Inflection points | Year 12

01 / LEARN

The idea, step by step

An inflection changes the bend

A point of inflection is a point on the curve where concavity changes. For a sufficiently smooth function, a zero second derivative gives a candidate, not an automatic answer.

After finding a candidate x-value, test the second derivative on each side. Substitute into the original function to find its y-coordinate.

Stationary is a separate property

An inflection can have a zero or nonzero first derivative. A stationary inflection has a horizontal tangent but is neither a local maximum nor a local minimum.

If the original function is not defined at a candidate, there is no point on its graph there. A jump between separate branches is not itself an inflection point.

A common trap

Declaring an inflection whenever $f^{''} = 0$ .

Instead: A sign change must be established; $x^{4}$ is a useful counterexample.

02 / WORKED EXAMPLES

See the thinking

Example 1

A genuine change of concavity

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

$f^{''} = 6 x - 6$ , so the candidate is $x = 1$ .
The sign changes from negative to positive at 1.
$f (1) = - 2$ .

(1, - 2)

Example 2

Why a zero test is insufficient

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

$y^{''} = 12 x^{2}$ , which is zero at 0.
It is positive on both sides, so concavity does not change.

No.

Check your understanding

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

Show answer

Yes.

The second derivative $6 x$ changes sign and the first derivative $3 x^{2}$ is zero there.

03 / WORKSHEET

Your turn

6 questions · 16 marks · Work at your own pace.

Worksheet PDF 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 ↗

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