# Monotonic functions, second derivative test

> IB Mathematics AI HL · IB Math AI HL
> Source: https://www.owlsprep.com/study/ib-math-ai-hl-u4-monotonic-functions-second-derivative-test/

This subtopic covers how to use first derivatives to identify monotonic (increasing/decreasing) functions, and apply the second derivative test to classify stationary points as local maxima, minima, or indeterminate cases for IB exams.

**Prerequisites:** [Basic differentiation rules](https://www.owlsprep.com/study/ib-math-ai-hl-u3-basic-differentiation-rules/); [Introduction to stationary points](https://www.owlsprep.com/study/ib-math-ai-hl-u4-stationary-points/)

## Learning objectives

- Identify increasing/decreasing monotonic functions using first derivatives
- Classify stationary points using the second derivative test
- Recognize cases where the second derivative test is inconclusive
- Apply the test to solve classification problems in IB exams

## Monotonic Functions and First Derivative Criterion

**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 $f(x) = 4x^3 - 12x + 5$ 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.

*Calculator:* forbidden

## The Second Derivative Test for Stationary Points

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

*Notation:* For a stationary point at $x=a$ where $f'(a)=0$

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 $f(x) = x^3 - 6x^2 + 9x + 1$.

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 $(3, 1)$.

**Check your understanding**

Test your understanding of the sign rule:

1. 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

   *Why:* 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 $x$.

*Calculator:* allowed

## Limitations and Inconclusive Cases

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.

> **Common misconception**
>
> $f''(a) = 0$ does not automatically mean the point is a point of inflection. A point of inflection requires a change in the sign of $f''(x)$ (change in concavity) around $x=a$, not just a value of zero at the point.

**Worked example:** Classify the stationary point at $x=0$ for $f(x) = x^4$ and $f(x) = x^3$.

1. For $f(x) = x^4$: $f'(x) = 4x^3$, so $f'(0) = 0$, meaning $x=0$ is stationary. $f''(x) = 12x^2$, 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 $f(x) = x^3$: $f'(x) = 3x^2$, so $f'(0) = 0$, stationary. $f''(x) = 6x$, 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.

*Calculator:* forbidden

## Common pitfalls

- **Wrong:** Assuming a stationary point where $f''(x)=0$ is automatically a point of inflection
  - Why it fails: $f''(x)=0$ alone is not sufficient; a change in concavity (sign change of $f''(x)$) is required
  - Correct: When $f''(x)=0$, test the sign of $f'(x)$ around the point to classify the stationary point
- **Wrong:** Forgetting to calculate the $y$-coordinate of classified stationary points
  - Why it fails: IB exam questions require full coordinates for full marks, and most students lose easy marks here
  - Correct: After finding the $x$-coordinate, substitute back into the original function $f(x)$ to get $y$
- **Wrong:** Mixing up the sign rule: claiming $f''(x) > 0$ is a local maximum
  - Why it fails: Confusion between concavity direction: positive $f''$ means concave up (curves upward)
  - Correct: Remember: negative = frown (maximum), positive = cup (minimum)
- **Wrong:** Claiming a function with $f'(x)=0$ at an isolated point is not monotonic
  - Why it fails: Isolated zeros of $f'(x)$ do not break monotonicity if the sign does not change
  - Correct: Check the sign of $f'(x)$ across the entire interval; if it stays non-positive or non-negative, the function is monotonic

## 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 $x=a$ |

## 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.

- [Optimization Problems](https://www.owlsprep.com/study/ib-math-ai-hl-u4-optimization-problems/)
- [Related rates](https://www.owlsprep.com/study/ib-math-ai-hl-u4-related-rates/)
- [Antiderivatives and indefinite integration](https://www.owlsprep.com/study/ib-math-ai-hl-u4-antiderivatives-and-indefinite-integration/)

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