IB Mathematics Applications & Interpretation Topic…

Why The Second Derivative Matters: Measuring How The Change Itself Changes

The first derivative $f'(x)$ tells you the gradient: whether a quantity is rising or falling, and how fast.
But often the interesting question is how the change itself is changing.
1. Is a car speeding up or easing off? Is an epidemic's daily case count still accelerating, or has the growth started to slow?
The second derivative answers exactly this, because it is the rate of change of the gradient.
It also gives a quick way to tell a peak from a trough, and to find the moment a model switches from bending one way to bending the other (the point of inflexion), which in a logistic growth model is the moment growth is fastest.

Analogy

Think of driving: position is where you are, the first derivative is your speed, and the second derivative is your acceleration.
A positive second derivative means you are speeding up; a negative one means you are braking, even if you are still moving forward.

The Second Derivative: The Rate Of Change Of The Gradient

To get the second derivative, differentiate the function once to get $f'(x)$, then differentiate again.
It is written $f''(x)$ or $\dfrac{d^2y}{dx^2}$, and higher derivatives follow the same pattern as $f^{(n)}(x)$ or $\dfrac{d^n y}{dx^n}$. $$ f''(x) = \frac{d}{dx}\big(f'(x)\big) $$

Note

The notation $\dfrac{d^n y}{dx^n}$ for higher derivatives is listed in the formula booklet, so you only need to recognise and use it.
Each derivative is found with the standard rules from 5.9, applied one more time.

Example question

Given $f(x) = x^4 - 2x^2$, find $f''(x)$ and state whether the curve is concave up or concave down at $x = 1$.

Solution

Differentiate once: $$f'(x) = 4x^3 - 4x$$
Differentiate again: $$f''(x) = 12x^2 - 4$$
At $x = 1$: $f''(1) = 12 - 4 = 8 > 0$, so the curve is concave up there.

Concavity: Which Way A Curve Bends

The sign of the second derivative tells you which way the curve is bending, independently of whether it is going up or down.
Where $f''(x) > 0$ the curve is concave up (it bends upward, like a valley), and the gradient is increasing.
Where $f''(x) < 0$ the curve is concave down (it bends downward, like a hill), and the gradient is decreasing.

Sign of f''(x)	Shape of the curve	What the gradient is doing
Positive	Concave up (valley)	Increasing
Negative	Concave down (hill)	Decreasing
Zero, with a sign change	Point of inflexion	Turning from one to the other

Tip

A quick memory aid: concave up holds water, concave down spills it.
On a GDC you can confirm concavity by graphing $f''(x)$ and reading off where it is positive or negative.

Points Of Inflexion: Where The Bending Switches Over

A point of inflexion is where the curve changes concavity, switching from bending one way to bending the other.
Finding one takes two conditions, not one:
1. Solve $f''(x) = 0$ to find candidate points.
2. Check that $f''(x)$ actually changes sign across each candidate; only then is it a point of inflexion.

Example

For $f(x) = x^3$, $f''(x) = 6x$, which is zero at $x = 0$ and changes from negative to positive there, so $(0,0)$ is a point of inflexion.

Common Mistake

$f''(x) = 0$ on its own is not enough; you must confirm a sign change.
For $f(x) = x^4$ we get $f''(x) = 12x^2$, which is zero at $x = 0$ but never negative, so there is no inflexion there, it is actually a minimum.

Example question

Find the coordinates of the point of inflexion of $f(x) = x^3 - 6x^2 + 9x$.

Solution

Second derivative: $$f''(x) = 6x - 12$$
Set $f''(x) = 0$: $6x - 12 = 0$, so $x = 2$.
Check the sign change: $f''$ is negative for $x < 2$ and positive for $x > 2$, so it is a genuine inflexion.
Find the $y$-value: $f(2) = 8 - 24 + 18 = 2$, so the point of inflexion is $(2, 2)$.

The Second-Derivative Test For Maximum And Minimum

Once you have a stationary point (where $f'(x) = 0$), the second derivative gives a fast way to say whether it is a peak or a trough.
If $f''(x) < 0$ at the stationary point, the curve is concave down there, so it is a local maximum.
If $f''(x) > 0$ at the stationary point, the curve is concave up there, so it is a local minimum.
If $f''(x) = 0$, the test is inconclusive, and you fall back to the sign of $f'(x)$ on either side (or read the shape off the GDC graph).

Example

For $f(x) = x^3 - 3x$:

Solve $f'(x) = 3x^2 - 3 = 0$ to get $x = 1$ and $x = -1$.
Since $f''(x) = 6x$: $f''(1) = 6 > 0$ gives a local minimum at $x = 1$, and $f''(-1) = -6 < 0$ gives a local maximum at $x = -1$.

Exam technique

The derivative rules you apply here are in the formula booklet, but the second-derivative test itself is a method, not a booklet formula, so learn the logic.
Because every AI paper allows a GDC, you may also justify a maximum or minimum straight from the calculator graph, but show the stationary-point set-up $f'(x) = 0$ to earn the method mark.
State the nature of each point in words (maximum or minimum), since a bare coordinate rarely earns full credit.

