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Points of inflection

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What an inflection is

A point of inflection is where a curve changes its concavity — where it stops bending one way and starts bending the other.
It is a change in how the curve bends, not in whether it is rising or falling.

At an inflection the gradient is at a local extreme — the curve is at its steepest or shallowest there, in the sense that $f'$ itself turns around.

The two types

Non-stationary inflection — $f''(a) = 0$ but $f'(a) \neq 0$ . The concavity flips while the curve is still climbing or falling. This is the usual case.
Stationary inflection — $f''(a) = 0$ and $f'(a) = 0$ . The curve flattens completely, then carries on in the same direction. $y = x^3$ at the origin is the standard example.

Finding points of inflection

Differentiate twice to get $f''(x)$ .
Solve $f''(x) = 0$ — these are the candidates, not the answers.
Test for a sign change in $f''$ either side of each candidate.
Substitute into the original function for the $y$ -coordinate.
The sign-change test is not optional. $f''(a) = 0$ alone proves nothing.

Why the test matters

$y = x^4$ has $f''(x) = 12x^2$ , which is zero at $x = 0$ .
But $12x^2$ is a perfect square — it is positive on both sides and merely touches zero. The concavity never changes, so the origin is a minimum, not an inflection.
Compare $y = x^3$ , where $f''(x) = 6x$ crosses zero, going from negative to positive. That is a genuine inflection.
The rule: $f''$ must cross the axis, not touch it.

Setting out the sign test

Choose a test value just below and just above the candidate.
Evaluate the sign of $f''$ at each.
State the conclusion in words.
Worked through for $y = x^3 - 6x^2 + 5x$ :
- $f'(x) = 3x^2 - 12x + 5$ and $f''(x) = 6x - 12$
- $f''(x) = 0$ at $x = 2$
- $f''(1) = -6 < 0$ (concave down) and $f''(3) = 6 > 0$ (concave up)
- The concavity changes, so there is an inflection at $x = 2$
- $f(2) = 8 - 24 + 10 = -6$ , so the point is $(2, -6)$

What the two derivatives say together

$f'$ tells you whether the curve is going up or down.
$f''$ tells you which way it bends.
An inflection is a change in $f''$ , which says nothing about whether $f'$ changes — and usually it does not.
Note that a curve can be increasing throughout and still have an inflection. $y = x^3$ rises everywhere and has one at the origin.

Inflections on a cubic

Every cubic has exactly one point of inflection.
Since $f''$ is linear for a cubic, it has exactly one root and always crosses the axis there.
That inflection is the centre of symmetry of the cubic — the curve has rotational symmetry of order 2 about it.
For $y = ax^3 + bx^2 + cx + d$ , it sits at $x = -\dfrac{b}{3a}$ , exactly midway between the two turning points when they exist.

Sketching with all the information

A full sketch uses everything on the last three pages:
- Intercepts — set $x = 0$ and $y = 0$
- Stationary points — from $f'(x) = 0$ , with their nature
- Inflections — from $f''(x) = 0$ with a sign change
- Asymptotes — from the denominator and from limits at infinity
- End behaviour — what happens as $x \to \pm\infty$

Tips and Tricks

$f''(a) = 0$ is a candidate, not a conclusion. You must show the concavity actually changes. Stating " $f''(2) = 0$ therefore there is an inflection at $x = 2$ " is incomplete and can lose the justification mark.

The quickest way to show the change is to evaluate $f''$ at one point either side and state both signs. You do not need the values — just "negative, then positive".

Do not confuse a stationary point with an inflection. They answer different questions: $f' = 0$ locates stationary points, $f'' = 0$ locates candidate inflections. A point can be both (a stationary inflection), one, or neither.

Give the $y$ -coordinate. Substitute into $f$ , not into $f'$ or $f''$ . The report lists "calculated the $x$ - and $y$ -coordinates correctly for any stationary points and inflection points" as an Achieved behaviour — both coordinates, both kinds of point.

For a cubic, you can use the shortcut $x = -\frac{b}{3a}$ as a check on your $f'' = 0$ answer, but show the differentiation as your method — the specification requires the derivatives to be visible.

Worked ExampleA full analysis of a quartic

For $y = x^4 - 4x^3 + 10$ , find all stationary points and all points of inflection, stating the nature of each.

Step 1 — Find both derivatives

$\frac{dy}{dx} = 4x^3 - 12x^2$

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

Step 2 — Find the stationary points

Set the first derivative to zero and factorise:

$4x^3 - 12x^2 = 0$

$4x^2(x - 3) = 0 \quad \Rightarrow \quad x = 0 \text{ or } x = 3$

Coordinates, substituting into the original function:

$f(0) = 10 \qquad f(3) = 81 - 108 + 10 = -17$

Stationary points: $(0, 10)$ and $(3, -17)$ .

