⛰️ Stationary Points and Optimisation

A Level Maths Year 1 · Differentiation

Level 3 · Ages 16–17

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1 Increasing and Decreasing Functions
The sign of the gradient tells you which way the curve is going. DfE content area G asks you to use the gradient to identify where a function is increasing or decreasing.
GradientFunction isGraph, left to right
$\mathrm{f}'(x) > 0$IncreasingGoing uphill
$\mathrm{f}'(x) < 0$DecreasingGoing downhill
$\mathrm{f}'(x) = 0$StationaryMomentarily flat
Worked Example 1 — Finding where a function increases

Find the values of $x$ for which $\mathrm{f}(x) = x^3 - 6x^2 + 9x + 1$ is increasing.

①$\mathrm{f}'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3)$
②Solve $\mathrm{f}'(x) > 0$: a positive quadratic with roots $1$ and $3$ is positive outside the roots.
③Increasing for $x < 1$ or $x > 3$. (Decreasing for $1 < x < 3$.)
Many textbooks include the end points and write $x \leqslant 1$ or $x \geqslant 3$; both are normally accepted. This is the quadratic inequality technique applied to $\mathrm{f}'(x)$.
Worked Example 2 — Increasing everywhere

Show that $\mathrm{f}(x) = x^3 + 3x - 5$ is an increasing function.

①$\mathrm{f}'(x) = 3x^2 + 3$
②$x^2 \geqslant 0$ for all $x$, so $3x^2 + 3 \geqslant 3 > 0$.
③$\mathrm{f}'(x) > 0$ for all $x$, so $\mathrm{f}$ is increasing for all $x$.
2 Finding and Classifying Stationary Points
Stationary point
A point where $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$
x y 1 2 3 4 2 4 6 O max (1, 5) min (3, 1) y = x³ − 6x² + 9x + 1 Gradient zero at both points; the second derivative tells them apart

There are two ways to decide the nature of a stationary point.

MethodLocal maximumLocal minimum
Second derivative$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} < 0$$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0$
Sign of gradient either side$+ \;\; 0 \;\; -$  (up, flat, down)$- \;\; 0 \;\; +$  (down, flat, up)
Why the second derivative works. At a maximum the gradient goes from positive, through zero, to negative — it is decreasing, so its derivative $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$ is negative. At a minimum the gradient is increasing, so $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0$.
If $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 0$ the test tells you nothing. The point could be a maximum, a minimum or a stationary point of inflection. Use the sign of the gradient either side instead.
Worked Example 3 — Using the second derivative

Find the stationary points of $y = x^3 - 6x^2 + 9x + 1$ and determine their nature.

①$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 12x + 9 = 3(x - 1)(x - 3) = 0$, so $x = 1$ or $x = 3$.
②$x = 1$: $y = 1 - 6 + 9 + 1 = 5$.   $x = 3$: $y = 27 - 54 + 27 + 1 = 1$.
③$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 6x - 12$.
④At $x = 1$: $-6 < 0$, so $(1, 5)$ is a local maximum.
⑤At $x = 3$: $6 > 0$, so $(3, 1)$ is a local minimum.
Worked Example 4 — A curve with negative powers

Find and classify the stationary points of $y = x + \dfrac{4}{x}$.

①$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1 - 4x^{-2} = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$.
②$x = 2$: $y = 4$.   $x = -2$: $y = -4$.
③$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 8x^{-3} = \dfrac{8}{x^3}$.
④At $x = 2$: $1 > 0$, minimum $(2, 4)$.   At $x = -2$: $-1 < 0$, maximum $(-2, -4)$.
The maximum is lower than the minimum. That is fine: "local" means highest or lowest nearby, and the curve has two separate branches.
Worked Example 5 — When the second derivative is zero

Find and classify the stationary points of $y = x^4 - 4x^3$.

①$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4x^3 - 12x^2 = 4x^2(x - 3) = 0$, so $x = 0$ or $x = 3$.
②Points: $(0, 0)$ and $(3, 81 - 108) = (3, -27)$.
③$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12x^2 - 24x$. At $x = 3$: $108 - 72 = 36 > 0$, so $(3, -27)$ is a minimum.
④At $x = 0$: $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 0$ — inconclusive. Test the gradient either side:
$x$$-0.1$$0$$0.1$
$4x^2(x-3)$$-0.124$$0$$-0.116$
Slope$\searrow$$\rightarrow$$\searrow$
⑤Negative on both sides, so $(0, 0)$ is a stationary point of inflection.
Choose test values close enough that no other stationary point lies between them and the point being tested.
3 Sketching the Gradient Function

Given the graph of $y = \mathrm{f}(x)$, you can sketch $y = \mathrm{f}'(x)$ without knowing the equation, by reading off the sign and size of the gradient.

