5.8 Maxima, minima, inflexion and optimisation · Educerie Mathematics: analysis and approaches

Educerie · IB Diploma · Mathematics: analysis and approaches

Topic 5 Calculus · 5.8 Maxima, minima, inflexion and optimisation

Level: SL and HL. Nothing here is HL only, so every section is examinable for both.
Themes (key concepts): change, relationships, modelling. Where a quantity stops increasing and starts decreasing, its rate of change is zero; that one fact finds the highest and lowest points of a graph, and, inside a model, the best box, the cheapest can or the largest profit.
The question this unit answers: where does a function reach its highest and lowest values, how do you prove which is which, and how do you turn that into the best possible answer to a real problem?
Where it is examined: Paper 1: find and classify the stationary points of a function (5 to 8 marks), often with its points of inflexion. Paper 2: optimisation in context, setting up a function from a diagram and maximising or minimising it, by hand or with the GDC (6 to 10 marks). Section B questions on both papers often end here.

What you must be able to do

You must be able to	Level	What it looks like in the exam
Find the coordinates of stationary points by solving f′(x) = 0	SL, HL	"Find the coordinates of the stationary points of the curve" (3 to 4 marks)
Classify a stationary point with the sign of f′ on either side	SL, HL	"Show that the point is a local maximum", with a sign diagram as the reason
Classify with the second derivative: f″(x) > 0 a minimum, f″(x) < 0 a maximum	SL, HL	"Use the second derivative to determine the nature of each stationary point" (2 to 3 marks)
Know that f″(x) = 0 at a stationary point decides nothing, and fall back on the sign of f′	SL, HL	A stationary point where f″ = 0: "determine its nature"
Find points of inflexion: f″(x) = 0 and f″ changes sign; say whether the gradient there is zero or not	SL, HL	"Find the coordinates of the point of inflexion. Justify your answer" (3 to 4 marks)
Use "concave up" for f″(x) > 0 and "concave down" for f″(x) < 0	SL, HL	"Write down the interval on which the graph is concave down"
Tell a local maximum from the greatest value, checking the ends of a domain	SL, HL	"Find the maximum value of f for 0 ≤ x ≤ 5" (2 to 4 marks)
Optimise in context: profit, area, volume and similar	SL, HL	"Show that A = …", then "find the value of x that minimises A" and "justify that it is a minimum" (6 to 10 marks)

Before you start

You need 5.7: the second derivative, and what concave up and concave down mean. You need 5.2: f′(x) > 0 means f is increasing. You need all the differentiation rules of 5.3 and 5.6. For optimisation you need prior-learning geometry: the area of a rectangle and a circle, the volume of a cuboid and a cylinder, and the curved surface area of a cylinder, which the formula booklet's prior learning section gives you.

1The idea in one paragraph

At the top of a hill the ground is momentarily flat, and at the bottom of a valley too. So the highest and lowest points of a smooth graph are found where the gradient is zero: solve f′(x) = 0. Those points are the stationary points. Each one is a local maximum, a local minimum or neither, and you decide which from the sign of f′ either side or the sign of f″ at the point. A point of inflexion is where the bending changes over: f″ is zero there and changes sign. Optimisation puts this to work: write the quantity to be made largest or smallest as a function of one variable, differentiate, set the derivative to zero, and prove you have the best value.

2Stationary points

A stationary point is a point on a graph where the gradient is zero, f′(x) = 0. The tangent there is horizontal. There are three kinds, drawn in Figure 1.

A local maximum: the graph rises to it and falls away after it. It is higher than every point close to it, though not necessarily the highest point of the whole graph.
A local minimum: the graph falls to it and rises after it.
A stationary point of inflexion: the graph flattens for an instant and carries on in the same direction.

The first two are turning points.

Figure 1 · Three kinds of stationary point

Finding them takes three steps.

Differentiate and factorise f′(x) if you can.
Solve f′(x) = 0 to find the x-coordinates.
Substitute each x into f(x), not f′(x), to get the y-coordinates.

Take f(x) = 2x³ − 3x² − 12x + 5.

f′(x) = 6x² − 6x − 12 = 6(x² − x − 2) = 6(x + 1)(x − 2)

f′(x) = 0 ⇔ x = −1 or x = 2

f(−1) = −2 − 3 + 12 + 5 = 12into f, not f′

f(2) = 16 − 12 − 24 + 5 = −15

The stationary points are (−1, 12) and (2, −15). Now decide what each one is.

3The first derivative test: watch the sign of f′

Figure 1 shows the rule under each picture. Read the sign of f′ just to the left of the point, at the point, and just to the right.

