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Stationary points and their nature

Covers AS91578

What a stationary point is

A stationary point is a point where the curve has a horizontal tangent:

$f'(x) = 0$

The three kinds:
- Local maximum — the curve rises to the point and falls away
- Local minimum — the curve falls to the point and rises away
- Stationary point of inflection — the curve flattens but carries on in the same direction

"Local" means within a neighbourhood. A local maximum need not be the highest point on the whole curve.

Finding stationary points

Differentiate to get $f'(x)$ .
Set $f'(x) = 0$ and solve — write this equation down explicitly.
Substitute each solution into the original function to get the $y$ -coordinates.
State the coordinates as points, not just $x$ -values.
Worked through for $y = x^3 - 3x^2 - 9x + 5$ :
- $f'(x) = 3x^2 - 6x - 9 = 3(x - 3)(x + 1)$
- $f'(x) = 0$ gives $x = 3$ and $x = -1$
- $f(3) = 27 - 27 - 27 + 5 = -22$ and $f(-1) = -1 - 3 + 9 + 5 = 10$
- Stationary points: $(3, -22)$ and $(-1, 10)$

Method 1 — the second derivative test

Differentiate again and substitute the $x$ -value:
- $f''(a) < 0$ — the curve is concave down, so a maximum
- $f''(a) > 0$ — the curve is concave up, so a minimum
- $f''(a) = 0$ — the test fails; use the sign test instead
A memory aid: a negative second derivative means a hill; a positive one means a valley.
Continuing the example: $f''(x) = 6x - 6$ .
- $f''(3) = 12 > 0$ , so $(3, -22)$ is a minimum
- $f''(-1) = -12 < 0$ , so $(-1, 10)$ is a maximum
Note the trap: $f''(a) = 0$ does not prove an inflection. For $y = x^4$ , $f''(0) = 0$ but the origin is a minimum.

Method 2 — the sign test on $f'$

Always works, including when the second derivative test fails.
Choose a value of $x$ just below and just above the stationary point.
Evaluate the sign of $f'$ at each — the sign is all you need, not the value.
Read the pattern:

Sign of $f'$ before	at the point	after	Nature
$+$	$0$	$-$	Maximum
$-$	$0$	$+$	Minimum
$+$	$0$	$+$	Inflection
$-$	$0$	$-$	Inflection

Choose test values carefully — they must lie between neighbouring stationary points, not beyond them.

Method 3 — an annotated sketch

The 2026 specification for Level 2 explicitly permits justifying the nature "with an annotated sketch of the shape of the curve", and the same reasoning is accepted at Level 3.
For a cubic with a positive leading coefficient, the shape is fixed: it rises, turns down at a maximum, turns up at a minimum, then rises. The left-hand turning point is the maximum.

Global maximum and minimum on an interval

On a closed interval $a \leq x \leq b$ , the largest and smallest values can occur either at a stationary point or at an endpoint.
Evaluate the function at every stationary point in the interval and at both endpoints, then compare.
Forgetting the endpoints is a standard error in optimisation questions with a restricted domain.

Reading stationary points from $f'$

If you are given the graph of $f'$ $f^{'}$ rather than $f$ $f$ :
- Stationary points of $f$ are where the graph of $f'$ crosses the $x$ -axis.
- A crossing from positive to negative is a maximum of $f$ ; negative to positive is a minimum.
- A touch of the axis without crossing is a stationary point of inflection.

Tips and Tricks

Write $f'(x) = 0$ down before you solve it. The 2025 report is unusually specific here: "Too many candidates remove the denominator before communicating that the gradient at these points is equal to zero. The marker needs to be shown every step of the process, with full justification given." For a rational derivative, state $\frac{\text{numerator}}{\text{denominator}} = 0$ first, then say the numerator must be zero.

Give coordinates, not just $x$ -values. A stationary point is a point. Substituting back into the original function is part of the answer, not an optional extra.

Substitute into $f$ for the $y$ -value, not into $f'$ . Putting the $x$ -value into the derivative gives 0 every time, which is not the height of the point.

If $f''(a) = 0$ , do not conclude "inflection" — switch to the sign test on $f'$ . The second derivative test is silent in that case, not confirmatory.

For a global maximum on a closed interval, check the endpoints as well as the stationary points. The biggest value is often at an end.

The specification notes that at Achieved and Merit, optimisation questions will not require you to prove the nature — the paper will tell you to assume it. Do not spend time proving what you have been told to assume; but at Excellence, expect to justify it.

Worked ExampleStationary points of a rational function

Find the coordinates and nature of the stationary points of $y = \dfrac{x^2 + 9}{x}$ .

