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

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

This guide covers the Extreme Value Theorem, critical point definition, and classification of local vs global extrema, core foundational concepts for all AP Calculus BC optimization and curve analysis problems.

**Prerequisites:** Computing derivatives of algebraic, transcendental, and composite functions; Understanding continuity of functions on intervals; Solving polynomial and rational equations

## Learning objectives

- State the definition of a critical point and Fermat's Theorem
- Apply the Extreme Value Theorem correctly by checking its conditions
- Classify extrema as local or global correctly
- Find all critical points and global extrema on a closed interval

## Critical Points

**Critical Point** — A point that satisfies either $f'(c) = 0$ or $f'(c)$ does not exist, while $c$ remains in the domain of the original function $f$.

*Notation:* Interior point $c$ of interval $I$

*Example:* For $f(x) = x^3$, $x=0$ is a critical point because $f'(0)=0$.

This definition follows from Fermat's Theorem, which states that if a function $f$ has a local extremum at an interior point $c$, then $c$ must be a critical point. A common misconception is that all critical points are extrema; this is not true: $f(x) = x^3$ has a critical point at $x=0$, but it is not an extremum.

**Worked example:** Find all critical points of $f(x) = x^{2/3}(4 - x)$ on the interval $[-1, 4]$.

1. Use the product rule to compute the first derivative:

   $$f'(x) = \frac{2}{3}x^{-1/3}(4 - x) + x^{2/3}(-1)$$
2. Simplify the derivative by factoring out $\frac{1}{3}x^{-1/3}$:

   $$f'(x) = \frac{2(4 - x) - 3x}{3x^{1/3}} = \frac{8 - 5x}{3x^{1/3}}$$
3. $f'(x) = 0$ when the numerator equals 0: $8 - 5x = 0 \implies x = \frac{8}{5} = 1.6$, which is an interior point of $[-1, 4]$ and in the domain of $f$. $f'(x)$ is undefined when the denominator equals 0: $x=0$, which is also an interior point and in the domain of $f$.
4. The critical points on $[-1, 4]$ are $x=0$ and $x=\frac{8}{5}$.

> **tip**
>
> Always confirm that a point where the derivative is undefined is actually in the domain of the original function before labeling it a critical point—if the original function is undefined there, it cannot be a critical point.

## The Extreme Value Theorem

**Extreme Value Theorem (EVT)** — An existence theorem that guarantees a function will have at least one global maximum and one global minimum on an interval, provided two conditions are both satisfied: (1) the function is continuous at every point on the interval, and (2) the interval is closed and bounded (has finite length and includes both endpoints). EVT only confirms extrema exist, it does not show how to find them.

If either condition fails, EVT does not guarantee global extrema exist. They may exist by chance, but you cannot rely on the theorem to confirm their existence. For example, $f(x) = 1/x$ on the open interval $(0, 1)$ is continuous but has no global maximum.

**Worked example:** Does the Extreme Value Theorem guarantee the existence of a global maximum and global minimum for $f(x) = \frac{1}{x^2 - 1}$ on (a) $[0, 2]$ (b) $[2, 4]$? Justify your answer.

1. Find all discontinuities of $f(x)$: as a rational function, it is discontinuous where the denominator equals 0:

   $$x^2 - 1 = 0 \implies x = 1, x = -1$$
2. For interval (a) $[0, 2]$: $x=1$ is an interior point of $[0, 2]$, so $f(x)$ is not continuous over the entire interval. One condition of EVT is violated, so EVT does not guarantee global maximum or minimum exist on $[0, 2]$.
3. For interval (b) $[2, 4]$: no discontinuities of $f(x)$ lie in this interval, and $[2, 4]$ is a closed, bounded interval. Both conditions of EVT are satisfied, so EVT guarantees a global maximum and global minimum exist on $[2, 4]$.

> **tip**
>
> On AP MCQ, EVT application questions always test whether you remember both conditions. Write both checks (continuity, closed bounded interval) explicitly for justification points on FRQ.

