# Stationary points, monotonicity and concavity

> IB Mathematics AA SL · IB Mathematics AA SL Calculus Unit 5
> Source: https://www.owlsprep.com/study/ib-math-aa-sl-u5-stationary-points-monotonicity-and-concavity/

This sub-topic connects first and second derivatives to the shape of a function’s graph. You will learn to identify where functions increase, decrease, flatten, or curve, core skills for graph sketching and optimization in IB exams.

**Prerequisites:** [Differentiation of basic functions](https://www.owlsprep.com/study/ib-math-aa-sl-u4-intro-to-differentiation/); [First and second derivatives](https://www.owlsprep.com/study/ib-math-aa-sl-u5-first-and-second-derivatives/)

## Learning objectives

- Identify and classify stationary points of functions
- Determine intervals of increase/decrease (monotonicity)
- Find intervals of concavity and points of inflection
- Apply first and second derivative tests correctly

## Monotonicity and Increasing/Decreasing Intervals

**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 $f(x) = x^3 - 3x^2 + 2x$ 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.

## Stationary Points and the First Derivative Test

**Stationary point** — 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.

*Notation:* (a, f(a))

1. If $f'(x)$ changes from positive to negative: local maximum
2. If $f'(x)$ changes from negative to positive: local minimum
3. If $f'(x)$ does not change sign: stationary point of inflection

**Worked example:** Classify the stationary points of $f(x) = x^3 - 12x$

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.

## Concavity and Points of Inflection

**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 $f(x) = x^3 - 3x^2 + 4x - 1$

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:

   $$f(1) = 1, \text{ so inflection point at } (1, 1)$$

> **warning**
>
> A common mistake is assuming all points where $f''(x) = 0$ are inflection points. For $f(x) = x^4$, $f''(0) = 0$, but $f''(x)$ is positive on both sides of 0, so there is no inflection point at $x=0$.

## Second Derivative Test for Stationary Points

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

1. If $f'(a) = 0$ and $f''(a) < 0$: local maximum at $x=a$
2. If $f'(a) = 0$ and $f''(a) > 0$: local minimum at $x=a$
3. 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 $f(x) = 2x^3 - 9x^2 + 12x$

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.

## Common pitfalls

- **Wrong:** Claiming a point is an inflection point just because $f''(x) = 0$
  - Why it fails: $f''(x)$ must change sign around the point for it to be an inflection point, which does not always happen when $f''(x)=0$
  - Correct: Always test the sign of $f''(x)$ on both sides of the point to confirm a change in concavity
- **Wrong:** Writing intervals of increase/decrease with closed brackets
  - Why it fails: At the endpoints of the interval, $f'(x)=0$, so the function is not increasing or decreasing at those points
  - Correct: Always write intervals of increase/decrease as open intervals
- **Wrong:** Assuming $f''(a)=0$ means the point is an inflection point
  - Why it fails: Students often misremember the rule for inconclusive second derivative tests
  - Correct: 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:** Mixing up the sign rules for maximum/minimum in the second derivative test
  - Why it fails: It is easy to confuse which sign corresponds to which classification
  - Correct: Remember: negative $f''$ is concave down (like a frown, maximum), positive $f''$ is concave up (like a cup, minimum)
- **Wrong:** Claiming all inflection points are stationary points
  - Why it fails: Inflection points only require a change in concavity, not a gradient of zero
  - Correct: Only stationary points of inflection have $f'(x)=0$; most inflection points have a non-zero gradient

## 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 |

## 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.

- [Optimization applications](https://www.owlsprep.com/study/ib-math-aa-sl-u5-optimization-applications/)
- [Indefinite integration and anti-differentiation](https://www.owlsprep.com/study/ib-math-aa-sl-u5-indefinite-integration-and-anti-differentiation/)
- [Definite integration and fundamental theorem of calculus](https://www.owlsprep.com/study/ib-math-aa-sl-u5-definite-integration-and-fundamental-theorem/)

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