+, 0, − is a local maximum. −, 0, + is a local minimum. +, 0, + or −, 0, − is a stationary point of inflexion.

For f′(x) = 6(x + 1)(x − 2), pick a test value in each interval.

x = −2: f′(−2) = 6(−1)(−4) = 24 > 0increasing

x = 0: f′(0) = 6(1)(−2) = −12 < 0decreasing

x = 3: f′(3) = 6(4)(1) = 24 > 0increasing

So f′ goes +, 0, − through x = −1, which is a local maximum at (−1, 12), and −, 0, + through x = 2, which is a local minimum at (2, −15). Figure 2 shows the graph with the sign diagram underneath. On an exam paper the sign diagram is your reason; draw it, or write the three test values, and the R1 mark is yours.

Figure 2 · Classifying the stationary points of f(x) = 2x³ − 3x² − 12x + 5

A test value must have no other zero of f′ between it and the point it tests; x = 0 is safe here because f′ has no zero between −1 and 2.

4The second derivative test: which way does it bend?

From 5.7, f″ tells you the concavity. A stationary point where the graph is concave down is at the top of a cap: a maximum. Where it is concave up it is at the bottom of a cup: a minimum.

At a stationary point: f″(x) < 0 means a local maximum. f″(x) > 0 means a local minimum. If f″(x) = 0, the test gives no information.

For the same function, f″(x) = 12x − 6.

f″(−1) = −12 − 6 = −18 < 0concave down: local maximum at (−1, 12)

f″(2) = 24 − 6 = 18 > 0concave up: local minimum at (2, −15)

Same answers, in two lines. The second derivative test is quicker when f″ is easy; the first derivative test is better when f″ is messy, and it is the only test that works when f″ = 0. If the question names a test, use that one.

Why f″ = 0 tells you nothing. In Figure 3, y = x⁴, y = −x⁴ and y = x³ all have f′(0) = 0 and f″(0) = 0, yet the origin is a minimum, a maximum and a stationary inflexion respectively. So when f″ = 0 at a stationary point, go back to the sign of f′. For y = x⁴, f′(x) = 4x³ is negative for x < 0 and positive for x > 0, so the origin is a minimum.

Figure 3 · f′(0) = 0 and f″(0) = 0 in all three

5Points of inflexion

A point of inflexion is a point where the graph changes concavity: from concave up to concave down, or the other way. Because f″ is positive on one side and negative on the other, at the point itself f″ = 0. The guide is careful about the converse, and so must you be.

At a point of inflexion, f″(x) = 0 and f″ changes sign. f″(x) = 0 on its own is not enough.

The guide's own example is y = x⁴ at (0, 0). There f″(x) = 12x², so f″(0) = 0. But 12x² is positive on both sides of 0, so the concavity never changes: the graph is concave up throughout, and (0, 0) is a minimum, not a point of inflexion. Anyone who solves f″ = 0 and stops will get this wrong.

Points of inflexion come in two kinds.

A stationary point of inflexion has f′ = 0 as well: a horizontal tangent, like y = x³ at the origin.
A non-stationary point of inflexion has f′ ≠ 0: the graph changes its bend while still climbing or falling. These are far more common. Every cubic has one.

A full example. f(x) = x⁴ − 4x³ has both kinds. Figure 4 draws it.

f′(x) = 4x³ − 12x² = 4x²(x − 3)

f″(x) = 12x² − 24x = 12x(x − 2)

Stationary points. f′(x) = 0 at x = 0 and x = 3, with f(0) = 0 and f(3) = 81 − 108 = −27.

At x = 3, f″(3) = 108 − 72 = 36 > 0, so (3, −27) is a local minimum.
At x = 0, f″(0) = 0: the second derivative test fails. Use the sign of f′: f′(−1) = 4(1)(−4) = −16 < 0 and f′(1) = 4(1)(−2) = −8 < 0. The sign goes −, 0, −, so (0, 0) is a stationary point of inflexion.

Points of inflexion. f″(x) = 0 at x = 0 and x = 2. Check that f″ changes sign at each.

f″(−1) = 36 > 0, f″(1) = −12 < 0changes sign at x = 0

f″(1) = −12 < 0, f″(3) = 36 > 0changes sign at x = 2

f(2) = 16 − 32 = −16, f′(2) = 4(4)(−1) = −16

So (0, 0) is a point of inflexion with zero gradient, and (2, −16) is a point of inflexion with non-zero gradient, −16. The graph is concave up for x < 0, concave down for 0 < x < 2, and concave up for x > 2.

