1 and (−1)² = 1. Then −9 × (−1) = +9. The point is (−1, 10).
9
$y = 3^{3} - 3 (3)^{2} - 9 (3) + 5 = 27 - 27 - 27 + 5 = - 22$ — Same move for x = 3. The point is (3, −22).

Common mistake

Stopping at the x values. “Find the coordinates” needs y too.

Key idea

“Show that the x-axis is a tangent” means find a stationary point with y = 0. The curve touches the axis there without crossing it.

When x Factors Out

Some derivatives have no number term, so x factors out. Take it out as a factor. Never divide by x, or you lose the answer x = 0.

Worked example

1
$y = x^{4} - 18 x^{2} + 7$ — The curve.
2
$\frac{d y}{d x} = 4 x^{3} - 36 x$ — Power rule on each term, and 7 → 0.
3
$4 x^{3} - 36 x = 0$ — Stationary means gradient 0.
4
$4 x (x^{2} - 9) = 0$ — Take out the common factor 4x, which divides both terms.
5
$4 x (x - 3) (x + 3) = 0$ — x² − 9 is a difference of two squares, so it splits into (x − 3)(x + 3).
6
$x = 0, x = 3, x = - 3$ — Each factor equal to 0 gives an x value. There are three.
7
$y = 0^{4} - 18 (0)^{2} + 7 = 7$ — Put x = 0 into the curve. The point is (0, 7).
8
$y = 3^{4} - 18 (3)^{2} + 7 = 81 - 162 + 7 = - 74$ — Put x = 3 into the curve. 3⁴ = 81 and 18 × 9 = 162.
9
$y = (- 3)^{4} - 18 (- 3)^{2} + 7 = - 74$ — An even power of −3 is positive, so x = −3 gives the same y as x = 3.

Common mistake

Dividing both sides of $4 x^{3} - 36 x = 0$ by $x$ . That throws away $x = 0$ , which is a turning point.

Maximum or Minimum?

Differentiate a second time. The sign of $\frac{d ^{2} y}{d x ^{2}}$ tells you which.

Worked example

1
$\frac{d y}{d x} = 3 x^{2} - 6 x - 9$ — From the first example, Finding the Points.
2
$\frac{d ^{2} y}{d x ^{2}} = 6 x - 6$ — Differentiate dy/dx again with the power rule.
3
$x = 3 : 6 (3) - 6$ — Put the first x value into the second derivative.
4
$18 - 6 = 12 > 0$ — Positive: the curve is a valley here, so it is a minimum.
5
$x = - 1 : 6 (- 1) - 6$ — Now the other x value, with brackets round the negative.
6
$- 6 - 6 = - 12 < 0$ — Negative: the curve is a hill here, so it is a maximum.

Memory trick

Positive is a smile (minimum). Negative is a frown (maximum).

Key idea

You can also sketch. A cubic with a positive $x^{3}$ goes up, makes a hill, then a valley, then goes up. So its left turning point is the maximum and its right one is the minimum.

The Biggest Value (Higher)

The 5-mark questions build a formula for a volume or an area, then ask for its maximum. Here a cuboid is x cm by 2x cm by (18 − 3x) cm.

Worked example

1
$V = x \times 2 x \times (18 - 3 x)$ — Volume is length × width × height.
2
$V = 2 x^{2} (18 - 3 x)$ — x × 2x = 2x².
3
$V = 36 x^{2} - 6 x^{3}$ — Multiply 2x² by each term: 2 × 18 = 36, and 2 × (−3) = −6, with x² × x = x³.
4
$\frac{d V}{d x} = 72 x - 18 x^{2}$ — Differentiate each term.
5
$72 x - 18 x^{2} = 0$ — The maximum is a stationary point, so set it to 0.
6
$18 x (4 - x) = 0$ — Take out the common factor 18x.
7
$x = 0 or x = 4$ — Each factor equal to 0 gives an x value.
8
$x = 4$ — x = 0 would make no box, so reject it. At x = 4 the height is 18 − 12 = 6, which is fine.
9
$\frac{d ^{2} V}{d x ^{2}} = 72 - 36 x$ — Differentiate again to check it is a maximum.
10
$72 - 36 (4) = - 72 < 0$ — Negative means a hill, so x = 4 gives a maximum.
11
$V = 2 (4)^{2} (18 - 3 (4))$ — Put x = 4 into V (the volume formula, not dV/dx).
12
$V = 2 (16) (6) = 192$ — The maximum volume is 192 cm³.

Key idea

Always ask: could this x really happen? A length cannot be 0 or negative, so reject any x that makes a side zero or negative.

A Missing Number (Higher)

If the question tells you where the stationary point is, put that x into $\frac{d y}{d x} = 0$ . Then use the curve to find y.

Worked example

1
$y = k x^{2} + \frac{8}{x}$ — Stationary at x = 1/2. Find k, then the y-coordinate there.
2
$y = k x^{2} + 8 x^{- 1}$ — Write 8/x as a power so the rule works.
3
$\frac{d y}{d x} = 2 k x - 8 x^{- 2}$ — kx² → 2kx, because k is just a number. 8x⁻¹ → −8x⁻².
4
$2 k (\frac{1}{2}) - \frac{8}{( \frac{1}{2} ) ^{2}} = 0$ — Stationary means gradient 0. Put x = 1/2 in.
5
$k - \frac{8}{1/4} = 0$ — 2k × 1/2 = k, and (1/2)² = 1/4.
6
$k - 32 = 0$ — 8 ÷ 1/4 = 8 × 4 = 32.
7
$k = 32$ — Add 32 to both sides.
8
$y = 32 (\frac{1}{2})^{2} + \frac{8}{1/2}$ — Put k = 32 and x = 1/2 into the curve (not dy/dx).
9
$y = 8 + 16 = 24$ — 32 × 1/4 = 8, and 8 ÷ 1/2 = 16. The point is (1/2, 24).

Key idea

If the question gives a gradient instead, such as “the gradient at x = 2 is 6”, put that x into dy/dx and set it equal to that number, not 0.

Which Method? Read the Question

Every question here starts with dy/dx = 0.

You see	You do
“Turning points”	dy/dx = 0, then find y
“Maximum or minimum”	Sign of the second derivative
“Maximum volume”	Build the formula, then dy/dx = 0
“Stationary at x = …”	Put that x into dy/dx = 0
“Find dy/dx” only	See Differentiating Polynomials

Key idea

Check your answer makes sense: a length must be positive, and the maximum must be bigger than nearby values.

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