Step 3 — Classify $x = 3$ with the second derivative

$f''(3) = 12(9) - 24(3) = 108 - 72 = 36 > 0$

Positive, so $(3, -17)$ is a minimum.

Step 4 — Classify $x = 0$

$f''(0) = 0 - 0 = 0$

The test fails, so use a sign test on $f'(x) = 4x^2(x-3)$ :

At $x = -1$ : $4(1)(-4) = -16$ — negative
At $x = 1$ : $4(1)(-2) = -8$ — negative

The gradient is negative on both sides, so the curve does not turn: $(0, 10)$ is a stationary point of inflection.

Step 5 — Find the candidate inflections

Set the second derivative to zero:

$12x^2 - 24x = 0$

$12x(x - 2) = 0 \quad \Rightarrow \quad x = 0 \text{ or } x = 2$

Step 6 — Test $x = 0$ for a concavity change

$f''(-1) = 12 + 24 = 36$ — positive (concave up)
$f''(1) = 12 - 24 = -12$ — negative (concave down)

The sign changes, so $x = 0$ is a genuine inflection — and since $f'(0) = 0$ as well, it is the stationary inflection already identified. The two methods agree ✓

Step 7 — Test $x = 2$ for a concavity change

$f''(1) = -12$ — negative (concave down)
$f''(3) = 36$ — positive (concave up)

The sign changes, so $x = 2$ is also an inflection. Its height:

$f(2) = 16 - 32 + 10 = -6$

Since $f'(2) = 32 - 48 = -16 \neq 0$ , this is a non-stationary inflection — the curve is falling as it passes through.

Step 8 — Summarise

Point	$f'$	$f''$	Nature
$(0, 10)$	$0$	$0$ , changes sign	Stationary point of inflection
$(2, -6)$	$-16$	$0$ , changes sign	Non-stationary point of inflection
$(3, -17)$	$0$	$36 > 0$	Local minimum

Points of inflection

Covers AS91578

What an inflection is

A point of inflection is where a curve changes its concavity — where it stops bending one way and starts bending the other.
It is a change in how the curve bends, not in whether it is rising or falling.

At an inflection the gradient is at a local extreme — the curve is at its steepest or shallowest there, in the sense that $f'$ itself turns around.

The two types

Non-stationary inflection — $f''(a) = 0$ but $f'(a) \neq 0$ . The concavity flips while the curve is still climbing or falling. This is the usual case.
Stationary inflection — $f''(a) = 0$ and $f'(a) = 0$ . The curve flattens completely, then carries on in the same direction. $y = x^3$ at the origin is the standard example.

Finding points of inflection

Differentiate twice to get $f''(x)$ .
Solve $f''(x) = 0$ — these are the candidates, not the answers.
Test for a sign change in $f''$ either side of each candidate.
Substitute into the original function for the $y$ -coordinate.
The sign-change test is not optional. $f''(a) = 0$ alone proves nothing.

Why the test matters

$y = x^4$ has $f''(x) = 12x^2$ , which is zero at $x = 0$ .
But $12x^2$ is a perfect square — it is positive on both sides and merely touches zero. The concavity never changes, so the origin is a minimum, not an inflection.
Compare $y = x^3$ , where $f''(x) = 6x$ crosses zero, going from negative to positive. That is a genuine inflection.
The rule: $f''$ must cross the axis, not touch it.

Setting out the sign test

Choose a test value just below and just above the candidate.
Evaluate the sign of $f''$ at each.
State the conclusion in words.
Worked through for $y = x^3 - 6x^2 + 5x$ :
- $f'(x) = 3x^2 - 12x + 5$ and $f''(x) = 6x - 12$
- $f''(x) = 0$ at $x = 2$
- $f''(1) = -6 < 0$ (concave down) and $f''(3) = 6 > 0$ (concave up)
- The concavity changes, so there is an inflection at $x = 2$
- $f(2) = 8 - 24 + 10 = -6$ , so the point is $(2, -6)$

What the two derivatives say together

$f'$ tells you whether the curve is going up or down.
$f''$ tells you which way it bends.
An inflection is a change in $f''$ , which says nothing about whether $f'$ changes — and usually it does not.
Note that a curve can be increasing throughout and still have an inflection. $y = x^3$ rises everywhere and has one at the origin.

Inflections on a cubic

Every cubic has exactly one point of inflection.
Since $f''$ is linear for a cubic, it has exactly one root and always crosses the axis there.
That inflection is the centre of symmetry of the cubic — the curve has rotational symmetry of order 2 about it.
For $y = ax^3 + bx^2 + cx + d$ , it sits at $x = -\dfrac{b}{3a}$ , exactly midway between the two turning points when they exist.