On $y = \mathrm{f}(x)$On $y = \mathrm{f}'(x)$
Stationary pointCrosses or touches the $x$-axis
Increasing sectionAbove the $x$-axis
Decreasing sectionBelow the $x$-axis
Steepest pointA maximum or minimum of $\mathrm{f}'$
x y -2 -1 1 2 -4 -2 2 4 6 O y = f(x) = x³ − 3x y = f′(x) = 3x² − 3 Stationary points of f line up with the roots of f′
Worked Example 6 — Reading a gradient function

For $\mathrm{f}(x) = x^3 - 3x$, describe the graph of $y = \mathrm{f}'(x)$ using only the graph of $\mathrm{f}$.

①$\mathrm{f}$ has stationary points at $x = -1$ and $x = 1$, so $\mathrm{f}'$ crosses the $x$-axis there.
②$\mathrm{f}$ is increasing for $x < -1$ and $x > 1$, so $\mathrm{f}'$ is positive there; between them $\mathrm{f}'$ is negative.
③$\mathrm{f}$ is falling most steeply at $x = 0$, so $\mathrm{f}'$ has its minimum at $x = 0$ — a U-shaped curve. Indeed $\mathrm{f}'(x) = 3x^2 - 3$.
4 Optimisation Problems
Worked Example 7 — The open box

A rectangular sheet of card measures $24$ cm by $15$ cm. Squares of side $x$ cm are cut from each corner and the sides folded up to make an open box. Find the maximum volume.

①$V = x(24 - 2x)(15 - 2x) = 4x^3 - 78x^2 + 360x$, with $0 < x < 7.5$.
②$\dfrac{\mathrm{d}V}{\mathrm{d}x} = 12x^2 - 156x + 360 = 12(x^2 - 13x + 30) = 12(x - 3)(x - 10)$
③$x = 3$ or $x = 10$. But $x = 10$ is impossible ($2x$ would exceed $15$), so $x = 3$.
④$\dfrac{\mathrm{d}^2V}{\mathrm{d}x^2} = 24x - 156 = -84 < 0$ at $x = 3$: a maximum.
⑤$V = 3 \times 18 \times 9 = 486$ cm³.
Worked Example 8 — Minimising surface area of a can

A closed cylindrical can has volume $250\pi$ cm³. Show that its surface area is $A = 2\pi r^2 + \dfrac{500\pi}{r}$, and find the minimum surface area.

①Volume: $\pi r^2h = 250\pi \Rightarrow h = \dfrac{250}{r^2}$.
②$A = 2\pi r^2 + 2\pi rh = 2\pi r^2 + 2\pi r \times \dfrac{250}{r^2} = 2\pi r^2 + \dfrac{500\pi}{r}$ ✓
③$\dfrac{\mathrm{d}A}{\mathrm{d}r} = 4\pi r - 500\pi r^{-2} = 0 \Rightarrow r^3 = 125 \Rightarrow r = 5$.
④$\dfrac{\mathrm{d}^2A}{\mathrm{d}r^2} = 4\pi + 1000\pi r^{-3} > 0$ for $r > 0$: a minimum.
⑤$h = \dfrac{250}{25} = 10$ cm, and $A = 50\pi + 100\pi = 150\pi \approx 471$ cm².
The optimal can has height equal to its diameter. Real cans are usually taller — a good "evaluate the model" talking point.
5 Quick Reference

Increasing

$\mathrm{f}'(x) > 0$

Decreasing

$\mathrm{f}'(x) < 0$

Stationary

$\mathrm{f}'(x) = 0$; find $y$ from $\mathrm{f}(x)$.

Maximum

$\mathrm{f}''(x) < 0$, or gradient $+ \; 0 \; -$

Minimum

$\mathrm{f}''(x) > 0$, or gradient $- \; 0 \; +$

$\mathrm{f}''(x) = 0$

Inconclusive: test the gradient either side.

Inflection

Same sign of gradient on both sides.

Gradient graph

Stationary points of $\mathrm{f}$ are roots of $\mathrm{f}'$.

Optimising

Formula, eliminate, differentiate, $= 0$, check, interpret.

6 Practice Questions
Question 1

Find the set of values of $x$ for which $\mathrm{f}(x) = x^2 - 6x + 2$ is increasing.

▶ Show solution

$\mathrm{f}'(x) = 2x - 6 > 0 \Rightarrow x > 3$.

Question 2

Find the set of values of $x$ for which $\mathrm{f}(x) = x^3 - 12x$ is decreasing.

▶ Show solution

$\mathrm{f}'(x) = 3x^2 - 12 = 3(x - 2)(x + 2) < 0$.

A positive quadratic is negative between its roots: $-2 < x < 2$.

Question 3

Find the coordinates of the stationary point of $y = x^2 - 8x + 3$ and state its nature.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x - 8 = 0 \Rightarrow x = 4$; $\;y = 16 - 32 + 3 = -13$.