Figure 4 · Two kinds of point of inflexion on y = x⁴ − 4x³

An exact-value example for Paper 1. Let f(x) = x e^−x.

f′(x) = e^−x − x e^−x = (1 − x)e^−x

f′(x) = 0 ⇔ x = 1e⁻ˣ is never 0

f(1) = e⁻¹

f″(x) = −e^−x − (1 − x)e^−x = (x − 2)e^−x

f″(1) = −e⁻¹ < 0local maximum (1, 1/e)

f″(x) = 0 ⇔ x = 2, and x − 2 changes sign theree⁻ˣ > 0

f(2) = 2e⁻², f′(2) = −e⁻²

The maximum is (1, 1/e) and the point of inflexion is (2, 2/e²), a non-stationary one with gradient −1/e². Leave the answers exact: 1/e, not 0.368, on Paper 1.

6Local or greatest? Check the ends

A local maximum is only the top of its own hill. The greatest value of a function on a domain (sometimes called the global or absolute maximum) may be somewhere else, and on a closed interval it is often at an end point, where the graph simply stops.

Take f(x) = 2x³ − 3x² − 12x + 5 again, on the domain −2 ≤ x ≤ 4. The candidates are the stationary points and the two ends.

x	−2 (end)	−1 (local max)	2 (local min)	4 (end)
f(x)	1	12	−15	37

The greatest value is 37, at the end x = 4, not the local maximum 12. The least value is −15, at the local minimum. Figure 5 shows why. Whenever a question gives a domain and asks for the "maximum value" or "minimum value", evaluate f at the ends too.

Figure 5 · Local is not the same as greatest

7Optimisation

Optimisation means finding the value of a variable that makes some quantity as large or as small as possible: the most volume, the least material, the greatest profit. The calculus is the easy part. The marks go to setting it up and to justifying the answer. Work through the same steps every time.

Draw and label a diagram. Name the variables.
Write the quantity to optimise, Q, in terms of the variables.
If Q has two variables, use the constraint (the fixed fact in the question) to eliminate one. Q must be a function of one variable before you differentiate.
State the domain: which values make physical sense.
Differentiate, set Q′ = 0 and solve.
Justify that it is a maximum or a minimum: second derivative, sign of Q′, or end points.
Answer the question asked, with units. If it asked for the maximum volume, a value of x is not the answer.

Example A, volume (Paper 1). A rectangular sheet of card measures 24 cm by 15 cm. A square of side x cm is cut from each corner and the sides are folded up to make an open box. Find the maximum volume.

Figure 6(a) shows the sheet: the base is (24 − 2x) by (15 − 2x) and the height is x.

V = x(24 − 2x)(15 − 2x)

= 4x³ − 78x² + 360xexpand before differentiating

domain: 0 < x < 7.515 − 2x must stay positive

dV/dx = 12x² − 156x + 360 = 12(x − 3)(x − 10)

dV/dx = 0 ⇔ x = 3 or x = 10reject x = 10: outside the domain

d²V/dx² = 24x − 156; at x = 3: 72 − 156 = −84 < 0maximum

V = 3 × 18 × 9 = 486

The maximum volume is 486 cm³, when x = 3 cm. The domain did real work: x = 10 would need a 20 cm cut from a 15 cm side. Figure 6(b) confirms the peak.

Figure 6 · An open box from a 24 cm by 15 cm sheet

Example B, surface area (Paper 2). A drinks can is a closed cylinder holding 330 cm³. Find the radius that uses the least metal, and that least area.

Two variables, r and h, and one constraint: the volume.

πr²h = 330 ⇒ h = 330 / (πr²)constraint

A = 2πr² + 2πrhtwo ends plus the curved side

= 2πr² + 2πr × 330 / (πr²) = 2πr² + 660/rnow one variable, r > 0

dA/dr = 4πr − 660/r²

dA/dr = 0 ⇒ r³ = 165/π ⇒ r = 3.745… ≈ 3.74 cm

d²A/dr² = 4π + 1320/r³ > 0 for all r > 0minimum

h = 330 / (π × 3.745…²) = 7.49 cm

A = 264 cm²3 s.f.

The least metal is about 264 cm², with r ≈ 3.74 cm and h ≈ 7.49 cm, so the height equals the diameter. Figure 7 shows the shape of A(r): the curved side dominates for small r, the ends for large r. On Paper 2 you may also graph A(r) on the GDC and use the minimum tool, but still write down the function you graphed; "from GDC, r = 3.74" with no function shown earns little.