Sketching with all the information

A full sketch uses everything on the last three pages:
- Intercepts — set $x = 0$ and $y = 0$
- Stationary points — from $f'(x) = 0$ , with their nature
- Inflections — from $f''(x) = 0$ with a sign change
- Asymptotes — from the denominator and from limits at infinity
- End behaviour — what happens as $x \to \pm\infty$

Tips and Tricks

The quickest way to show the change is to evaluate $f''$ at one point either side and state both signs. You do not need the values — just "negative, then positive".

Worked ExampleA full analysis of a quartic

For $y = x^4 - 4x^3 + 10$ , find all stationary points and all points of inflection, stating the nature of each.

Step 1 — Find both derivatives

$\frac{dy}{dx} = 4x^3 - 12x^2$

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

Step 2 — Find the stationary points

Set the first derivative to zero and factorise:

$4x^3 - 12x^2 = 0$

$4x^2(x - 3) = 0 \quad \Rightarrow \quad x = 0 \text{ or } x = 3$

Coordinates, substituting into the original function:

$f(0) = 10 \qquad f(3) = 81 - 108 + 10 = -17$

Stationary points: $(0, 10)$ and $(3, -17)$ .

Step 3 — Classify $x = 3$ with the second derivative

$f''(3) = 12(9) - 24(3) = 108 - 72 = 36 > 0$

Positive, so $(3, -17)$ is a minimum.

Step 4 — Classify $x = 0$

$f''(0) = 0 - 0 = 0$

The test fails, so use a sign test on $f'(x) = 4x^2(x-3)$ :

At $x = -1$ : $4(1)(-4) = -16$ — negative
At $x = 1$ : $4(1)(-2) = -8$ — negative

The gradient is negative on both sides, so the curve does not turn: $(0, 10)$ is a stationary point of inflection.

Step 5 — Find the candidate inflections

Set the second derivative to zero:

$12x^2 - 24x = 0$

$12x(x - 2) = 0 \quad \Rightarrow \quad x = 0 \text{ or } x = 2$

Step 6 — Test $x = 0$ for a concavity change

$f''(-1) = 12 + 24 = 36$ — positive (concave up)
$f''(1) = 12 - 24 = -12$ — negative (concave down)

The sign changes, so $x = 0$ is a genuine inflection — and since $f'(0) = 0$ as well, it is the stationary inflection already identified. The two methods agree ✓

Step 7 — Test $x = 2$ for a concavity change

$f''(1) = -12$ — negative (concave down)
$f''(3) = 36$ — positive (concave up)

The sign changes, so $x = 2$ is also an inflection. Its height:

$f(2) = 16 - 32 + 10 = -6$

Since $f'(2) = 32 - 48 = -16 \neq 0$ , this is a non-stationary inflection — the curve is falling as it passes through.

Step 8 — Summarise

Point	$f'$	$f''$	Nature
$(0, 10)$	$0$	$0$ , changes sign	Stationary point of inflection
$(2, -6)$	$-16$	$0$ , changes sign	Non-stationary point of inflection
$(3, -17)$	$0$	$36 > 0$	Local minimum

Points of inflection

What an inflection is

The two types

Finding points of inflection

Why the test matters

Setting out the sign test

What the two derivatives say together

Inflections on a cubic

Sketching with all the information

Step 1 — Find both derivatives

Step 2 — Find the stationary points

Step 3 — Classify x=3x = 3x=3 with the second derivative

Step 4 — Classify x=0x = 0x=0

Step 5 — Find the candidate inflections

Step 6 — Test x=0x = 0x=0 for a concavity change

Step 7 — Test x=2x = 2x=2 for a concavity change

Step 8 — Summarise

Points of inflection

What an inflection is

The two types

Finding points of inflection

Why the test matters

Setting out the sign test

What the two derivatives say together

Inflections on a cubic

Sketching with all the information

Step 1 — Find both derivatives

Step 2 — Find the stationary points

Step 3 — Classify x=3x = 3x=3 with the second derivative

Step 4 — Classify x=0x = 0x=0

Step 5 — Find the candidate inflections

Step 6 — Test x=0x = 0x=0 for a concavity change

Step 7 — Test x=2x = 2x=2 for a concavity change

Step 8 — Summarise

Step 3 — Classify $x = 3$ with the second derivative

Step 4 — Classify $x = 0$

Step 6 — Test $x = 0$ for a concavity change

Step 7 — Test $x = 2$ for a concavity change

Step 3 — Classify $x = 3$ with the second derivative

Step 4 — Classify $x = 0$

Step 6 — Test $x = 0$ for a concavity change

Step 7 — Test $x = 2$ for a concavity change