Resources For Teachers For Tutors For Students & Parents Pricing
Create Free Account
Year 12 Maths - Specialist (Unit 3 & Unit 4) Differential calculus

Second derivatives, concavity and points of inflection

20 practice questions 0 video lessons Theory + worked examples

Master second derivatives, concavity and points of inflection in Year 12 VCE Specialist Mathematics. The second derivative \(f''(x)\), or \(\dfrac{d^2y}{dx^2}\), measures how a curve bends: it is concave up where \(f''>0\) and concave down where \(f''<0\), with a point of inflection wherever \(f''\) is zero and changes sign. It sits in the Calculus area of study of the VCE Mathematics Study Design (VCAA), within the Differential calculus topic of Unit 3.

You will learn to differentiate twice, read concavity from the sign of \(f''\), locate genuine points of inflection, and classify stationary points with the second-derivative test — the graphing tools that sharpen curve sketching and optimisation across the calculus course.

Practice 20 questions
Practice questions

Every question with a fully worked solution.

Start practising
Create a free accountTrack your progress and save your work as you go.
Create free account

Theory

The second derivative \(f''(x)\) measures how a curve bends. In Year 12 Specialist Mathematics you use \(f''(x)\) and \(\dfrac{d^2y}{dx^2}\) to decide where a graph is concave up (\(f''>0\)) or concave down (\(f''<0\)), to locate points of inflection (where \(f''=0\) and changes sign), and to classify stationary points with the second-derivative test.

The second derivative is the derivative of the first derivative: differentiate \(f(x)\) once to get \(f'(x)\), then again to get \(f''(x)\). It is written \(f''(x)\) or \(\dfrac{d^2y}{dx^2}\), and it measures the rate of change of the gradient — how the curve is bending.

A graph is concave up on an interval where \(f''(x)>0\): the gradient is increasing and the curve holds water like a cup. It is concave down where \(f''(x)<0\): the gradient is decreasing and the curve spills. Concavity is about \(f''\), which is a separate question from whether the curve itself is rising or falling (that is about \(f'\)).

A point of inflection is a point where the concavity changes — from concave up to concave down, or the reverse. At such a point \(f''(x)=0\), but that alone is not enough: \(f''\) must also change sign there. For example \(f(x)=x^4\) has \(f''(0)=0\) yet stays concave up on both sides, so \(x=0\) is not a point of inflection.

The second-derivative test classifies a stationary point \(x=a\) (where \(f'(a)=0\)) by its concavity: if \(f''(a)>0\) the curve is concave up there, giving a local minimum; if \(f''(a)<0\) it is concave down, giving a local maximum; if \(f''(a)=0\) the test is inconclusive and you check the sign of \(f''\) on each side.

Concavity changing at a point of inflectionThe cubic y equals x cubed minus 3 x squared plus 4. To the left of x equals 1 the curve is concave down (bending downwards); to the right it is concave up (bending upwards). A red dot at the point (1, 2) marks the point of inflection, where the concavity changes and the second derivative is zero and changes sign. x y concave down concave up (1, 2)
Concavity changing at a point of inflection. The cubic is concave down (\(f''<0\)) for \(x<1\) and concave up (\(f''>0\)) for \(x>1\); the inflection point \((1,\ 2)\) is where \(f''=0\) and changes sign.
The second-derivative test on y = x^3 - 3xThe curve y equals x cubed minus 3 x. Where the second derivative is negative (x less than 0) the curve is concave down and holds a local maximum at (-1, 2). Where the second derivative is positive (x greater than 0) the curve is concave up and holds a local minimum at (1, -2). A red dot at the origin marks the point of inflection where f'' equals 0. x y f'' < 0 f'' > 0 local max local min
The second-derivative test. Where \(f''<0\) the curve is concave down and holds a local maximum \((-1,\ 2)\); where \(f''>0\) it is concave up and holds a local minimum \((1,\ -2)\); the inflection is at the origin.