## Local vs Global (Absolute) Extrema

Extrema are classified by the interval over which they are the maximum or minimum value. The AP exam frequently tests the key distinctions between these two classifications:

- **Global (Absolute) Maximum**: A value $f(c)$ such that $f(c) \geq f(x)$ for *all* $x$ in the interval $I$.
- **Global (Absolute) Minimum**: A value $f(c)$ such that $f(c) \leq f(x)$ for *all* $x$ in the interval $I$.
- **Local (Relative) Maximum**: A value $f(c)$ such that $f(c) \geq f(x)$ for all $x$ in some small open interval around $c$ (only nearby points, not the entire interval).
- **Local (Relative) Minimum**: A value $f(c)$ such that $f(c) \leq f(x)$ for all $x$ in some small open interval around $c$.

Key AP-tested distinctions:

- Global extrema can occur at critical points *or endpoints*, but local extrema only occur at interior critical points (endpoints cannot be local extrema, since you cannot have an open interval around the endpoint within the domain).
- A function can have multiple local extrema, but only one global maximum value and one global minimum value (though these values can occur at multiple points).
- Any global extremum at an interior point is automatically a local extremum, but global extrema at endpoints are never local.

**Worked example:** Given $f(x) = x^3 - 3x^2 + 1$ on $[-1, 4]$, with critical points at $x=0$ and $x=2$, classify all extrema as local or global.

1. Evaluate $f(x)$ at all critical points and endpoints, per the closed interval method:

   $$f(-1) = -3, \quad f(0) = 1, \quad f(2) = -3, \quad f(4) = 17$$
2. Identify global extrema: The largest value across all points is 17 at $x=4$ (endpoint), so this is the global maximum. The smallest value is $-3$ at $x=-1$ (endpoint) and $x=2$ (interior critical point), so these are both global minima.
3. Classify local extrema: Only interior points can be local extrema. $f(0) = 1$ is larger than all nearby values, so it is a local maximum. $f(2) = -3$ is smaller than all nearby values, so it is a local minimum (and also global). Endpoints $x=-1$ and $x=4$ are not classified as local extrema.
4. Final classification: Global max = 17 at $x=4$; Global min = $-3$ at $x=-1, x=2$; Local max = 1 at $x=0$; Local min = $-3$ at $x=2$.

> **tip**
>
> When asked for global extrema on a closed interval, always evaluate at endpoints—AP exam problems often place the global extremum at an endpoint to test if you forget to check it.

## AP-Style Practice Worked Examples

**Check your understanding**

Test your understanding of core concepts with this AP-style multiple choice question:

1. Which of the following functions satisfies both conditions of the Extreme Value Theorem on $[-2, 2]$ and has exactly two critical points on this interval?

   - $f(x) = \frac{1}{x^2 - 1}$
   - $f(x) = x^3 - 3x + 2$
   - $f(x) = e^{-x^2}$
   - $f(x) = \sqrt{x + 2}$

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

   *Why:* Correct. $f(x)$ is a polynomial (continuous everywhere, so EVT applies) and has critical points at $x=1$ and $x=-1$, giving exactly two critical points. Option A is discontinuous at $x=\pm 1$, so EVT does not apply; Option C has only one critical point at $x=0$; Option D has zero interior critical points.

**Worked example:** Let $f(x) = x \ln x$ defined for $x > 0$, and consider the interval $[\frac{1}{2}, 4]$. (a) Find all critical points, (b) Justify if EVT applies, (c) Classify all extrema.

1. Find critical points: Use product rule to compute the derivative:

   $$f'(x) = \ln x + x \cdot \frac{1}{x} = \ln x + 1$$
2. Set $f'(x)=0$: $\ln x = -1 \implies x = 1/e \approx 0.368$, which is outside the interval $[\frac{1}{2}, 4]$. $f'(x)$ is defined for all $x>0$, so there are no interior critical points on the interval.
3. Check EVT conditions: $f(x) = x \ln x$ is continuous for all $x>0$, so it is continuous over the entire closed bounded interval $[\frac{1}{2}, 4]$. Both conditions are satisfied, so EVT guarantees global extrema exist.
4. Final result: Global maximum = $8\ln 2$ at $x=4$, global minimum = $-\frac{\ln 2}{2}$ at $x=1/2$, no local extrema.

