Stationary Points, Second Derivatives and Points of Inflection — Edexcel A-level Mathematics (Pure)

SPEC 7.3, 7.1 • DIFFERENTIATION

Stationary Points, Second Derivatives & Points of Inflection

Finding and classifying turning points, understanding concavity, and identifying inflection points

What are Stationary Points?

Definition

A stationary point is a point on a curve where the gradient is zero:

f^{'} (x) = 0

At a stationary point, the tangent to the curve is horizontal.

How to Find Stationary Points

Differentiate to find

f^{'} (x)

Solve

f^{'} (x) = 0

to find x-coordinates

Substitute back into

f (x)

to find y-coordinates

Three Types

There are three types of stationary points:

Local Maximum — highest point in local region

Local Minimum — lowest point in local region

Stationary Point of Inflection — curve flattens but doesn't turn

Types of Stationary Points

Local Maximum

↗ increasing → peak → ↘ decreasing

f'(x): + → 0 → −

Local Minimum

↘ decreasing → trough → ↗ increasing

f'(x): − → 0 → +

↗ increasing → flat → ↗ increasing

f'(x): + → 0 → +

OR − → 0 → −

(no sign change)

Key Question

How do we determine which type of stationary point we have? We use either the First Derivative Test or the Second Derivative Test.

The Second Derivative

Definition

The second derivative is the derivative of the first derivative:

f^{''} (x) = \frac{d ^{2} y}{d x ^{2}} = \frac{d}{d x} [f^{'} (x)]

What It Tells Us

The second derivative measures the rate of change of the gradient.

If

f^{''} (x) > 0

: gradient is increasing (curve bending upward)

If

f^{''} (x) < 0

: gradient is decreasing (curve bending downward)

Notation

Three equivalent ways to write the second derivative:

f^{''} (x)

— "f double-dash of x"

\frac{d ^{2} y}{d x ^{2}}

— "d-two-y by dx-squared"

y^{''}

— "y double-dash"

The Second Derivative Test

Once you've found a stationary point at

x = a

(where

f^{'} (a) = 0

), evaluate

f^{''} (a)

:

Local Minimum

f^{'} (a) = 0

AND

f^{''} (a) > 0

Curve is concave UP at stationary point

Think: "smile shape" ∪ = minimum

Local Maximum

f^{'} (a) = 0

AND

f^{''} (a) < 0

Curve is concave DOWN at stationary point

Think: "frown shape" ∩ = maximum

Inconclusive Case

If

f^{'} (a) = 0

AND

f^{''} (a) = 0

, the test is INCONCLUSIVE.

You must use the First Derivative Test instead.

The First Derivative Test

Check the sign of $f^{'} (x)$ just before and just after the stationary point:

Local Maximum

f^{'} (x)

: + → 0 → −

Gradient changes from positive to negative

Local Minimum

f^{'} (x)

: − → 0 → +

Gradient changes from negative to positive

Inflection Point

f^{'} (x)

: + → 0 → + or − → 0 → −

Gradient does NOT change sign

When to Use Each Test

Second Derivative Test: Quick and efficient when

f^{''} (a) \neq = 0

First Derivative Test: Essential when

f^{''} (a) = 0

(second derivative test fails)

Worked Example01

Find and classify the stationary points of $y = x^{3} - 6 x^{2} + 9 x + 1$ using the second derivative test.

1

Find the first derivative:

\frac{d y}{d x} = 3 x^{2} - 12 x + 9

2

Set

\frac{d y}{d x} = 0

:

3 x^{2} - 12 x + 9 = 0

3

Factorise:

3 (x^{2} - 4 x + 3) = 3 (x - 1) (x - 3) = 0

4

Stationary points at

x = 1

and

x = 3

5

Find second derivative:

\frac{d ^{2} y}{d x ^{2}} = 6 x - 12

6

At

x = 1

:

f^{''} (1) = 6 (1) - 12 = - 6 < 0

→ Local Maximum

7

At

x = 3

:

f^{''} (3) = 6 (3) - 12 = 6 > 0

→ Local Minimum

8

Find y-coordinates: At

x = 1

:

y = 1 - 6 + 9 + 1 = 5

9

At

x = 3

:

y = 27 - 54 + 27 + 1 = 1

AnswerLocal maximum at (1, 5), local minimum at (3, 1)

Worked Example02

For $y = x^{4} - 4 x^{3}$ , find the stationary points and classify them using appropriate tests.

