Inflection points | Year 12

01 / LEARN

Understand the idea.

An inflection changes the bend

A point of inflection lies on the curve where concavity changes. For a smooth function, solving $f^{''} (x) = 0$ finds possible locations, but each candidate needs a sign change of $f^{''}$ on its two sides.

After verifying the change in bend, substitute the input into $f (x)$ to find the point's height. A solution of the second-derivative equation supplies an input, not a complete coordinate pair.

Stationary is a separate property

Stationary describes the first derivative; inflection describes concavity. An inflection is stationary only when $f^{'} = 0$ there. A horizontal tangent can therefore occur without a maximum or minimum.

A candidate must belong to the original curve. An excluded input separating two branches gives no point of inflection there, even if the signs of the second derivative differ between those branches.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A genuine change of concavity

Find the inflection point of $f (x) = x^{3} - 3 x^{2}$ .

Worked solution

1
Differentiate twice and find the candidate.
$\begin{aligned} f^{'} (x) & = 3 x^{2} - 6 x \\ f^{''} (x) & = 6 x - 6 = 6 (x - 1) \\ f^{''} (x) = 0 & ⟹ x = 1 \end{aligned}$
A zero second derivative identifies a possible change in concavity.
2
Establish a change of sign.
$\begin{aligned} x < 1 & ⟹ f^{''} (x) < 0 \\ x > 1 & ⟹ f^{''} (x) > 0 \end{aligned}$
The linear factor changes from negative to positive across 1.
3
Find and state the point.
$f (1) = 1^{3} - 3 {(1)}^{2} = - 2, inflection point (1, - 2)$
The polynomial is defined at 1 and changes from concave down to concave up there.

Answer

$(1, - 2)$

EXAMPLE 02

Why a zero test is insufficient

Does $y = x^{4}$ have an inflection at the origin?

Worked solution

1
Calculate the two derivatives.
$y^{'} = 4 x^{3}, y^{''} = 12 x^{2}$
The second derivative determines concavity.
2
Test the zero as a candidate.
$y^{''} (0) = 0, 12 x^{2} > 0 for every x \neq 0$
A nonzero square is positive on both sides of the origin.
3
Decide whether concavity changes.
$Concave up on both sides of 0; no inflection at (0, 0)$
A zero second derivative is insufficient when its sign does not change.

Answer

No.

03 / YOUR TURN

Try it yourself.

Check your understanding

Does $y = x^{3}$ have a stationary inflection at zero?

Show answer & reasoning

Yes.

For $y = x^{3}$ , $y^{''} = 6 x$ changes from negative to positive at zero, so the origin is an inflection. Also $y^{'} (0) = 3 {(0)}^{2} = 0$ , so the inflection is stationary.

Keep practising

6 worksheet questions · 16 marks

Open worksheet Worked solutions

Read the questions online

Find the inflection point of $y = x^{3} + 2 x$ .
2 marks
Explain why $y = x^{4} + 1$ has no inflection point.
2 marks
Find the inflection points of $y = x^{4} - 4 x^{3}$ .
4 marks
Find the tangent at the inflection of $y = x^{3} - 3 x^{2}$ .
3 marks
Does $y = e^{x}$ have an inflection? Justify.
2 marks
Classify the origin of $y = x^{5}$ as stationary/nonstationary and maximum/minimum/inflection.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the inflection point of $y = x^{3} + 2 x$ .
2 marks
The second derivative is $6 x$ .
It changes sign at zero; substitute into the original function.
$(0, 0)$
Explain why $y = x^{4} + 1$ has no inflection point.
2 marks
Zero at the origin is only a candidate.
The curve is concave up on both sides.
Its second derivative $12 x^{2}$ never changes sign.
Find the inflection points of $y = x^{4} - 4 x^{3}$ .
4 marks
The second derivative is $12 x (x - 2)$ .
Its sign changes at both 0 and 2.
Substitute both values into the original function.
$(0, 0)$ and $(2, - 16)$
Find the tangent at the inflection of $y = x^{3} - 3 x^{2}$ .
3 marks
The inflection is $(1, - 2)$ .
The derivative there is $3 - 6 = - 3$ .
Use $y + 2 = - 3 (x - 1)$ .
$y = - 3 x + 1$
Does $y = e^{x}$ have an inflection? Justify.
2 marks
Its second derivative is $e^{x}$ .
This is positive everywhere, so concavity does not change.
No.
Classify the origin of $y = x^{5}$ as stationary/nonstationary and maximum/minimum/inflection.
3 marks
The first derivative $5 x^{4}$ is zero at 0.
The second derivative $20 x^{3}$ changes sign.
The function continues increasing through zero.
Stationary inflection.

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Curriculum reference & lesson scope

SACE Stage 2 Mathematical Methods

Further differentiation and applications from the current SACE Methods outline. Logarithmic derivatives have their own chapter; integration follows later. Trigonometric calculus uses radians.

Official curriculum source ↗

In this lesson

Understand the idea.

An inflection changes the bend

Stationary is a separate property

See the method in action.

A genuine change of concavity

Differentiate twice and find the candidate.

Establish a change of sign.

Find and state the point.

Why a zero test is insufficient

Calculate the two derivatives.

Test the zero as a candidate.

Decide whether concavity changes.

Try it yourself.

Check your understanding

Keep practising