# Extreme Value Theorem, Global vs Local Extrema, Critical Points

> AP Calculus AB · Unit 5: Analytical Applications of Differentiation
> Source: https://www.owlsprep.com/study/ap-calculus-ab-u5-extreme-value-theorem-global-vs/

This guide covers the Extreme Value Theorem, definitions of local vs global extrema, and critical point identification for AP Calculus AB. You will learn to locate candidate extrema and justify existence for exam problems.

**Prerequisites:** Compute derivatives of algebraic, trigonometric, and composite functions; Properties of continuous functions; Interval notation for domains

## Learning objectives

- Verify conditions for the Extreme Value Theorem
- Identify critical points per AP CED definitions
- Distinguish between local and global extrema
- Locate and justify global extrema on closed intervals

## The Extreme Value Theorem

The Extreme Value Theorem (EVT) is an existence theorem: it tells you when you are guaranteed to find a global maximum and global minimum, but it does not tell you how to find them.

**Extreme Value Theorem** — If a function $f(x)$ is continuous on a closed, bounded interval $[a, b]$, then $f(x)$ must have at least one global maximum value and at least one global minimum value on $[a, b]$.

*Example:* A continuous function on $[0, 5]$ is guaranteed to have a global maximum and minimum by EVT, but a continuous function on $(0,5)$ is not.

Intuition: If you draw a continuous curve from $(a, f(a))$ to $(b, f(b))$ without lifting your pen, the curve must reach a highest point and a lowest point. The guarantee breaks if either condition fails: open intervals let you approach an extremum without reaching it, and discontinuities (like vertical asymptotes) let the function grow without bound.

**Worked example:** For which of the following functions does the Extreme Value Theorem guarantee a global maximum and global minimum on the given interval?
(A) $f(x) = \tan(x)$ on $[0, \pi/2]$
(B) $f(x) = x^2 - 3x$ on $(-2, 4)$
(C) $f(x) = \dfrac{x^2}{x-2}$ on $[-1, 1]$
(D) $f(x) = \dfrac{\sin(x)}{x-\pi}$ on $[\pi/2, 3\pi/2]$

1. Check the first requirement of EVT: the interval must be closed. Option B uses an open interval $(-2, 4)$, so EVT cannot apply. Eliminate B.
2. Check the second requirement: the function must be continuous on the entire interval. For A: $\tan(x)$ has a vertical asymptote at $x=\pi/2$, so it is discontinuous at the endpoint. Eliminate A. For D: $\sin(x)/(x-\pi)$ has a vertical asymptote at $x=\pi$, which lies inside the interval, so it is discontinuous. Eliminate D.
3. Verify option C: $[-1, 1]$ is a closed interval, and the only discontinuity of $f(x)$ is at $x=2$, which lies outside the interval. $f(x)$ is continuous at every point in $[-1, 1]$, so both conditions of EVT are satisfied.
4. EVT applies only to option C.

> **tip**
>
> On AP FRQs, if you are asked to justify that a global extremum exists, you must explicitly state two things: (1) the function is continuous on the closed interval, (2) therefore by the Extreme Value Theorem, global extrema exist. You will lose points if you skip this justification step.

> **Exam tip:** Always state both EVT conditions for full justification points

## Critical Points

**Critical Point** — A point where either $f'(c) = 0$ or $f'(c)$ does not exist. Per AP CED definition, endpoints are never critical points, and $c$ must lie in the domain of the original function $f$.

*Notation:* Interior point $c$ of domain of $f$

*Example:* For $f(x) = x^{2/3}$, $x=0$ is a critical point because $f'(0)$ does not exist, and $x=0$ is in the domain of $f$.

By Fermat's Theorem, if $f(x)$ has a local extremum at an interior point $c$, then $c$ must be a critical point. The converse is *not* true: not all critical points are local extrema.

> **note**
>
> Two common edge cases: (1) Endpoints are never critical points, but they are always candidates for global extrema. (2) A point where the original function is undefined can never be a critical point, even if the derivative is also undefined there.

**Worked example:** Find all critical points of $f(x) = x^{2/3}(4 - x)$ on $(-\infty, \infty)$.

1. Rewrite $f(x)$ for easier differentiation and compute the derivative using the power rule:
2. $$f(x) = 4x^{2/3} - x^{5/3} \\ f'(x) = \frac{8}{3x^{1/3}} - \frac{5x^{2/3}}{3} = \frac{8 - 5x}{3x^{1/3}}$$
3. Find where $f'(x) = 0$: a rational derivative equals zero when its numerator is zero (and denominator is non-zero):
4. Set $8 - 5x = 0 \implies x = 8/5$. $x=8/5$ is in the domain of $f(x)$, so this is a critical point.
5. Find where $f'(x)$ does not exist: the denominator is zero when $x=0$. Check if $x=0$ is in the domain of $f(x)$: $f(0) = 0$, which is defined, so $x=0$ is also a critical point.
6. Final answer: Critical points are $x=0$ and $x=8/5$.