Figure 7 · Surface area of a 330 cm³ can against its radius

Example C, profit, with end points. A small workshop makes x hundred chairs a week, where 0 ≤ x ≤ 10, and its weekly profit is P(x) = −2x³ + 27x² − 84x + 100 hundred dollars.

P′(x) = −6x² + 54x − 84 = −6(x − 2)(x − 7)

P′(x) = 0 ⇔ x = 2 or x = 7

P″(x) = −12x + 54; P″(7) = −30 < 0 (max), P″(2) = 30 > 0 (min)

P(7) = 149, P(0) = 100, P(10) = −40compare with the end points

The largest profit is 149 hundred dollars, $14 900 a week, from 700 chairs. Had P(0) been bigger than 149, making nothing would have been the answer: that is why the ends are checked.

8Where marks are lost

Substituting into f′ for the y-coordinate. f′ is zero at every stationary point, by definition. The y-coordinate comes from f.

Calling a zero of f″ a point of inflexion without checking the sign change. y = x⁴ has f″(0) = 0 and no inflexion. Show f″ on both sides.

Using the second derivative test when f″ = 0. It gives no answer at all. "f″ = 0, so it is a point of inflexion" scores zero; use the sign of f′.

Giving the local maximum as the greatest value. On a closed domain, evaluate the ends too.

Differentiating before reducing to one variable. A = 2πr² + 2πrh has two variables. Use the constraint first, or the derivative means nothing.

Keeping a solution outside the domain. x = 10 in the box example solves V′ = 0 but is impossible. Say so in words when you reject it.

Answering the wrong question. If the question asks for the maximum area, a value of x earns the method marks but not the final A1.

9Work it right

Solve f′(x) = 0 and give each stationary point as coordinates, with y from f(x).
Classify every stationary point with a stated reason: the sign of f″, or a sign diagram for f′ (which you must use if f″ = 0).
For a point of inflexion, show f″ = 0 and show f″ changing sign, then say whether f′ is zero there.
With a domain, list the values at the stationary points and at the end points before choosing.
In optimisation, write the constraint, reduce to one variable, and state the domain before differentiating.
Justify the maximum or minimum, then answer the question with units and to 3 significant figures unless told otherwise.
On Paper 2, write the function you gave the GDC and the value it returned.

10Try it

Marks in brackets. Q1 to Q4 are Paper 1 style, no calculator. Q5 is Paper 2 style, with a GDC.

Q1. Let f(x) = x³ − 3x² − 9x + 2.

(a) Find the coordinates of the stationary points of the graph of f. 4 marks

(b) Use the second derivative to determine the nature of each stationary point. 3 marks

Q2. Let f(x) = x² eˣ.

(a) Find the x-coordinates of the stationary points of the graph of f. 4 marks

(b) Show that f″(x) = (x² + 4x + 2)eˣ. 2 marks

(d) Find the x-coordinates of the points of inflexion of the graph of f. Justify your answer. 3 marks

Q3. Let f(x) = 3x⁵ − 5x³.

(a) Find the coordinates of the three stationary points of the graph of f. 4 marks

(b) Explain why the second derivative cannot be used to determine the nature of the stationary point at the origin, and determine its nature. 4 marks

Q4. An open-topped tank has a square base of side x metres and a height of h metres. Its volume is 4 m³.

(a) Show that the area of metal used, A m², is given by A = x² + 16/x. 3 marks

(b) Find the dimensions of the tank that use the least metal, and justify that your answer gives a minimum. 5 marks

Q5. A rectangle is drawn under the curve y = cos x, for −π/2 ≤ x ≤ π/2, with its base on the x-axis and its two upper vertices on the curve at (−x, cos x) and (x, cos x), where 0 < x < π/2.

(a) Show that the area of the rectangle is A = 2x cos x. 1 mark

(b) Find the value of x for which the area is greatest, and the greatest area. 4 marks

11In one breath

Stationary points are where f′(x) = 0; solve it and put each x back into f for the y-coordinate. Classify each one: by the sign of f′ either side (+ 0 − is a maximum, − 0 + a minimum, the same sign both sides a stationary inflexion), or by f″ (negative is a maximum, positive a minimum), remembering that f″ = 0 decides nothing, as y = x⁴, y = −x⁴ and y = x³ show. A point of inflexion is where the concavity changes, so f″ = 0 and changes sign; it may have zero gradient or not. Concave up means f″ > 0 and concave down f″ < 0. A local maximum need not be the greatest value, so check the ends of any domain. To optimise, draw it, write the quantity, use the constraint to reduce it to one variable with a sensible domain, differentiate and solve, justify maximum or minimum, and answer the question actually asked, with units.