Choosing Between The Two Tests

Both tests classify a stationary point; pick whichever is less work for the function in front of you.

Feature	First-derivative (sign) test	Second-derivative test
What you check	Sign of f' each side	Sign of f'' at the point
Extra differentiation	None	One more derivative
When f''=0	Still works	Inconclusive
Best for	Awkward or piecewise f	Quick polynomial checks

Example question

For $f(x) = 2x^3 - 3x^2 - 12x + 5$, find the stationary points and use the second-derivative test to classify each.

Solution

$f'(x) = 6x^2 - 6x - 12 = 6(x-2)(x+1)$, so stationary points at $x = 2$ and $x = -1$.
Second derivative: $f''(x) = 12x - 6$.
At $x = 2$: $f''(2) = 18 > 0$, a local minimum; $f(2) = -15$, so minimum $(2, -15)$.
At $x = -1$: $f''(-1) = -18 < 0$, a local maximum; $f(-1) = 12$, so maximum $(-1, 12)$.

Active recall

What does the sign of $f''(x)$ tell you about the shape of a curve?
State the two conditions needed for a point of inflexion, not just one.
In the second-derivative test, what do $f''(c) > 0$ and $f''(c) < 0$ each indicate at a stationary point $x = c$?
Why does $f''(x) = 0$ not guarantee a point of inflexion? Give the standard counterexample.
Which classification test would you choose when the second derivative is also zero, and why?

Why The Second Derivative Matters: Measuring How The Change Itself Changes

The first derivative $f'(x)$ tells you the gradient: whether a quantity is rising or falling, and how fast.
But often the interesting question is how the change itself is changing.
1. Is a car speeding up or easing off? Is an epidemic's daily case count still accelerating, or has the growth started to slow?
The second derivative answers exactly this, because it is the rate of change of the gradient.
It also gives a quick way to tell a peak from a trough, and to find the moment a model switches from bending one way to bending the other (the point of inflexion), which in a logistic growth model is the moment growth is fastest.

Analogy

Think of driving: position is where you are, the first derivative is your speed, and the second derivative is your acceleration.
A positive second derivative means you are speeding up; a negative one means you are braking, even if you are still moving forward.

The Second Derivative: The Rate Of Change Of The Gradient

To get the second derivative, differentiate the function once to get $f'(x)$, then differentiate again.
It is written $f''(x)$ or $\dfrac{d^2y}{dx^2}$, and higher derivatives follow the same pattern as $f^{(n)}(x)$ or $\dfrac{d^n y}{dx^n}$. $$ f''(x) = \frac{d}{dx}\big(f'(x)\big) $$

Note

The notation $\dfrac{d^n y}{dx^n}$ for higher derivatives is listed in the formula booklet, so you only need to recognise and use it.
Each derivative is found with the standard rules from 5.9, applied one more time.

Example question

Given $f(x) = x^4 - 2x^2$, find $f''(x)$ and state whether the curve is concave up or concave down at $x = 1$.

Solution

Differentiate once: $$f'(x) = 4x^3 - 4x$$
Differentiate again: $$f''(x) = 12x^2 - 4$$
At $x = 1$: $f''(1) = 12 - 4 = 8 > 0$, so the curve is concave up there.

Concavity: Which Way A Curve Bends

The sign of the second derivative tells you which way the curve is bending, independently of whether it is going up or down.
Where $f''(x) > 0$ the curve is concave up (it bends upward, like a valley), and the gradient is increasing.
Where $f''(x) < 0$ the curve is concave down (it bends downward, like a hill), and the gradient is decreasing.

Sign of f''(x)	Shape of the curve	What the gradient is doing
Positive	Concave up (valley)	Increasing
Negative	Concave down (hill)	Decreasing
Zero, with a sign change	Point of inflexion	Turning from one to the other

Tip

A quick memory aid: concave up holds water, concave down spills it.
On a GDC you can confirm concavity by graphing $f''(x)$ and reading off where it is positive or negative.

Points Of Inflexion: Where The Bending Switches Over

A point of inflexion is where the curve changes concavity, switching from bending one way to bending the other.
Finding one takes two conditions, not one:
1. Solve $f''(x) = 0$ to find candidate points.
2. Check that $f''(x)$ actually changes sign across each candidate; only then is it a point of inflexion.

Example

For $f(x) = x^3$, $f''(x) = 6x$, which is zero at $x = 0$ and changes from negative to positive there, so $(0,0)$ is a point of inflexion.

Common Mistake

$f''(x) = 0$ on its own is not enough; you must confirm a sign change.
For $f(x) = x^4$ we get $f''(x) = 12x^2$, which is zero at $x = 0$ but never negative, so there is no inflexion there, it is actually a minimum.

Example question

Find the coordinates of the point of inflexion of $f(x) = x^3 - 6x^2 + 9x$.

