Monotonic functions, second derivative test

IB Mathematics AI HL· 30 min read

1. Monotonic Functions and First Derivative Criterion★★☆☆☆⏱ 15 min

🚫 No Calculator

📘 Definition

Monotonic Function

A function is monotonic on an interval if it is either entirely non-increasing or entirely non-decreasing on that interval. Strictly monotonic functions are strictly increasing or decreasing throughout the interval.

Example:

$f (x) = x^{3}$ is strictly monotonic over all real numbers

To determine if a function is increasing or decreasing on an interval, we use the sign of the first derivative. The rule is: if $f^{'} (x) > 0$ for all $x$ in the interval, $f$ is strictly increasing; if $f^{'} (x) < 0$ for all $x$ in the interval, $f$ is strictly decreasing. Isolated points where $f^{'} (x) = 0$ do not break monotonicity if the sign does not change.

📐 Worked Example

Prove that $f (x) = 4 x^{3} - 12 x + 5$ is strictly increasing on $(- \infty, - 1) \cup (1, \infty)$ and strictly decreasing on $(- 1, 1)$ .

1
Compute the first derivative of $f (x)$ :
2
$f'(x) = 12x^2 - 12 = 12(x^2 - 1) = 12(x-1)(x+1)$
3
For $x \in (- \infty, - 1)$ : both $(x - 1)$ and $(x + 1)$ are negative, so their product is positive. Thus $f^{'} (x) > 0$ .
4
For $x \in (- 1, 1)$ : $(x - 1)$ is negative, $(x + 1)$ is positive, so product is negative. Thus $f^{'} (x) < 0$ .
5
For $x \in (1, \infty)$ : both factors are positive, so product is positive. Thus $f^{'} (x) > 0$ .
6
By the first derivative criterion, the claim holds as required.

Exam tip:

Always test the sign of $f^{'} (x)$ on either side of critical values to confirm monotonicity, don't just rely on the value at the critical point.

2. The Second Derivative Test for Stationary Points★★★☆☆⏱ 20 min

✓ Calculator OK

📘 Definition

Second Derivative Test

For a stationary point at $x = a$ where $f^{'} (a) = 0$

A method to classify stationary points of a twice-differentiable function by evaluating the sign of the second derivative at the stationary point.

The rule for classification is straightforward: if $f^{''} (a) < 0$ , the point is a local maximum; if $f^{''} (a) > 0$ , the point is a local minimum. This test is faster than the first derivative test when the second derivative is easy to compute.

📐 Worked Example

Find and classify all stationary points of $f (x) = x^{3} - 6 x^{2} + 9 x + 1$ .

1
First, compute $f^{'} (x)$ and solve for stationary points:
2
$f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$
3
Set $f^{'} (x) = 0$ , so stationary points occur at $x = 1$ and $x = 3$ .
4
Next, compute the second derivative:
5
$f''(x) = 6x - 12$
6
Evaluate at $x = 1$ : $f^{''} (1) = 6 (1) - 12 = - 6 < 0$ . A negative second derivative means $x = 1$ is a local maximum. Calculate $y$ : $f (1) = 5$ , so local maximum at $(1, 5)$ .
7
Evaluate at $x = 3$ : $f^{''} (3) = 6 (3) - 12 = 6 > 0$ . A positive second derivative means $x = 3$ is a local minimum. Calculate $y$ : $f (3) = 1$ , so local minimum at $(3, 1)$ .

✓ Quick check

Test your understanding of the sign rule:

If $f^{'} (2) = 0$ and $f^{''} (2) = - 4$ , what is the classification of the stationary point at $x = 2$ ?
- Local maximum
- Local minimum
- Point of inflection
- Cannot be determined
Reveal answer
Local maximum —
Correct! A negative second derivative at a stationary point means the function is concave down (curves downward), so it forms a local maximum.

Exam tip:

IB exam questions almost always ask for full coordinates of stationary points, don't forget to calculate the $y$ -value after finding $x$ .