## Common pitfalls

- **Wrong:** Calling an endpoint of the interval a critical point.
  - Why it fails: Students confuse the requirement to check endpoints for global extrema with the definition of a critical point, which requires the point to be interior to the interval.
  - Correct: Always maintain two separate lists when finding global extrema: one for interior critical points, one for endpoints; never add endpoints to the critical point list.
- **Wrong:** Claiming EVT applies to a continuous function on an open interval, or a discontinuous function on a closed interval.
  - Why it fails: Students memorize 'EVT gives extrema' but forget that both conditions are required to apply the theorem.
  - Correct: For any EVT application question, explicitly check 'continuous on entire interval' and 'closed bounded interval' one after the other before concluding.
- **Wrong:** Only finding points where $f'(x) = 0$ when searching for critical points, ignoring points where $f'(x)$ is undefined.
  - Why it fails: Most introductory examples use polynomials that are differentiable everywhere, so students forget the second half of the critical point definition.
  - Correct: After solving $f'(x) = 0$, always check where $f'(x)$ is undefined, then confirm those points are interior and in the domain of $f$ to add them to your critical point list.
- **Wrong:** Classifying a global extremum at an endpoint as a local extremum.
  - Why it fails: Students assume all global extrema are automatically local, but the definition of local extrema requires an open interval around the point within the domain.
  - Correct: If an extremum is at an endpoint of the given interval, only label it as a global extremum (if it qualifies), never call it local.
- **Wrong:** Assuming every critical point is automatically a local extremum.
  - Why it fails: Students misremember Fermat’s Theorem as working both directions, when Fermat’s only says all local extrema are at critical points, not the reverse.
  - Correct: Always test a critical point with the first or second derivative test to confirm it is an extremum, do not skip this step.

## Cheatsheet

| Category | Rule/Definition | Notes |
| --- | --- | --- |
| Critical Point Definition | Interior point $c$ in domain of $f$: $f'(c) = 0$ OR $f'(c)$ does not exist | Only interior points count, endpoints are never critical points |
| Fermat's Theorem | If $f$ has a local extremum at interior $c$, then $c$ is a critical point | Reverse is not true: not all critical points are extrema |
| Extreme Value Theorem | If $f$ is continuous on closed bounded $[a,b]$, $f$ has at least one global max and one global min on $[a,b]$ | Fails if $f$ is discontinuous anywhere on the interval, or interval is open/unbounded |
| Global Maximum | $f(c) \geq f(x)$ for all $x \in I$ | Can occur at critical points or endpoints; one global max value (can be at multiple points) |
| Global Minimum | $f(c) \leq f(x)$ for all $x \in I$ | Same as global maximum, can occur at endpoints or critical points |
| Local Maximum | $f(c) \geq f(x)$ for all $x$ in an open interval around $c$ | Only occurs at interior critical points, never at endpoints |
| Local Minimum | $f(c) \leq f(x)$ for all $x$ in an open interval around $c$ | Same as local maximum, never at endpoints |
| Closed Interval Method | 1. Find all critical points in $(a,b)$ 2. Evaluate $f$ at critical points and endpoints 3. Largest = global max, smallest = global min | Standard method for AP exam problems asking for global extrema on a closed interval |

## What's next

This topic is the foundational first step for all further work in analytical differentiation applications for AP Calculus BC. Immediately after mastering these concepts, you will move to the Mean Value Theorem and Rolle’s Theorem, which also rely on correctly checking continuity conditions for a function on an interval, just like the Extreme Value Theorem. Next, you will learn the First Derivative Test and Second Derivative Test to classify critical points as local extrema, which you cannot do without correctly identifying critical points and understanding the difference between local and global extrema first. This entire unit builds to constrained optimization problems, where you will use the methods from this sub-topic to find the maximum or minimum value of a real-world function on an interval.

- [Determining Intervals Where a Function Is Increasing/Decreasing](https://www.owlsprep.com/study/ap-calculus-bc-u5-determining-intervals-where-a-function/)
- [First derivative test for relative extrema](https://www.owlsprep.com/study/ap-calculus-bc-u5-first-derivative-test-for-relative/)
- [Candidates Test for Absolute Extrema](https://www.owlsprep.com/study/ap-calculus-bc-u5-candidates-test-for-absolute-extrema/)

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