Step 1 — Simplify before differentiating

The denominator is a single term, so split the fraction rather than reaching for the quotient rule:

$y = \frac{x^2}{x} + \frac{9}{x} = x + 9x^{-1}$

Note the domain: $x \neq 0$ .

Step 2 — Differentiate

$\frac{dy}{dx} = 1 - 9x^{-2} = 1 - \frac{9}{x^2}$

Step 3 — State the stationary condition explicitly

Write the equation down before manipulating it — this is what the assessor report asks for:

$\frac{dy}{dx} = 0 \quad \Rightarrow \quad 1 - \frac{9}{x^2} = 0$

Step 4 — Solve

$\frac{9}{x^2} = 1$

$x^2 = 9$

$x = \pm 3$

Both roots are kept. The square root has two values, and both are in the domain since neither is 0.

Step 5 — Find the $y$ -coordinates

Substitute into the original function:

$f(3) = \frac{9 + 9}{3} = \frac{18}{3} = 6$

$f(-3) = \frac{9 + 9}{-3} = \frac{18}{-3} = -6$

The stationary points are $(3, 6)$ and $(-3, -6)$ .

Step 6 — Find the second derivative

$\frac{dy}{dx} = 1 - 9x^{-2} \quad \Rightarrow \quad \frac{d^2y}{dx^2} = 18x^{-3} = \frac{18}{x^3}$

Step 7 — Test each point

At $x = 3$ :

$\frac{d^2y}{dx^2} = \frac{18}{27} = \frac{2}{3} > 0$

Positive, so concave up — a minimum at $(3, 6)$ .

At $x = -3$ :

$\frac{d^2y}{dx^2} = \frac{18}{-27} = -\frac{2}{3} < 0$

Negative, so concave down — a maximum at $(-3, -6)$ .

Step 8 — Sense-check

A "maximum" lower than a "minimum" looks wrong, but it is correct here: the curve has a vertical asymptote at $x = 0$ separating two entirely separate branches. Each turning point is a local extremum on its own branch, and neither is global.

Stationary points and their nature

Covers AS91578

What a stationary point is

A stationary point is a point where the curve has a horizontal tangent:

$f'(x) = 0$

The three kinds:
- Local maximum — the curve rises to the point and falls away
- Local minimum — the curve falls to the point and rises away
- Stationary point of inflection — the curve flattens but carries on in the same direction

"Local" means within a neighbourhood. A local maximum need not be the highest point on the whole curve.

Finding stationary points

Differentiate to get $f'(x)$ .
Set $f'(x) = 0$ and solve — write this equation down explicitly.
Substitute each solution into the original function to get the $y$ -coordinates.
State the coordinates as points, not just $x$ -values.
Worked through for $y = x^3 - 3x^2 - 9x + 5$ :
- $f'(x) = 3x^2 - 6x - 9 = 3(x - 3)(x + 1)$
- $f'(x) = 0$ gives $x = 3$ and $x = -1$
- $f(3) = 27 - 27 - 27 + 5 = -22$ and $f(-1) = -1 - 3 + 9 + 5 = 10$
- Stationary points: $(3, -22)$ and $(-1, 10)$

Method 1 — the second derivative test

Differentiate again and substitute the $x$ -value:
- $f''(a) < 0$ — the curve is concave down, so a maximum
- $f''(a) > 0$ — the curve is concave up, so a minimum
- $f''(a) = 0$ — the test fails; use the sign test instead
A memory aid: a negative second derivative means a hill; a positive one means a valley.
Continuing the example: $f''(x) = 6x - 6$ .
- $f''(3) = 12 > 0$ , so $(3, -22)$ is a minimum
- $f''(-1) = -12 < 0$ , so $(-1, 10)$ is a maximum
Note the trap: $f''(a) = 0$ does not prove an inflection. For $y = x^4$ , $f''(0) = 0$ but the origin is a minimum.

Method 2 — the sign test on $f'$

Always works, including when the second derivative test fails.
Choose a value of $x$ just below and just above the stationary point.
Evaluate the sign of $f'$ at each — the sign is all you need, not the value.
Read the pattern:

Sign of $f'$ before	at the point	after	Nature
$+$	$0$	$-$	Maximum
$-$	$0$	$+$	Minimum
$+$	$0$	$+$	Inflection
$-$	$0$	$-$	Inflection

Choose test values carefully — they must lie between neighbouring stationary points, not beyond them.

Method 3 — an annotated sketch

The 2026 specification for Level 2 explicitly permits justifying the nature "with an annotated sketch of the shape of the curve", and the same reasoning is accepted at Level 3.
For a cubic with a positive leading coefficient, the shape is fixed: it rises, turns down at a maximum, turns up at a minimum, then rises. The left-hand turning point is the maximum.