> **tip**
>
> Always confirm that a point where the derivative is undefined is in the domain of the original function before counting it as a critical point. Half of all incorrect critical point answers come from forgetting this check.

## Local vs Global (Absolute) Extrema

Extrema are classified by the interval over which they are the maximum or minimum. The key distinctions tested on the AP exam are:

- **Global (Absolute) Extremum**: A global maximum on an interval $I$ is a value $f(c)$ such that $f(c) \geq f(x)$ for *all* $x$ in $I$. A global minimum follows the reverse inequality. Global extrema can occur at critical points *or* endpoints. There can be only one global maximum value and one global minimum value per interval, though the same value can occur at multiple $x$-locations.
- **Local (Relative) Extremum**: A local maximum at $c$ is a value $f(c)$ such that $f(c) \geq f(x)$ for all $x$ in some small open interval around $c$. Local extrema only occur at interior points, so endpoints can never be local extrema. Any global extremum at an interior point is always also a local extremum.

**Worked example:** Given $f(x) = x^3 - 3x + 1$ on $[-3, 2]$, identify all local extrema and state the global maximum and global minimum.

1. Find critical points:
2. $$f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$$
3. Critical points are at interior points $x=-1$ and $x=1$, both in the domain.
4. Classify local extrema via the first derivative test: $f'(x)$ changes from positive to negative at $x=-1$, so $f(-1) = 3$ is a local maximum. $f'(x)$ changes from negative to positive at $x=1$, so $f(1) = -1$ is a local minimum.
5. Evaluate all candidates for global extrema (critical points and endpoints):
6. $$f(-3) = -17, \quad f(-1) = 3, \quad f(1) = -1, \quad f(2) = 3$$
7. Compare values: The largest value is 3, and the smallest value is $-17$.
8. Final answer: Local extrema are a local maximum of 3 at $x=-1$ and a local minimum of -1 at $x=1$. The global maximum value is 3, and the global minimum value is $-17$.

> **tip**
>
> Always read the question carefully: if it asks for the location of an extremum, give the $x$-value; if it asks for the value of the extremum, give the $y$-value. AP exam writers intentionally test this distinction.

## AP-Style Worked Practice

**Worked example:** Which of the following statements about $f(x) = \dfrac{x^2}{x - 3}$ on the closed interval $[0, 5]$ is true?
A) $f(x)$ has a global minimum but no global maximum on $[0,5]$ by EVT.
B) $f(x)$ has both a global minimum and global maximum on $[0,5]$ by EVT.
C) $f(x)$ has neither a global minimum nor global maximum on $[0,5]$, and EVT does not apply.
D) $f(x)$ has both a global minimum and global maximum on $[0,5]$, but EVT does not guarantee this.

1. Check EVT conditions first: the interval $[0,5]$ is closed, but $f(x)$ has a vertical asymptote and discontinuity at $x=3$, which lies inside the interval. $f(x)$ is not continuous on the entire interval, so EVT does not apply, eliminating options A and B.
2. Because $f(x)$ approaches $-\infty$ as $x$ approaches 3 from the left and $+\infty$ as $x$ approaches 3 from the right, there is no lower bound or upper bound for $f(x)$ on $[0,5]$, so no global minimum or maximum exists.
3. The correct answer is C.

**Worked example:** Let $f(x) = 2x^3 - 15x^2 + 24x + 4$, defined on the closed interval $[0, 5]$. (a) Find all critical points of $f(x)$ on $[0,5]$. (b) Identify all local extrema of $f(x)$ on $[0,5]$, classifying each as a local maximum or local minimum. (c) Find the global maximum value and global minimum value of $f(x)$ on $[0,5]$. Justify your answer.