The second derivative is the derivative of the first derivative, written two equivalent ways:

\[ f''(x) = \dfrac{d}{dx}\big(f'(x)\big), \qquad \dfrac{d^2y}{dx^2} = \dfrac{d}{dx}\!\left(\dfrac{dy}{dx}\right) \]
f″(x)=d2ydx2

The sign of \(f''\) fixes the concavity of the graph:

\[ f''(x) > 0 \ \Rightarrow\ \text{concave up}, \qquad f''(x) < 0 \ \Rightarrow\ \text{concave down} \]
f″(x)>0⇒concave up

A point of inflection needs both conditions — a zero and a change of sign:

\[ f''(x) = 0 \ \text{ and }\ f'' \text{ changes sign at } x \]
f″(x)=0

The second-derivative test classifies a stationary point \(x=a\), where \(f'(a)=0\):

At a stationary point \(x=a\)\(f''(a)>0\)\(f''(a)<0\)\(f''(a)=0\)
Conclusionlocal minimumlocal maximuminconclusive
A zero second derivative is not enough. \(f''(x)=0\) only flags a possible point of inflection. You must confirm that \(f''\) actually changes sign there — \(f(x)=x^4\) has \(f''(0)=0\) but stays concave up, so it has no inflection.

How to analyse a graph with the second derivative

  1. Differentiate twice to find \(f''(x)\) (or \(\dfrac{d^2y}{dx^2}\)).
  2. Read off concavity: the graph is concave up where \(f''(x)>0\) and concave down where \(f''(x)<0\).
  3. Find points of inflection: solve \(f''(x)=0\), confirm \(f''\) changes sign at each solution, then substitute back into \(f\) for the \(y\)-coordinate.
  4. Classify a stationary point \(x=a\) (where \(f'(a)=0\)) with the second-derivative test: \(f''(a)>0\) gives a local minimum, \(f''(a)<0\) a local maximum, and \(f''(a)=0\) is inconclusive — fall back to checking the sign of \(f''\) either side.
Example 1 — The second derivative of a cubic
For \(f(x) = x^3 - 6x^2 + 5x - 2\), find the second derivative \(f''(x)\).
Solution

Differentiate once for \(f'(x)\):

\(f'(x)\)\(=\)\(3x^2 - 12x + 5\)

Differentiate again for \(f''(x)\):

\(f''(x)\)\(=\)\(6x - 12\)

\(f''(x) = 6x - 12\).

Example 2 — Concavity and a point of inflection
For \(f(x) = x^3 - 3x^2 + 4\), state where the graph is concave up and concave down, and find the point of inflection.
Solution

Find the second derivative:

\(f'(x)\)\(=\)\(3x^2 - 6x\)
\(f''(x)\)\(=\)\(6x - 6\)

Set \(f''(x)=0\) for a possible inflection:

\(6x - 6\)\(=\)\(0\)
\(x\)\(=\)\(1\)

Check the sign of \(f''\) on each side of \(x=1\):

\(f''(0)\)\(=\)\(-6 < 0 \quad (\text{concave down})\)
\(f''(2)\)\(=\)\(6 > 0 \quad (\text{concave up})\)

The sign changes, so it is a genuine inflection. Find its \(y\)-coordinate:

\(f(1)\)\(=\)\(1 - 3 + 4\)
\(=\)\(2\)

Concave down for \(x<1\), concave up for \(x>1\); the point of inflection is \((1,\ 2)\).

Concavity changing at a point of inflectionThe cubic y equals x cubed minus 3 x squared plus 4. To the left of x equals 1 the curve is concave down (bending downwards); to the right it is concave up (bending upwards). A red dot at the point (1, 2) marks the point of inflection, where the concavity changes and the second derivative is zero and changes sign. x y concave down concave up (1, 2)
Example 3 — The second-derivative test
For \(f(x) = x^3 - 3x\), classify each stationary point with the second-derivative test, and state the point of inflection.
Solution

Stationary points where \(f'(x)=0\):

\(f'(x)\)\(=\)\(3x^2 - 3\)
\(3x^2 - 3\)\(=\)\(0\)
\(x\)\(=\)\(-1 \ \text{ or } \ 1\)

Second derivative:

\(f''(x)\)\(=\)\(6x\)

Apply the second-derivative test at each stationary point:

\(f''(-1)\)\(=\)\(-6 < 0 \quad (\text{local max})\)
\(f''(1)\)\(=\)\(6 > 0 \quad (\text{local min})\)

Inflection where \(f''=0\) with a sign change:

\(6x\)\(=\)\(0\)
\(x\)\(=\)\(0\)
\(f(0)\)\(=\)\(0\)

Local maximum at \(x=-1\), local minimum at \(x=1\); the point of inflection is \((0,\ 0)\).