1

\frac{d y}{d x} = 4 x^{3} - 12 x^{2} = 4 x^{2} (x - 3)

2

Set

\frac{d y}{d x} = 0

:

4 x^{2} (x - 3) = 0

→

x = 0

or

x = 3

3

\frac{d ^{2} y}{d x ^{2}} = 12 x^{2} - 24 x

4

At

x = 3

:

f^{''} (3) = 12 (9) - 24 (3) = 108 - 72 = 36 > 0

→ Local Min

5

At

x = 0

:

f^{''} (0) = 0

— INCONCLUSIVE! Use first derivative test.

6

Check sign of

f^{'} (x)

either side of

x = 0

:

•

At

x = - 1

:

f^{'} (- 1) = 4 (1) (- 4) = - 16 < 0

•

At

x = 1

:

f^{'} (1) = 4 (1) (- 2) = - 8 < 0

7

Sign is − → 0 → − (no change) → Stationary point of inflection

8

Coordinates: (0, 0) inflection, (3, -27) local min

AnswerStationary point of inflection at (0, 0), local minimum at (3, -27)

Concavity

Definition

Concavity describes how a curve bends. It is determined by the sign of the second derivative.

Concave Up

f^{''} (x) > 0

Gradient is increasing

Curve "holds water" like a bowl

Tangent lies BELOW curve

Concave Down

f^{''} (x) < 0

Gradient is decreasing

Curve "sheds water" like umbrella

Tangent lies ABOVE curve

Important

Concavity is about how the gradient is changing — not whether the function is increasing or decreasing. A function can be increasing AND concave down, or decreasing AND concave up.

Worked Example03

Find the intervals where $y = x^{3} - 3 x^{2} + 2$ is concave up and concave down.

1

Find the first derivative:

\frac{d y}{d x} = 3 x^{2} - 6 x

2

Find the second derivative:

\frac{d ^{2} y}{d x ^{2}} = 6 x - 6

3

Find where

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

:

6 x - 6 = 0

→

x = 1

4

Test sign of

\frac{d ^{2} y}{d x ^{2}}

in each region:

•

For

x < 1

(e.g.,

x = 0

):

6 (0) - 6 = - 6 < 0

→ Concave Down

•

For

x > 1

(e.g.,

x = 2

):

6 (2) - 6 = 6 > 0

→ Concave Up

AnswerConcave down for

x < 1

, concave up for

x > 1

Points of Inflection

Definition

A point of inflection is where the curve changes from concave up to concave down, or vice versa.

How to Find Points of Inflection

Find

f^{''} (x)

Solve

f^{''} (x) = 0

to find potential inflection points

VERIFY that

f^{''} (x)

changes sign at each point

Find the y-coordinate by substituting into

f (x)

Critical Warning

f^{''} (x) = 0

is NECESSARY but NOT SUFFICIENT for a point of inflection!

You MUST verify that

f^{''} (x)

changes sign either side of the point.

Example:

y = x^{4}

has

f^{''} (0) = 0

, but the origin is NOT an inflection point because

f^{''} (x) > 0

on both sides.

Types of Inflection Points

$f^{'} (x) = 0$ AND $f^{''} (x) = 0$

Tangent is horizontal

Example:

y = x^{3}

at origin

$f^{'} (x) \neq = 0$ but $f^{''} (x) = 0$

Tangent is NOT horizontal

Example:

y = x^{3} - 3 x

at

x = 0

Verification Method

After finding

f^{''} (a) = 0

, check the sign of

f^{''} (x)

at points either side:

If signs are different (e.g., + then −) → it IS a point of inflection ✓

If signs are the same (e.g., + then +) → it is NOT a point of inflection ✗

Worked Example04

Find all points of inflection of $y = x^{4} - 6 x^{2} + 8 x$ and verify each one.

