Extended. At the top of a hill or the bottom of a valley the curve is momentarily flat. Setting the gradient to zero finds these turning points exactly, and a second check says which is which.
Stationary points are where \(\dfrac{dy}{dx} = 0\). Solve for \(x\), then find \(y\) from the original equation. If \(\dfrac{d^2y}{dx^2} > 0\) it is a minimum; if \(\dfrac{d^2y}{dx^2} < 0\) it is a maximum.
Fig. 1: y = x³ − 3x + 2 has horizontal tangents at the maximum (−1, 4) and the minimum (1, 0); the gradient goes + then − at the maximum and − then + at the minimum
Section 01
Where the gradient is zero.
In the lab's stationary points tab, slide the point along the curve and watch the tangent. Stop where the tangent is exactly horizontal: the gradient readout shows 0. Find both such points on a cubic.
Step 1: differentiate to get \(\dfrac{dy}{dx}\).
Step 2: solve \(\dfrac{dy}{dx} = 0\) to find the \(x\)-coordinates.
Step 3: substitute each \(x\) into the ORIGINAL equation to find \(y\).
Worked example A quadratic
Find the stationary point of \(y = x^2 - 8x + 3\).
\(\dfrac{dy}{dx} = 2x - 8\).
\(2x - 8 = 0\), so \(x = 4\).
\(y = 4^2 - 8(4) + 3 = 16 - 32 + 3 = -13\).
Answer: \((4, -13)\). The curve is U-shaped, so this is a minimum.
Worked example A cubic
Find the stationary points of \(y = x^3 - 3x + 2\).
\(\dfrac{dy}{dx} = 3x^2 - 3\).
\(3x^2 - 3 = 0\), so \(x^2 = 1\) and \(x = 1\) or \(x = -1\).
\(x = 1\): \(y = 1 - 3 + 2 = 0\).
\(x = -1\): \(y = -1 + 3 + 2 = 4\).
Answer: \((1, 0)\) and \((-1, 4)\) (Fig. 1)
Examiner trap
\(x^2 = 1\) has TWO solutions, \(x = 1\) and \(x = -1\). Missing the negative root loses a whole stationary point.
Section 02
Maximum or minimum?
The syllabus accepts any correct method. The second derivative is quickest; checking the gradient either side always works.
Method
Maximum
Minimum
Second derivative \(\dfrac{d^2y}{dx^2}\)
negative
positive
Gradient just before and just after
\(+\), 0, \(-\)
\(-\), 0, \(+\)
Accurate sketch
top of a hill
bottom of a valley
Worked example Using the second derivative
Determine the nature of the stationary points \((1, 0)\) and \((-1, 4)\) of \(y = x^3 - 3x + 2\).
So the gradient goes \(+, 0, -\) through \(x = 1\) and \(-, 0, +\) through \(x = 2\).
Answer: \((1, 5)\) is a maximum; \((2, 4)\) is a minimum
Your turn
Find the stationary point of \(y = 5 + 4x - x^2\) and show that it is a maximum.
\(\dfrac{dy}{dx} = 4 - 2x = 0\), so \(x = 2\), \(y = 5 + 8 - 4 = 9\). \(\dfrac{d^2y}{dx^2} = -2 < 0\), so \((2, 9)\) is a maximum.
Examiner trap
Decide using the second derivative or the gradient, not the size of \(y\). Here the maximum \((1, 5)\) is only slightly higher than the minimum \((2, 4)\), and on other curves a local maximum can even be LOWER than a minimum.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
What is true at every stationary point of a curve?
Quick check+10 XP
Find the \(x\)-coordinate of the stationary point of \(y = x^2 - 6x\).
Quick check+10 XP
At a stationary point, \(\dfrac{d^2y}{dx^2} = -4\). What kind of point is it?
Quick check+10 XP
The gradient of a curve is negative just before a stationary point and positive just after it. What kind of point is it?
Section 03
The biggest and the smallest.
Many problems ask for the largest area, the greatest volume or the least cost. Write the quantity as a formula in one variable, differentiate, set the derivative to zero and check that you have the kind of turning point you want.
Worked example Largest area
A farmer has 40 m of fencing to make three sides of a rectangular pen against a long wall. The two sides at right angles to the wall are \(x\) m long. Find the greatest possible area.
The side parallel to the wall is \(40 - 2x\) m, so the area is \(A = x(40 - 2x) = 40x - 2x^2\).
\(\dfrac{dA}{dx} = 40 - 4x = 0\), so \(x = 10\).
\(\dfrac{d^2A}{dx^2} = -4 < 0\), so this is a maximum.
\(A = 10 \times (40 - 20) = 10 \times 20 = 200\).
Answer: Greatest area \(200\ \text{m}^2\), when \(x = 10\) m
Worked example Greatest volume
Squares of side \(x\) cm are cut from the corners of a 12 cm by 12 cm sheet, and the sides are folded up to make an open box. Show that \(V = 4x^3 - 48x^2 + 144x\) and find the greatest volume.
The base is \((12 - 2x)\) by \((12 - 2x)\) and the height is \(x\), so \(V = x(12 - 2x)^2 = x(144 - 48x + 4x^2) = 4x^3 - 48x^2 + 144x\).
Answer: Greatest volume \(128\ \text{cm}^3\), when \(x = 2\) cm
Examiner trap
Answer the question that was asked. Finding \(x = 10\) is not the end: the question wants the AREA, \(200\ \text{m}^2\). Also reject values that make no physical sense, such as \(x = 6\) for the box.
Stage 2 · Exam
Exam-style questions
Unlocks once the four checks above are done. Worth more XP, written to IGCSE Paper 2 and Paper 4 standard.
Finish the four checks above to unlock the exam questions
Exam style+20 XP
Find the coordinates of the stationary point of \(y = x^2 - 8x + 3\).
Exam style+20 XP
Find the coordinates of the maximum point of \(y = x^3 - 12x + 5\).
Exam style+20 XP
\(y = 2x^3 - 9x^2 + 12x\) has a stationary point at \((2, 4)\). Which statement is correct?
Exam style+20 XP
A rectangle has a perimeter of 36 cm. One side is \(x\) cm, so its area is \(A = x(18 - x)\). Find the greatest possible area.
Skill unlocked
Turning points found and classified.
This skill is now recorded on your profile. Keep the chain going.