The second-derivative test on y = x^3 - 3xThe curve y equals x cubed minus 3 x. Where the second derivative is negative (x less than 0) the curve is concave down and holds a local maximum at (-1, 2). Where the second derivative is positive (x greater than 0) the curve is concave up and holds a local minimum at (1, -2). A red dot at the origin marks the point of inflection where f'' equals 0. x y f'' < 0 f'' > 0 local max local min
Example 4 — When \(f''=0\) is not an inflection
Does the graph of \(f(x) = x^4\) have a point of inflection at \(x = 0\)?
Solution

Find \(f''\) and evaluate it at \(x=0\):

\(f'(x)\)\(=\)\(4x^3\)
\(f''(x)\)\(=\)\(12x^2\)
\(f''(0)\)\(=\)\(0\)

Check the sign of \(f''\) on each side of \(x=0\):

\(f''(-1)\)\(=\)\(12 > 0\)
\(f''(1)\)\(=\)\(12 > 0\)

The second derivative is zero at \(x=0\) but does not change sign — the curve is concave up on both sides.

No: \(f''(0)=0\) but \(f''\) does not change sign, so there is no point of inflection.

Common pitfalls

Assuming \(f''=0\) means an inflection. A zero second derivative only flags a possible inflection. You must check that \(f''\) changes sign there. \(f(x)=x^4\) has \(f''(0)=0\) but stays concave up, so \(x=0\) is not an inflection.
Confusing concavity with increasing/decreasing. Concave up (\(f''>0\)) is about how the curve bends, not whether it is rising. A curve can be concave up while falling — concavity is set by \(f''\), direction by \(f'\).
Forgetting the second-derivative test can fail. If \(f''(a)=0\) at a stationary point, the test is inconclusive — the point could be a max, a min, or an inflection. Check the sign of \(f''\) either side instead.
Mixing up the sign convention. \(f''>0\) is concave up (holds water); \(f''<0\) is concave down (spills). A quick sketch of a cup and a cap keeps them straight.

Frequently asked questions

What is the second derivative?

The second derivative \(f''(x)\), also written \(\dfrac{d^2y}{dx^2}\), is the derivative of the first derivative. You differentiate \(f(x)\) twice. It measures the rate of change of the gradient, which tells you how the curve is bending.

What do concave up and concave down mean?

A graph is concave up where \(f''(x)>0\): it bends upwards and holds water like a cup. It is concave down where \(f''(x)<0\): it bends downwards like a cap. Concavity is decided by the sign of the second derivative.

What is a point of inflection?

A point of inflection is where a curve changes concavity — from concave up to concave down, or the reverse. At an inflection \(f''(x)=0\) and \(f''\) changes sign there.

Does f''(x) = 0 always give a point of inflection?

No. \(f''(x)=0\) is necessary but not sufficient. \(f''\) must also change sign at that \(x\). For \(f(x)=x^4\), \(f''(0)=0\) but the curve stays concave up, so there is no inflection.

What is the second-derivative test?

It classifies a stationary point \(x=a\) (where \(f'(a)=0\)) by concavity: \(f''(a)>0\) means concave up, a local minimum; \(f''(a)<0\) means concave down, a local maximum; \(f''(a)=0\) is inconclusive.

How is concavity related to the first derivative?

Where a graph is concave up (\(f''>0\)) the gradient \(f'\) is increasing; where it is concave down (\(f''<0\)) the gradient \(f'\) is decreasing. The second derivative is the rate of change of the first derivative.

More in Differential calculus
Derivatives of inverse circular functions
20 questions
Implicit differentiation
20 questions
Related rates of change
20 questions