3. Limitations and Inconclusive Cases★★★★☆⏱ 15 min

🚫 No Calculator

The second derivative test only works if $f^{''} (x)$ exists and is not equal to zero at the stationary point. If $f^{''} (a) = 0$ at a stationary point $x = a$ , the test is inconclusive, and you must use the first derivative test (checking for a sign change of $f^{'} (x)$ around the point) to classify it.

📐 Worked Example

Classify the stationary point at $x = 0$ for $f (x) = x^{4}$ and $f (x) = x^{3}$ .

1
For $f (x) = x^{4}$ : $f^{'} (x) = 4 x^{3}$ , so $f^{'} (0) = 0$ , meaning $x = 0$ is stationary. $f^{''} (x) = 12 x^{2}$ , so $f^{''} (0) = 0$ , so the second derivative test is inconclusive.
2
Check the sign of $f^{'} (x)$ around $0$ : for $x < 0$ , $f^{'} (x) < 0$ ; for $x > 0$ , $f^{'} (x) > 0$ . $f^{'} (x)$ changes from negative to positive, so $x = 0$ is a local minimum.
3
For $f (x) = x^{3}$ : $f^{'} (x) = 3 x^{2}$ , so $f^{'} (0) = 0$ , stationary. $f^{''} (x) = 6 x$ , so $f^{''} (0) = 0$ , test is inconclusive.
4
Check the sign of $f^{'} (x)$ around $0$ : $f^{'} (x) > 0$ for both $x < 0$ and $x > 0$ . No sign change, so $x = 0$ is a stationary point of inflection, not a maximum or minimum.

4. Common Pitfalls

Wrong move:

Assuming a stationary point where $f^{''} (x) = 0$ is automatically a point of inflection

Why:

$f^{''} (x) = 0$ alone is not sufficient; a change in concavity (sign change of $f^{''} (x)$ ) is required

Correct move:

When $f^{''} (x) = 0$ , test the sign of $f^{'} (x)$ around the point to classify the stationary point

Wrong move:

Forgetting to calculate the $y$ -coordinate of classified stationary points

Why:

IB exam questions require full coordinates for full marks, and most students lose easy marks here

Correct move:

After finding the $x$ -coordinate, substitute back into the original function $f (x)$ to get $y$

Wrong move:

Mixing up the sign rule: claiming $f^{''} (x) > 0$ is a local maximum

Why:

Confusion between concavity direction: positive $f^{''}$ means concave up (curves upward)

Correct move:

Remember: negative = frown (maximum), positive = cup (minimum)

Wrong move:

Claiming a function with $f^{'} (x) = 0$ at an isolated point is not monotonic

Why:

Isolated zeros of $f^{'} (x)$ do not break monotonicity if the sign does not change

Correct move:

Check the sign of $f^{'} (x)$ across the entire interval; if it stays non-positive or non-negative, the function is monotonic

5. Quick Reference Cheatsheet

Concept	Condition	Conclusion
Strictly increasing	$f^{'} (x) > 0$ on interval	Function is monotonic increasing
Strictly decreasing	$f^{'} (x) < 0$ on interval	Function is monotonic decreasing
Local maximum (2nd test)	$f^{'} (a) = 0, f^{''} (a) < 0$	$(a, f (a))$ is local maximum
Local minimum (2nd test)	$f^{'} (a) = 0, f^{''} (a) > 0$	$(a, f (a))$ is local minimum
Inconclusive 2nd test	$f^{'} (a) = 0, f^{''} (a) = 0$ or undefined	Use first derivative test to classify
Point of inflection	$f^{''} (x)$ changes sign at $a$	Concavity changes at $x = a$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 1
Classify stationary points of cubic
2021 · 2
Monotonicity of rational function
2023 · 1
Inconclusive second derivative test

Going deeper

syllabusIB AI HL Calculus SyllabusTopic 5.6, 5.7

What's Next

Understanding monotonic functions and the second derivative test is foundational for solving optimization problems, one of the most common long-response question types in IB Math AI HL exams. These concepts also underpin curve sketching, where you need to identify all key features of a function (maxima, minima, inflection points) before drawing it. Mastering this subtopic will set you up to earn full marks on most calculus-based questions, which make up a large portion of your final exam grade.