Stationary points, monotonicity and concavity

IB Mathematics AA SL· 5.7 Applications of differentiation· 20 min read

1. Monotonicity and Increasing/Decreasing Intervals★★☆☆☆⏱ 6 min

📘 Definition

Monotonic function

A function is monotonic on an interval if it is entirely non-increasing or non-decreasing. A strictly monotonic function is strictly increasing when $f^{'} (x) > 0$ and strictly decreasing when $f^{'} (x) < 0$ .

Example:

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

To find intervals of increase/decrease, you first calculate the first derivative $f^{'} (x)$ , then find critical points where $f^{'} (x) = 0$ or undefined. You then test the sign of $f^{'} (x)$ in each interval between critical points.

📐 Worked Example

Find the intervals where $f (x) = x^{3} - 3 x^{2} + 2 x$ is increasing or decreasing

1
First compute the first derivative:
$f'(x) = 3x^2 - 6x + 2$
2
Find critical points by solving $f^{'} (x) = 0$ :
$3x^2 - 6x + 2 = 0 \implies x = 1 \pm \frac{\sqrt{3}}{3} \approx 0.423, 1.577$
3
Test the sign of $f^{'} (x)$ in each interval:
- For $x < 0.423$ : $f^{'} (0) = 2 > 0$ (increasing)
- Between $0.423$ and $1.577$ : $f^{'} (1) = - 1 < 0$ (decreasing)
- For $x > 1.577$ : $f^{'} (2) = 2 > 0$ (increasing)
4
Final result:
$\text{Increasing: } (- infty, 1 - \frac{\sqrt{3}}{3}) \cup (1 + \frac{\sqrt{3}}{3}, \infty) \\ \text{Decreasing: } (1 - \frac{\sqrt{3}}{3}, 1 + \frac{\sqrt{3}}{3})$

Exam tip:

Always write intervals of increase/decrease as open intervals, since $f^{'} (x) = 0$ at endpoints, so the function is not increasing/decreasing there.

2. Stationary Points and the First Derivative Test★★★☆☆⏱ 7 min

📘 Definition

Stationary point

(a, f(a))

A point where $f^{'} (a) = 0$ , so the tangent gradient is zero and the function is temporarily stationary. Stationary points can be local maxima, local minima, or stationary points of inflection.

If $f^{'} (x)$ changes from positive to negative: local maximum
If $f^{'} (x)$ changes from negative to positive: local minimum
If $f^{'} (x)$ does not change sign: stationary point of inflection

📐 Worked Example

Classify the stationary points of $f (x) = x^{3} - 12 x$

1
Calculate first derivative:
$f'(x) = 3x^2 - 12$
2
Solve for stationary points:
$3x^2 - 12 = 0 \implies x = -2, x = 2$
3
Test sign change around $x = - 2$ : $f^{'} (- 3) = 15 > 0$ , $f^{'} (0) = - 12 < 0$ . Sign changes + to -, so local maximum.
4
Test sign change around $x = 2$ : $f^{'} (0) = - 12 < 0$ , $f^{'} (3) = 15 > 0$ . Sign changes - to +, so local minimum.
5
Find coordinates:
$\text{Local maximum at } (-2, 16), \text{ local minimum at } (2, -16)$

Exam tip:

The first derivative test always works, even when the second derivative is zero or undefined, so it's a safe fallback if you're unsure.

3. Concavity and Points of Inflection★★★☆☆⏱ 6 min

📘 Definition

Concavity

A function is concave up (convex) when $f^{''} (x) > 0$ , and curves upward. It is concave down when $f^{''} (x) < 0$ , and curves downward.

A point of inflection is where the concavity of the function changes. Two conditions must be satisfied: (1) $f^{''} (x) = 0$ (or undefined), and (2) the sign of $f^{''} (x)$ changes around the point.

📐 Worked Example

Find intervals of concavity and inflection points for $f (x) = x^{3} - 3 x^{2} + 4 x - 1$

1
Calculate first and second derivatives:
$f'(x) = 3x^2 - 6x + 4, \quad f''(x) = 6x - 6$
2
Solve $f^{''} (x) = 0$ :
$6x - 6 = 0 \implies x = 1$
3
Check for sign change: For $x < 1$ , $f^{''} (0) = - 6 < 0$ (concave down). For $x > 1$ , $f^{''} (2) = 6 > 0$ (concave up). Concavity changes, so $x = 1$ is an inflection point.
4
Find coordinate of inflection point:
$f(1) = 1, \text{ so inflection point at } (1, 1)$

4. Second Derivative Test for Stationary Points★★★☆☆⏱ 5 min

The second derivative test is a faster alternative to the first derivative test for classifying stationary points. The rules are:

If $f^{'} (a) = 0$ and $f^{''} (a) < 0$ : local maximum at $x = a$
If $f^{'} (a) = 0$ and $f^{''} (a) > 0$ : local minimum at $x = a$
If $f^{'} (a) = 0$ and $f^{''} (a) = 0$ : test is inconclusive, use first derivative test

📐 Worked Example

Use the second derivative test to classify stationary points of $f (x) = 2 x^{3} - 9 x^{2} + 12 x$

1
Find first derivative and stationary points:
$f'(x) = 6x^2 - 18x + 12 = 6(x-1)(x-2) \implies x=1, x=2$
2
Calculate second derivative:
$f''(x) = 12x - 18$
3
Evaluate $f^{''}$ at each stationary point:
$f''(1) = -6 < 0 \implies \text{ local maximum} \\ f''(2) = 6 > 0 \implies \text{ local minimum}$
4
Final result:
$\text{Local maximum at } (1, 5), \text{ local minimum at } (2, 4)$

Exam tip:

Never skip checking the sign change if the second derivative test is inconclusive. Examiners frequently test this case to catch students who assume $f^{''} (a) = 0$ means it's an inflection point automatically.

5. Common Pitfalls

Wrong move:

Claiming a point is an inflection point just because $f^{''} (x) = 0$

Why:

$f^{''} (x)$ must change sign around the point for it to be an inflection point, which does not always happen when $f^{''} (x) = 0$

Correct move:

Always test the sign of $f^{''} (x)$ on both sides of the point to confirm a change in concavity

Wrong move:

Writing intervals of increase/decrease with closed brackets

Why:

At the endpoints of the interval, $f^{'} (x) = 0$ , so the function is not increasing or decreasing at those points

Correct move:

Always write intervals of increase/decrease as open intervals

Wrong move:

Assuming $f^{''} (a) = 0$ means the point is an inflection point

Why:

Students often misremember the rule for inconclusive second derivative tests

Correct move:

If $f^{''} (a) = 0$ , test for a sign change in $f^{''} (x)$ to confirm it is an inflection point, and use the first derivative test to classify the stationary point

Wrong move:

Mixing up the sign rules for maximum/minimum in the second derivative test

Why:

It is easy to confuse which sign corresponds to which classification

Correct move:

Remember: negative $f^{''}$ is concave down (like a frown, maximum), positive $f^{''}$ is concave up (like a cup, minimum)

Wrong move:

Claiming all inflection points are stationary points

Why:

Inflection points only require a change in concavity, not a gradient of zero

Correct move:

Only stationary points of inflection have $f^{'} (x) = 0$ ; most inflection points have a non-zero gradient

6. Quick Reference Cheatsheet

Concept	Key Condition	Result
Strictly increasing	$f^{'} (x) > 0$	Monotonic
Strictly decreasing	$f^{'} (x) < 0$	Monotonic
Local maximum (1st test)	$f^{'} (x)$ changes + → -	Local maximum
Local minimum (1st test)	$f^{'} (x)$ changes - → +	Local minimum
Stationary inflection	$f^{'} (a) = 0$ , $f^{'} (x)$ sign unchanged	Stationary inflection
Local maximum (2nd test)	$f^{'} (a) = 0$ , $f^{''} (a) < 0$	Local maximum
Local minimum (2nd test)	$f^{'} (a) = 0$ , $f^{''} (a) > 0$	Local minimum
Concave up	$f^{''} (x) > 0$	Curves upward
Concave down	$f^{''} (x) < 0$	Curves downward
Point of inflection	$f^{''} (x)$ changes sign	Change in concavity

7. Frequently Asked

Do I have to use the second derivative test to classify stationary points?

No, you can use the first derivative test for any stationary point. The second derivative test is just a quicker alternative when it works.

Is every inflection point a stationary point?

No. Only stationary points of inflection have $f^{'} (x) = 0$ . Most inflection points have a non-zero gradient, only the concavity changes.

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.

2025 · 1
Classify 3 stationary points
2023 · 2
Find intervals of concavity
2021 · 1
Determine monotonicity of function

What's Next

This sub-topic is the foundation for two core assessed areas of IB AA SL calculus: graph sketching and applied optimization. Understanding how derivatives relate to the shape of a graph lets you accurately sketch any polynomial, rational, or trigonometric function, a common 5-7 mark question in both Paper 1 and Paper 2. It also enables you to find maximum and minimum values in optimization problems, which are almost guaranteed to appear in every exam. Mastery of these concepts is essential for all further calculus topics in IB.