Answers

Q1. (a) f′(x) = 3x² − 6x − 9 = 3(x + 1)(x − 3), so x = −1 or x = 3. f(−1) = −1 − 3 + 9 + 2 = 7 and f(3) = 27 − 27 − 27 + 2 = −25. The stationary points are (−1, 7) and (3, −25). A1 for f′(x), M1 for setting f′(x) = 0, A1 for both x-values, A1 for both points.

(b) f″(x) = 6x − 6. f″(−1) = −12 < 0, so (−1, 7) is a local maximum; f″(3) = 12 > 0, so (3, −25) is a local minimum. A1 for f″(x), R1 for evaluating the sign at each point, A1 for both natures. A correct nature with no f″ value shown scores R0.

Q2. (a) Product rule: f′(x) = 2x eˣ + x² eˣ = x(x + 2)eˣ. Since eˣ > 0, f′(x) = 0 when x = 0 or x = −2. M1 for the product rule, A1 for f′(x), R1 for eˣ ≠ 0, A1 for both values.

(b) f″(x) = (2x + 2)eˣ + (x² + 2x)eˣ = (x² + 4x + 2)eˣ, as required. M1 for the product rule on f′ written as (x² + 2x)eˣ, A1 for collecting to the given answer.

(c) f″(0) = 2 > 0, so (0, 0) is a local minimum. f″(−2) = (4 − 8 + 2)e⁻² = −2e⁻² < 0, so (−2, 4e⁻²) is a local maximum. M1 for substituting both x-values into f″, A1 for (0, 0) minimum, A1 for (−2, 4e⁻²) maximum. Accept 4/e².

(d) f″(x) = 0 when x² + 4x + 2 = 0, so x = −2 ± √2. These are simple roots of the quadratic and eˣ > 0, so f″ changes sign at each, and both are points of inflexion. M1 for solving x² + 4x + 2 = 0, A1 for −2 ± √2, R1 for the sign change. Stopping at f″ = 0 with no reason loses the R1.

Q3. (a) f′(x) = 15x⁴ − 15x² = 15x²(x − 1)(x + 1), so x = −1, 0, 1. f(−1) = −3 + 5 = 2, f(0) = 0, f(1) = 3 − 5 = −2. The points are (−1, 2), (0, 0) and (1, −2). A1 for f′(x), M1 for factorising and solving f′ = 0, A1 for the three x-values, A1 for the three points.

(b) f″(x) = 60x³ − 30x, so f″(0) = 0 and the second derivative test gives no information. Using f′ instead: f′(−0.5) = 15(0.25)(−0.75) < 0 and f′(0.5) = 15(0.25)(−0.75) < 0. The sign goes −, 0, −, so (0, 0) is a stationary point of inflexion. A1 for f″(0) = 0, R1 for saying the test is inconclusive, M1 for testing the sign of f′ on both sides, A1 for stationary point of inflexion. "f″ = 0 so it is a point of inflexion" scores A1 R0 M0 A0.

Q4. (a) Volume: x²h = 4, so h = 4/x². The tank has a base and four sides but no top:

A = x² + 4xh

= x² + 4x × 4/x²

= x² + 16/x

A1 for h = 4/x², M1 for the area of a base and four sides, A1 for substituting to reach the given answer. Including a lid scores M0.

(b) dA/dx = 2x − 16/x² = 0 gives x³ = 8, so x = 2 and h = 4/2² = 1. d²A/dx² = 2 + 32/x³ = 6 > 0 at x = 2, so this is a minimum. The tank is 2 m by 2 m by 1 m high, using 12 m² of metal. A1 for dA/dx, M1 for setting it to 0, A1 for x = 2, R1 for d²A/dx² > 0 or a sign change, A1 for all three dimensions.

Q5. (a) The width is x − (−x) = 2x and the height is cos x, so A = 2x cos x. A1, the width and height must both be shown.

(b) Graph y = 2x cos x for 0 < x < π/2 and use the maximum tool (or solve A′ = 2 cos x − 2x sin x = 0). x = 0.860 and the greatest area is A = 1.12 (3 s.f.). M1 for a valid method, stated: the function graphed or the equation A′ = 0, A1 for x = 0.860, M1 for substituting their x into A, A1 for 1.12.

Educerie · written from the published IB Diploma Programme Mathematics: analysis and approaches guide, first assessment 2021, section 5.8 Maxima, minima, inflexion and optimisation. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.

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