1

\frac{d y}{d x} = 4 x^{3} - 12 x + 8

2

\frac{d ^{2} y}{d x ^{2}} = 12 x^{2} - 12 = 12 (x^{2} - 1) = 12 (x - 1) (x + 1)

3

Set

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

:

x = 1

or

x = - 1

4

Verify sign change at

x = - 1

:

•

At

x = - 2

:

12 (4 - 1) = 36 > 0

•

At

x = 0

:

12 (0 - 1) = - 12 < 0

✓ Sign change

5

Verify sign change at

x = 1

:

•

At

x = 0

:

- 12 < 0

•

At

x = 2

:

12 (4 - 1) = 36 > 0

✓ Sign change

6

Find y-coordinates:

•

At

x = - 1

:

y = 1 - 6 - 8 = - 13

•

At

x = 1

:

y = 1 - 6 + 8 = 3

AnswerPoints of inflection at

(- 1, - 13)

and

(1, 3)

Practice

Question 1: Find and classify the stationary points of

y = 2 x^{3} - 9 x^{2} + 12 x - 3

using the second derivative test.

\frac{d y}{d x} = 6 x^{2} - 18 x + 12 = 6 (x^{2} - 3 x + 2) = 6 (x - 1) (x - 2)

Stationary points at

x = 1

and

x = 2

\frac{d ^{2} y}{d x ^{2}} = 12 x - 18

At

x = 1

:

f^{''} (1) = 12 - 18 = - 6 < 0

→ Local Maximum

At

x = 2

:

f^{''} (2) = 24 - 18 = 6 > 0

→ Local Minimum

At

x = 1

:

y = 2 - 9 + 12 - 3 = 2

At

x = 2

:

y = 16 - 36 + 24 - 3 = 1

AnswerLocal maximum at (1, 2), local minimum at (2, 1)

Practice

Question 2: Find any points of inflection of

y = x^{3} - 6 x^{2} + 12 x - 5

and verify your answer.

\frac{d y}{d x} = 3 x^{2} - 12 x + 12 = 3 (x^{2} - 4 x + 4) = 3 (x - 2)^{2}

\frac{d ^{2} y}{d x ^{2}} = 6 x - 12 = 6 (x - 2)

Set

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

:

x = 2

Verify sign change:

At

x = 1

:

6 (1 - 2) = - 6 < 0

(concave down)

At

x = 3

:

6 (3 - 2) = 6 > 0

(concave up)

Sign change confirmed ✓

At

x = 2

:

y = 8 - 24 + 24 - 5 = 3

Note: Since $f^{'} (2) = 3 (0)^{2} = 0$ , this is a stationary point of inflection.

AnswerPoint of inflection at (2, 3)

Practice

Question 3: For

y = x^{4} - 8 x^{2}

, find all stationary points and all points of inflection.

\frac{d y}{d x} = 4 x^{3} - 16 x = 4 x (x^{2} - 4) = 4 x (x - 2) (x + 2)

Stationary points at

x = 0, 2, - 2

\frac{d ^{2} y}{d x ^{2}} = 12 x^{2} - 16

At

x = 0

:

f^{''} (0) = - 16 < 0

→ Local Max at (0, 0)

At

x = 2

:

f^{''} (2) = 48 - 16 = 32 > 0

→ Local Min

At

x = - 2

:

f^{''} (- 2) = 48 - 16 = 32 > 0

→ Local Min

y-coords:

y (2) = 16 - 32 = - 16

,

y (- 2) = 16 - 32 = - 16

Inflection points:

12 x^{2} - 16 = 0

→

x^{2} = \frac{4}{3}

→

x = \pm \frac{2}{3} = \pm \frac{2 3}{3}

Sign change verified (+ → − → + as x increases through these points)

AnswerLocal max at (0, 0); Local min at (2, -16) and (-2, -16); Inflection points at

x = \pm \frac{2 3}{3}

Summary

Classifying Stationary Points

TestConditionClassification

Second Derivative Test

f^{'} (a) = 0

,

f^{''} (a) > 0

Local Minimum

f^{'} (a) = 0

,

f^{''} (a) < 0

Local Maximum

f^{'} (a) = 0

,

f^{''} (a) = 0

Inconclusive → use 1st deriv test

First Derivative Test

f^{'} (x)

: + → 0 → −Local Maximum

f^{'} (x)

: − → 0 → +Local Minimum

f^{'} (x)

: no sign changeStationary Inflection

Key Exam Tips

Always give full coordinates of stationary points

If

f^{''} (a) = 0

, you must use the first derivative test

f^{''} (x) = 0

alone does NOT guarantee an inflection point — verify sign change!

Concave up (∪) means

f^{''} > 0

; Concave down (∩) means

f^{''} < 0