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Learn7357Paper 2 · Pure and MechanicsPure content (sections A–J)Stationary points and increasing functions
Pure content (sections A–J) · Subtopic 25 of 66

Stationary Points and Increasing Functions

Where a curve turns: finding stationary points with dydx=0\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 dydx=0\dfrac{dy}{dx}=0.
  • Use d2ydx2\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)>0f'(x)>0) or decreasing (f′(x)<0f'(x)<0).
  • Show that a function is increasing for all xx.
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 dydx>0\dfrac{dy}{dx}>0 the function increases; where <0<0 it decreases. Stationary points satisfy dydx=0\dfrac{dy}{dx}=0. Classify them with d2ydx2\dfrac{d^2y}{dx^2}: negative → maximum, positive → minimum, zero → check the gradient either side.

1. Increasing and decreasing

Gradient along y = x³ − 3x
−2−112−4−3−2−112340xymax (−1, 2)min (1, −2)gradient 3x² − 3 is zero at x = ±1
The dashed tangent tilts up, flattens at the maximum, tilts down, flattens again at the minimum, then tilts up. Where it is flat, 3x2−3=03x^2-3=0.
Worked exampleWhere is f(x)=x3−3xf(x)=x^3-3x decreasing?

Find the interval.

Goal
Where f′(x)<0f'(x)<0.
Strategy
Factorise f′f'; solve the quadratic inequality.
  1. f′(x)=3x2−3=3(x−1)(x+1)f'(x)=3x^2-3=3(x-1)(x+1)
    Differentiate; factorise.
  2. 3(x−1)(x+1)<0⇒−1<x<13(x-1)(x+1)<0\Rightarrow-1<x<1
    ∪-shaped quadratic negative between roots.
Answer−1<x<1-1<x<1.
✓ Checkf(0)=0f(0)=0, f(0.5)=−1.375f(0.5)=-1.375: going down ✓.

2. Classifying stationary points

Why the second derivative works
  1. Minimum: gradient goes −→0→+\text{Minimum: gradient goes }-\to0\to+
    Down into the valley, then up.
  2. ⇒dydx is increasing⇒d2ydx2>0\Rightarrow\frac{dy}{dx}\text{ is increasing}\Rightarrow\frac{d^2y}{dx^2}>0
    An increasing function has positive derivative.
  3. Maximum: +→0→−⇒d2ydx2<0\text{Maximum: }+\to0\to-\Rightarrow\frac{d^2y}{dx^2}<0
    Same reasoning, reversed.
If d2ydx2=0\frac{d^2y}{dx^2}=0 the test is inconclusive: look at the sign of dydx\frac{dy}{dx} just either side.
Worked exampley=2x3−9x2+12x−4y=2x^3-9x^2+12x-4

Find and classify the stationary points.

Goal
Coordinates and nature.
Strategy
Solve y′=0y'=0; find yy; use y′′y''.
  1. y′=6x2−18x+12=6(x−1)(x−2)y'=6x^2-18x+12=6(x-1)(x-2)
    Differentiate; factorise.
  2. x=1: y=1;x=2: y=0x=1:\ y=1;\qquad x=2:\ y=0
    Substitute into yy (not y′y').
  3. y′′=12x−18y''=12x-18
    Second derivative.
  4. y′′(1)=−6<0 (max),y′′(2)=6>0 (min)y''(1)=-6<0\ \text{(max)},\quad y''(2)=6>0\ \text{(min)}
    Classify.
AnswerMaximum (1,1)(1,1), minimum (2,0)(2,0).
✓ Checky(1.5)=6.75−20.25+18−4=0.5y(1.5)=6.75-20.25+18-4=0.5, between 11 and 00 ✓.
Worked exampleWhen the test fails: y=x4y=x^4 and y=x3y=x^3

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

Goal
Nature.
Strategy
y′′(0)=0y''(0)=0 in both, so test gradients either side.
  1. x4: y′=4x3; y′(−0.1)<0, y′(0.1)>0x^4:\ y'=4x^3;\ y'(-0.1)<0,\ y'(0.1)>0
    −- then ++.

  2. So x4x^4 has a minimum at 00.

  3. x3: y′=3x2; y′(±0.1)>0x^3:\ y'=3x^2;\ y'(\pm0.1)>0
    ++ then ++.

  4. So x3x^3 has a stationary point of inflection at 00.

Answerx4x^4: minimum; x3x^3: point of inflection.
✓ CheckGraphs: x4x^4 is a flat-bottomed ∪; x3x^3 levels off then keeps rising ✓.
Edge casey′′=0y''=0 doesn't mean inflection As x4x^4 shows, y′′=0y''=0 can happen at a minimum. Always check either side.
Pause and think

Find kk if y=x2+kxy=x^2+\frac kx has a stationary point at x=2x=2.

Summary

  • y′>0y'>0 increasing; <0<0 decreasing.
  • Stationary: y′=0y'=0.
  • y′′<0y''<0 max; >0>0 min; =0=0 check signs.
Key words
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
−2−112−4−3−2−112340xyy = x³ − 3xmaximum (−1, 2)minimum (1, −2)
At a stationary point the tangent is horizontal, dydx=0\dfrac{dy}{dx}=0. The function is decreasing between the two stationary points, where dydx<0\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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