Solution

Second derivative: $$f''(x) = 6x - 12$$
Set $f''(x) = 0$: $6x - 12 = 0$, so $x = 2$.
Check the sign change: $f''$ is negative for $x < 2$ and positive for $x > 2$, so it is a genuine inflexion.
Find the $y$-value: $f(2) = 8 - 24 + 18 = 2$, so the point of inflexion is $(2, 2)$.

The Second-Derivative Test For Maximum And Minimum

Once you have a stationary point (where $f'(x) = 0$), the second derivative gives a fast way to say whether it is a peak or a trough.
If $f''(x) < 0$ at the stationary point, the curve is concave down there, so it is a local maximum.
If $f''(x) > 0$ at the stationary point, the curve is concave up there, so it is a local minimum.
If $f''(x) = 0$, the test is inconclusive, and you fall back to the sign of $f'(x)$ on either side (or read the shape off the GDC graph).

Example

For $f(x) = x^3 - 3x$:

Solve $f'(x) = 3x^2 - 3 = 0$ to get $x = 1$ and $x = -1$.
Since $f''(x) = 6x$: $f''(1) = 6 > 0$ gives a local minimum at $x = 1$, and $f''(-1) = -6 < 0$ gives a local maximum at $x = -1$.

Exam technique

The derivative rules you apply here are in the formula booklet, but the second-derivative test itself is a method, not a booklet formula, so learn the logic.
Because every AI paper allows a GDC, you may also justify a maximum or minimum straight from the calculator graph, but show the stationary-point set-up $f'(x) = 0$ to earn the method mark.
State the nature of each point in words (maximum or minimum), since a bare coordinate rarely earns full credit.

Choosing Between The Two Tests

Both tests classify a stationary point; pick whichever is less work for the function in front of you.

Feature	First-derivative (sign) test	Second-derivative test
What you check	Sign of f' each side	Sign of f'' at the point
Extra differentiation	None	One more derivative
When f''=0	Still works	Inconclusive
Best for	Awkward or piecewise f	Quick polynomial checks

Example question

For $f(x) = 2x^3 - 3x^2 - 12x + 5$, find the stationary points and use the second-derivative test to classify each.

Solution

$f'(x) = 6x^2 - 6x - 12 = 6(x-2)(x+1)$, so stationary points at $x = 2$ and $x = -1$.
Second derivative: $f''(x) = 12x - 6$.
At $x = 2$: $f''(2) = 18 > 0$, a local minimum; $f(2) = -15$, so minimum $(2, -15)$.
At $x = -1$: $f''(-1) = -18 < 0$, a local maximum; $f(-1) = 12$, so maximum $(-1, 12)$.

Active recall

What does the sign of $f''(x)$ tell you about the shape of a curve?
State the two conditions needed for a point of inflexion, not just one.
In the second-derivative test, what do $f''(c) > 0$ and $f''(c) < 0$ each indicate at a stationary point $x = c$?
Why does $f''(x) = 0$ not guarantee a point of inflexion? Give the standard counterexample.
Which classification test would you choose when the second derivative is also zero, and why?

IB Mathematics Applications & Interpretation AHL 5.10—second Derivatives, Testing for Max and Min Notes

IB Mathematics Applications & Interpretation AHL 5.10—second Derivatives, Testing for Max and Min Notes

Why The Second Derivative Matters: Measuring How The Change Itself Changes

The Second Derivative: The Rate Of Change Of The Gradient

Concavity: Which Way A Curve Bends

Points Of Inflexion: Where The Bending Switches Over

The Second-Derivative Test For Maximum And Minimum

Choosing Between The Two Tests

Why The Second Derivative Matters: Measuring How The Change Itself Changes

The Second Derivative: The Rate Of Change Of The Gradient

Concavity: Which Way A Curve Bends

Points Of Inflexion: Where The Bending Switches Over

The Second-Derivative Test For Maximum And Minimum

Choosing Between The Two Tests

IB Mathematics Applications & Interpretation AHL 5.10—second Derivatives, Testing for Max and Min Notes

Number and Algebra

Functions

Geometry and Trigonometry

Statistics and Probability

Calculus

IB Mathematics Applications & Interpretation AHL 5.10—second Derivatives, Testing for Max and Min Notes

Number and Algebra

Functions

Geometry and Trigonometry

Statistics and Probability

Calculus

Why The Second Derivative Matters: Measuring How The Change Itself Changes

The Second Derivative: The Rate Of Change Of The Gradient

Concavity: Which Way A Curve Bends

Points Of Inflexion: Where The Bending Switches Over

The Second-Derivative Test For Maximum And Minimum

Choosing Between The Two Tests

Why The Second Derivative Matters: Measuring How The Change Itself Changes

The Second Derivative: The Rate Of Change Of The Gradient

Concavity: Which Way A Curve Bends

Points Of Inflexion: Where The Bending Switches Over

The Second-Derivative Test For Maximum And Minimum

Choosing Between The Two Tests