Global maximum and minimum on an interval

On a closed interval $a \leq x \leq b$ , the largest and smallest values can occur either at a stationary point or at an endpoint.
Evaluate the function at every stationary point in the interval and at both endpoints, then compare.
Forgetting the endpoints is a standard error in optimisation questions with a restricted domain.

Reading stationary points from $f'$

If you are given the graph of $f'$ $f^{'}$ rather than $f$ $f$ :
- Stationary points of $f$ are where the graph of $f'$ crosses the $x$ -axis.
- A crossing from positive to negative is a maximum of $f$ ; negative to positive is a minimum.
- A touch of the axis without crossing is a stationary point of inflection.

Tips and Tricks

Give coordinates, not just $x$ -values. A stationary point is a point. Substituting back into the original function is part of the answer, not an optional extra.

Substitute into $f$ for the $y$ -value, not into $f'$ . Putting the $x$ -value into the derivative gives 0 every time, which is not the height of the point.

If $f''(a) = 0$ , do not conclude "inflection" — switch to the sign test on $f'$ . The second derivative test is silent in that case, not confirmatory.

For a global maximum on a closed interval, check the endpoints as well as the stationary points. The biggest value is often at an end.

Worked ExampleStationary points of a rational function

Find the coordinates and nature of the stationary points of $y = \dfrac{x^2 + 9}{x}$ .

Step 1 — Simplify before differentiating

The denominator is a single term, so split the fraction rather than reaching for the quotient rule:

$y = \frac{x^2}{x} + \frac{9}{x} = x + 9x^{-1}$

Note the domain: $x \neq 0$ .

Step 2 — Differentiate

$\frac{dy}{dx} = 1 - 9x^{-2} = 1 - \frac{9}{x^2}$

Step 3 — State the stationary condition explicitly

Write the equation down before manipulating it — this is what the assessor report asks for:

$\frac{dy}{dx} = 0 \quad \Rightarrow \quad 1 - \frac{9}{x^2} = 0$

Step 4 — Solve

$\frac{9}{x^2} = 1$

$x^2 = 9$

$x = \pm 3$

Both roots are kept. The square root has two values, and both are in the domain since neither is 0.

Step 5 — Find the $y$ -coordinates

Substitute into the original function:

$f(3) = \frac{9 + 9}{3} = \frac{18}{3} = 6$

$f(-3) = \frac{9 + 9}{-3} = \frac{18}{-3} = -6$

The stationary points are $(3, 6)$ and $(-3, -6)$ .

Step 6 — Find the second derivative

$\frac{dy}{dx} = 1 - 9x^{-2} \quad \Rightarrow \quad \frac{d^2y}{dx^2} = 18x^{-3} = \frac{18}{x^3}$

Step 7 — Test each point

At $x = 3$ :

$\frac{d^2y}{dx^2} = \frac{18}{27} = \frac{2}{3} > 0$

Positive, so concave up — a minimum at $(3, 6)$ .

At $x = -3$ :

$\frac{d^2y}{dx^2} = \frac{18}{-27} = -\frac{2}{3} < 0$

Negative, so concave down — a maximum at $(-3, -6)$ .

Stationary points and their nature

What a stationary point is

Finding stationary points

Method 1 — the second derivative test

Method 2 — the sign test on f′f'f′

Method 3 — an annotated sketch

Global maximum and minimum on an interval

Reading stationary points from f′f'f′

Step 1 — Simplify before differentiating

Step 2 — Differentiate

Step 3 — State the stationary condition explicitly

Step 4 — Solve

Step 5 — Find the yyy-coordinates

Step 6 — Find the second derivative

Step 7 — Test each point

Step 8 — Sense-check

Stationary points and their nature

What a stationary point is

Finding stationary points

Method 1 — the second derivative test

Method 2 — the sign test on f′f'f′

Method 3 — an annotated sketch

Global maximum and minimum on an interval

Reading stationary points from f′f'f′

Step 1 — Simplify before differentiating

Step 2 — Differentiate

Step 3 — State the stationary condition explicitly

Step 4 — Solve

Step 5 — Find the yyy-coordinates

Step 6 — Find the second derivative

Step 7 — Test each point

Step 8 — Sense-check

Method 2 — the sign test on $f'$

Reading stationary points from $f'$

Step 5 — Find the $y$ -coordinates

Method 2 — the sign test on $f'$

Reading stationary points from $f'$

Step 5 — Find the $y$ -coordinates