$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2 > 0$: minimum at $(4, -13)$.

Question 4

Find and classify the stationary points of $y = 2x^3 - 3x^2 - 12x + 4$.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 - 6x - 12 = 6(x - 2)(x + 1) = 0 \Rightarrow x = 2$ or $x = -1$.

$x = 2$: $y = 16 - 12 - 24 + 4 = -16$.   $x = -1$: $y = -2 - 3 + 12 + 4 = 11$.

$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12x - 6$. At $x = 2$: $18 > 0$, minimum $(2, -16)$. At $x = -1$: $-18 < 0$, maximum $(-1, 11)$.

Question 5

Find and classify the stationary points of $y = x^4 - 8x^2$.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4x^3 - 16x = 4x(x - 2)(x + 2) = 0 \Rightarrow x = 0, \pm 2$.

Points: $(0, 0)$, $(2, -16)$, $(-2, -16)$.

$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12x^2 - 16$. At $x = 0$: $-16$, maximum. At $x = \pm 2$: $32$, minima.

Question 6

Find the stationary point of $y = 4\sqrt x - x$ for $x > 0$ and show that it is a maximum.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x^{-1/2} - 1 = 0 \Rightarrow \sqrt x = 2 \Rightarrow x = 4$; $\;y = 8 - 4 = 4$.

$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -x^{-3/2} = -\dfrac18 < 0$ at $x = 4$: maximum at $(4, 4)$.

Question 7

Show that $y = x^3$ has a stationary point at the origin, explain why the second derivative test fails, and determine its nature.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 = 0 \Rightarrow x = 0$, $y = 0$.

$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 6x = 0$ at $x = 0$, which is inconclusive.

Gradient at $x = -0.1$: $0.03 > 0$; at $x = 0.1$: $0.03 > 0$. Positive on both sides, so it is a stationary point of inflection.

Question 8

The graph of $y = \mathrm{f}(x)$ is a cubic with a local maximum at $(-2, 5)$ and a local minimum at $(1, -3)$, and positive coefficient of $x^3$. Describe the graph of $y = \mathrm{f}'(x)$.

▶ Show solution

$\mathrm{f}'$ is a quadratic that crosses the $x$-axis at $x = -2$ and $x = 1$.

$\mathrm{f}$ is increasing for $x < -2$ and $x > 1$, so $\mathrm{f}'$ is positive there and negative between: a U-shaped parabola.

Its minimum is midway between the roots, at $x = -\tfrac12$ — where $\mathrm{f}$ is decreasing most steeply.

(The $y$-values $5$ and $-3$ do not appear on the graph of $\mathrm{f}'$.)

Question 9

An open-topped box has a square base of side $x$ cm and volume $32$ cm³. Show that its external surface area is $A = x^2 + \dfrac{128}{x}$, and find the dimensions that minimise $A$.

▶ Show solution

Height $h$: $x^2h = 32 \Rightarrow h = \dfrac{32}{x^2}$.

Area = base + four sides $= x^2 + 4xh = x^2 + \dfrac{128}{x}$ ✓

$\dfrac{\mathrm{d}A}{\mathrm{d}x} = 2x - 128x^{-2} = 0 \Rightarrow x^3 = 64 \Rightarrow x = 4$.

$\dfrac{\mathrm{d}^2A}{\mathrm{d}x^2} = 2 + 256x^{-3} = 6 > 0$: minimum.

Base $4$ cm by $4$ cm, height $2$ cm; minimum area $16 + 32 = 48$ cm².

Question 10

A farmer has $600$ m of fencing to make a rectangular enclosure, divided into two pens by an extra fence parallel to one pair of sides. Let $x$ m be the length of each of the three parallel fences and $y$ m the length of each of the other two sides.

(a) Show that the total area is $A = 300x - 1.5x^2$.   (b) Find the value of $x$ that maximises $A$, and the maximum area.   (c) Prove that this is a maximum.   (d) State the range of values of $x$ for which the model makes sense.

▶ Show solution

(a) Fencing: $3x + 2y = 600 \Rightarrow y = 300 - 1.5x$.

$A = xy = x(300 - 1.5x) = 300x - 1.5x^2$ ✓

(b) $\dfrac{\mathrm{d}A}{\mathrm{d}x} = 300 - 3x = 0 \Rightarrow x = 100$.

$y = 300 - 150 = 150$, and $A = 100 \times 150 = 15\,000$ m².

(c) $\dfrac{\mathrm{d}^2A}{\mathrm{d}x^2} = -3 < 0$, so this is a maximum.

(d) Need $x > 0$ and $y > 0$, i.e. $300 - 1.5x > 0 \Rightarrow x < 200$. So $0 < x < 200$.

Stationary Points and Optimisation · A Level Maths Year 1 · Created with MathJax