Learn9MA0Paper 2 · Pure Mathematics 2DifferentiationStationary points and increasing functions

Differentiation · Subtopic 3 of 10

Stationary Points and Increasing Functions

Where a curve turns: finding stationary points with $\dfrac{dy}{dx}=0$ , deciding whether each is a maximum or minimum with the second derivative, and finding where a function is increasing or decreasing.

Cambridge 9709 Paper 1Edexcel YMA01 WMA12Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 45 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Find stationary points by solving $\dfrac{dy}{dx}=0$ .
Use $\dfrac{d^2y}{dx^2}$ (or the gradient either side) to decide whether a point is a maximum or a minimum.
Find the intervals where a function is increasing ( $f'(x)>0$ ) or decreasing ( $f'(x)<0$ ).
Show that a function is increasing for all $x$ .

Pause and think

At the very top of a hill, which way is the path sloping?

Section 2 of 14

Core idea

Start hereThe top of the hill is flat

At the exact top of a hill, you're neither going up nor down: for that instant, the ground is level. The same is true at the bottom of a valley. So to find highest and lowest points of a curve, look for where its gradient is zero.

The big idea

Where $\dfrac{dy}{dx}>0$ the function increases; where $<0$ it decreases. Stationary points satisfy $\dfrac{dy}{dx}=0$ . Classify them with $\dfrac{d^2y}{dx^2}$ : negative → maximum, positive → minimum, zero → check the gradient either side.

1. Increasing and decreasing

Gradient along y = x³ − 3x

x = -2.20

The dashed tangent tilts up, flattens at the maximum, tilts down, flattens again at the minimum, then tilts up. Where it is flat,

3x^2-3=0

Worked exampleWhere is $f(x)=x^3-3x$ decreasing?

Find the interval.

Goal: Where $f'(x)<0$ .
Strategy: Factorise $f'$ ; solve the quadratic inequality.

$f'(x)=3x^2-3=3(x-1)(x+1)$
Differentiate; factorise.
$3(x-1)(x+1)<0\Rightarrow-1<x<1$
∪-shaped quadratic negative between roots.

Answer

-1<x<1

✓ Check

f(0)=0

f(0.5)=-1.375

: going down ✓.

2. Classifying stationary points

Why the second derivative works

$\text{Minimum: gradient goes }-\to0\to+$
Down into the valley, then up.
$\Rightarrow\frac{dy}{dx}\text{ is increasing}\Rightarrow\frac{d^2y}{dx^2}>0$
An increasing function has positive derivative.
$\text{Maximum: }+\to0\to-\Rightarrow\frac{d^2y}{dx^2}<0$
Same reasoning, reversed.

\frac{d^2y}{dx^2}=0

the test is inconclusive: look at the sign of

\frac{dy}{dx}

just either side.

Worked example $y=2x^3-9x^2+12x-4$

Find and classify the stationary points.

Goal: Coordinates and nature.
Strategy: Solve $y'=0$ ; find $y$ ; use $y''$ .

$y'=6x^2-18x+12=6(x-1)(x-2)$
Differentiate; factorise.
$x=1:\ y=1;\qquad x=2:\ y=0$
Substitute into $y$ (not $y'$ ).
$y''=12x-18$
Second derivative.
$y''(1)=-6<0\ \text{(max)},\quad y''(2)=6>0\ \text{(min)}$
Classify.

AnswerMaximum

(1,1)

, minimum

(2,0)

✓ Check

y(1.5)=6.75-20.25+18-4=0.5

, between

1

and

0

✓.

Worked exampleWhen the test fails: $y=x^4$ and $y=x^3$

Classify the stationary point at $x=0$ for each.

Goal: Nature.
Strategy: $y''(0)=0$ in both, so test gradients either side.

$x^4:\ y'=4x^3;\ y'(-0.1)<0,\ y'(0.1)>0$
$-$ then $+$ .
So $x^4$ has a minimum at $0$ .
$x^3:\ y'=3x^2;\ y'(\pm0.1)>0$
$+$ then $+$ .
So $x^3$ has a stationary point of inflection at $0$ .

Answer

x^4

: minimum;

x^3

: point of inflection.

✓ CheckGraphs:

x^4

is a flat-bottomed ∪;

x^3

levels off then keeps rising ✓.

Edge case $y''=0$ doesn't mean inflection As

x^4

shows,

y''=0

can happen at a minimum. Always check either side.

Pause and think

Find $k$ if $y=x^2+\frac kx$ has a stationary point at $x=2$ .

Summary

$y'>0$ increasing; $<0$ decreasing.
Stationary: $y'=0$ .
$y''<0$ max; $>0$ min; $=0$ check signs.

Stationary point: Where the gradient is zero.
Point of inflection: Where the curve changes from bending one way to the other.

Stationary points of y = x³ − 3x

At a stationary point the tangent is horizontal,

\dfrac{dy}{dx}=0

. The function is decreasing between the two stationary points, where

\dfrac{dy}{dx}<0

HistoryFermat's insight. Around 1636, Pierre de Fermat noticed that near a maximum or minimum a function changes very little, so its "slope" must vanish there. This idea, now called Fermat's theorem on stationary points, is the basis of every optimisation problem in the next lesson.

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