1. (a) Compute derivative and factor:
2. $$f'(x) = 6x^2 - 30x + 24 = 6(x-1)(x-4)$$
3. Both $x=1$ and $x=4$ are interior points in the domain, and there are no points where $f'(x)$ is undefined. Critical points are $x=1$ and $x=4$.
4. (b) First derivative test: For $0 < x < 1$, $f'(x) > 0$ (f increasing); for $1 < x < 4$, $f'(x) < 0$ (f decreasing); for $4 < x < 5$, $f'(x) > 0$ (f increasing). $f'$ changes from positive to negative at $x=1$, so $f(1) = 15$ is a local maximum at $x=1$. $f'$ changes from negative to positive at $x=4$, so $f(4) = -12$ is a local minimum at $x=4$.
5. (c) Justification: $f(x)$ is a polynomial, so it is continuous on the closed interval $[0,5]$. By the Extreme Value Theorem, global extrema exist at critical points or endpoints. Evaluate $f$ at all candidates:
6. $$f(0) = 4, \quad f(1) = 15, \quad f(4) = -12, \quad f(5) = -1$$
7. Comparing values, the largest value is 15 and the smallest is -12. Global maximum value = 15, global minimum value = -12.

## Common pitfalls

- **Wrong:** Claiming EVT applies to a continuous function on an open interval
  - Why it fails: Students remember EVT requires continuity but forget the interval must also be closed and bounded
  - Correct: Always check both conditions (continuous on the entire interval and the interval is closed) before invoking EVT
- **Wrong:** Counting endpoints of an interval as critical points
  - Why it fails: Many introductory resources simplify definitions, but AP CED defines critical points as interior points only
  - Correct: When asked for critical points, never list endpoints; only add endpoints to the candidate list for global extrema
- **Wrong:** Counting a point where $f(x)$ is undefined as a critical point just because $f'(x)$ is also undefined there
  - Why it fails: Students stop at checking the derivative and forget to confirm the point is in the original function's domain
  - Correct: For any point where $f'(c)$ does not exist, always check that $f(c)$ exists before calling it a critical point
- **Wrong:** Assuming every critical point is a local extremum
  - Why it fails: Students reverse Fermat's Theorem, incorrectly assuming all critical points must be extrema. For example, $f(x)=x^3$ has a critical point at $x=0$ that is not an extremum
  - Correct: Always test if the derivative changes sign around a critical point before classifying it as a local extremum
- **Wrong:** Forgetting to check endpoints of a closed interval when finding global extrema
  - Why it fails: Students only check critical points and assume the global extremum is interior
  - Correct: When finding global extrema on a closed interval, always evaluate $f(x)$ at all critical points and both endpoints, then compare all values
- **Wrong:** Claiming a global maximum has multiple different values when it occurs at multiple $x$-values
  - Why it fails: Students confuse the location ($x$-value) with the value of the extremum ($y$-value)
  - Correct: If asked for the global maximum value, it is a single $y$-value; if asked for locations, list all $x$-values where it occurs

## Cheatsheet

| Category | Rule / Condition | Notes |
| --- | --- | --- |
| Extreme Value Theorem | 1. $f(x)$ continuous on $I$ <br> 2. $I = [a,b]$ (closed, bounded) | If both conditions met, $f$ has at least one global max and one global min; only guarantees existence, not location |
| Critical Point Definition | Interior point $c$ in domain of $f$: <br> $f'(c) = 0$ OR $f'(c)$ does not exist | Endpoints are not critical points; $c$ must be in domain of original $f$ |
| Fermat's Theorem | If $f$ has local extremum at interior $c$, then $c$ is a critical point | Converse is false: not all critical points are local extrema |
| Global (Absolute) Extremum | Global max at $c$: $f(c) \geq f(x) \; \forall x \in I$ <br> Global min at $c$: $f(c) \leq f(x) \; \forall x \in I$ | Only one global max value and one global min value per interval; can occur at endpoints or critical points |
| Local (Relative) Extremum | Local max at $c$: $f(c) \geq f(x)$ for all $x$ in open interval around $c$ <br> Local min at $c$: $f(c) \leq f(x)$ for all $x$ in open interval around $c$ | Only occur at interior points of the domain / interval |
| Global Extrema on Closed Interval | Candidate list = critical points + endpoints | Evaluate $f$ at each candidate, select largest (max) and smallest (min) |

## What's next

This topic is the non-negotiable foundation for all remaining topics in Unit 5: Analytical Applications of Differentiation. Next, you will use critical points and the EVT framework to apply the First Derivative Test and Second Derivative Test, which let you classify local extrema and sketch the shape of a function's graph. Without correctly identifying critical points and verifying EVT conditions, you cannot correctly solve global optimization problems, which make up a large portion of FRQ points on the AP exam.

- [Second Derivative Test](https://www.owlsprep.com/study/ap-calculus-ab-u5-second-derivative-test/)
- [Determining intervals where a function is increasing/decreasing](https://www.owlsprep.com/study/ap-calculus-ab-u5-determining-intervals-where-a-function/)
- [First Derivative Test for Relative Extrema](https://www.owlsprep.com/study/ap-calculus-ab-u5-first-derivative-test-for-relative/)

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