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Stationary Points

Q: What is a stationary point?

Any point on a curve where the gradient is exactly zero — momentarily flat. It could be a local maximum, a local minimum, or a point of inflection.

Q: How does the second derivative tell max from min?

It measures how the gradient itself is changing. Positive means the gradient is increasing through zero (curve bends upward, a minimum); negative means the gradient is decreasing through zero (curve bends downward, a maximum).

Q: What do I do if the second derivative is zero?

The test is inconclusive. Check the sign of the gradient just before and just after the stationary point — if it changes sign, it is a max or min; if it stays the same sign, it is a point of inflection.

At the very top of a hill or the very bottom of a valley, the ground is momentarily flat — the gradient there is exactly zero. Finding these flat points on a curve, and telling a peak from a trough, is one of the most powerful applications of differentiation: optimisation.

32 min Video by TLMaths Differentiation

The big picture

A $stationary point$ is any point where $\frac{d y}{d x} = 0$ — the curve is momentarily flat. Setting the derivative to zero and solving turns calculus into an algebra problem, and the SECOND derivative then tells you, without ever sketching the curve, whether you have found a maximum, a minimum, or something stranger. This is the calculus behind every "find the maximum profit" or "minimise the material used" exam question.

A-Level Maths G3-07: Introducing Stationary PointsWatch the walkthrough, then work through the notes below.

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What you'll be able to do

Find stationary points by solving $\frac{d y}{d x} = 0$
Use the second derivative test to classify a stationary point as a maximum or minimum
Recognise a point of inflection where the second derivative test is inconclusive
Apply stationary points to optimisation problems

Finding stationary points

A $stationary point$ occurs where the gradient of the curve is zero. Since $\frac{d y}{d x}$ gives the gradient at every point, setting it equal to zero and solving the resulting equation gives the $x$ -coordinates of every stationary point.

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

Solve this equation to locate stationary points.

1Differentiate:

\frac{d y}{d x} = 3 x^{2} - 6 x - 9

2Set to zero:

3 x^{2} - 6 x - 9 = 0

, divide by 3:

x^{2} - 2 x - 3 = 0

3Factorise:

(x - 3) (x + 1) = 0

, so

x = 3

x = - 1

4Find

y

: at

x = 3

y = 27 - 27 - 27 + 5 = - 22

. At

x = - 1

y = - 1 - 3 + 9 + 5 = 10

AnswerStationary points at

(3, - 22)

and

(- 1, 10)

Tip — A stationary point on its own is just an $x$ -value with zero gradient — you must substitute back into the ORIGINAL equation to find the matching $y$ -coordinate, using the derivative for classification instead.

Classifying with the second derivative

The $second derivative$ , $\frac{d ^{2} y}{d x ^{2}}$ , is found by differentiating the derivative again — it measures how the GRADIENT itself is changing. At a stationary point: if $\frac{d ^{2} y}{d x ^{2}} > 0$ , the gradient is increasing through zero — the curve curls upward, a $minimum$ . If $\frac{d ^{2} y}{d x ^{2}} < 0$ , it is a $maximum$ . If $\frac{d ^{2} y}{d x ^{2}} = 0$ , the test is inconclusive and you must check the gradient just either side instead.

\frac{d ^{2} y}{d x ^{2}} > 0 \Rightarrow minimum; \frac{d ^{2} y}{d x ^{2}} < 0 \Rightarrow maximum

The second derivative test.

1Second derivative:

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

2At

x = 3

6 (3) - 6 = 12 > 0

, so a minimum.

3At

x = - 1

6 (- 1) - 6 = - 12 < 0

, so a maximum.

Answer

(3, - 22)

is a minimum;

(- 1, 10)

is a maximum

Why this works

The second derivative tracks concavity: positive means the curve bends upward like a smile (∪, a minimum sits at the bottom); negative means it bends downward like a frown (∩, a maximum sits at the top).

When the test fails: points of inflection

If $\frac{d ^{2} y}{d x ^{2}} = 0$ at a stationary point, it could still be a maximum or minimum, or it could be a $point of inflection$ (where the curve flattens briefly but continues in the same general direction, like $y = x^{3}$ at the origin). In this case, check the sign of the gradient just before and just after the point.

\frac{d y}{d x} = 3 x^{2} = 0

x = 0

\frac{d ^{2} y}{d x ^{2}} = 6 x = 0

x = 0

too — inconclusive.

3Check the gradient either side: at

x = - 1

3 (- 1)^{2} = 3 > 0

; at

x = 1

3 (1)^{2} = 3 > 0

4The gradient is positive on BOTH sides — it never changes sign, so this is a point of inflection, not a max or min.

AnswerA stationary point of inflection at

(0, 0)

— gradient is positive both sides.

Think like an examiner

Always find BOTH coordinates of a stationary point — the $x$ -value from solving $\frac{d y}{d x} = 0$ and the matching $y$ -value from the original equation.
When the second derivative test gives zero, do not assume it is automatically a point of inflection — you must check the gradient sign either side to be certain.
State your classification clearly (minimum / maximum / point of inflection) with the reasoning shown, not just the number.

Common misconceptions

Only finding the

x

-coordinate and stopping.

Substitute back into the ORIGINAL equation to find the

y

-coordinate too — a stationary "point" needs both coordinates.

Assuming

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

always means a point of inflection.

It means the test is INCONCLUSIVE — you must check the gradient sign either side of the point to confirm what type it is.

Mixing up the sign convention: thinking positive second derivative means maximum.

Positive second derivative means MINIMUM (curve bends upward, like ∪); negative means maximum (∩).

Stationary point classification

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

Locates stationary points.

\frac{d ^{2} y}{d x ^{2}} > 0 \Rightarrow min

Curve bends upward.

\frac{d ^{2} y}{d x ^{2}} < 0 \Rightarrow max

Curve bends downward.

Stretch yourself

Find and classify all stationary points of $y = x^{4} - 4 x^{3}$ .

Hint — Solve $\frac{d y}{d x} = 0$ by factorising fully — expect two solutions — then use the second derivative on each.

Questions students ask

What is a stationary point?

Any point on a curve where the gradient is exactly zero — momentarily flat. It could be a local maximum, a local minimum, or a point of inflection.

How does the second derivative tell max from min?

It measures how the gradient itself is changing. Positive means the gradient is increasing through zero (curve bends upward, a minimum); negative means the gradient is decreasing through zero (curve bends downward, a maximum).

What do I do if the second derivative is zero?

The test is inconclusive. Check the sign of the gradient just before and just after the stationary point — if it changes sign, it is a max or min; if it stays the same sign, it is a point of inflection.

Key takeaways

Stationary points occur where $\frac{d y}{d x} = 0$ ; find the $x$ -value then the $y$ -value from the original equation.
Second derivative positive $\Rightarrow$ minimum; negative $\Rightarrow$ maximum.
Second derivative zero is inconclusive — check the gradient sign either side.
A point of inflection has the same gradient sign on both sides of the stationary point.

How this fits the course

Build on

Tangents and Normals

The Chain, Product and Quotient Rules

Leads to

The Chain, Product and Quotient Rules

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Stationary Points

32 min Video by TLMaths Differentiation