Extreme Value Theorem, global vs local extrema, critical points — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: This chapter covers the Extreme Value Theorem (EVT), definitions of local vs global (absolute) extrema, rules for identifying critical points, and how to use critical points to locate candidate extrema for continuous functions.

You should already know: How to compute derivatives of algebraic, trigonometric, and composite functions; properties of continuous functions; interval notation for domains.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Extreme Value Theorem, global vs local extrema, critical points?

This topic is the foundational framework for all optimization, curve sketching, and applied maximum/minimum problems in AP Calculus AB. It accounts for roughly 4-6% of the total AP exam score, within Unit 5’s 15-18% overall weight. It appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often as a required stepping stone for larger optimization or curve analysis questions.

At its core, this set of concepts answers a basic question: given a function, how do we find its highest and lowest values, both over its entire domain and in small local regions? The Extreme Value Theorem gives us a formal guarantee that global extrema exist under specific conditions, critical points are the only locations where interior extrema can occur, and distinguishing local vs global extrema helps us correctly interpret what extrema mean for a function. Unlike many procedural derivative rules, this topic tests both conceptual understanding (verifying theorem conditions) and procedural skill (finding extrema), both of which are heavily tested on the AP exam.

2. The Extreme Value Theorem (EVT)

The Extreme Value Theorem 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. The formal statement is: 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 $[a,b]$.

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 somewhere along the segment. The guarantee breaks if either condition fails: if the interval is open (e.g., $(a,b)$) you can approach an extremum but never reach it, and if the function has a discontinuity (e.g., a vertical asymptote) on the interval, it can grow without bound and never reach a maximum or minimum. EVT is the first step in justifying any global extremum calculation on a closed interval, which is a common requirement for AP FRQ points.

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) $f(x)=\frac{x^2}{x-2}$ on $[-1,1]$ (D) $f(x)=\frac{\sin (x)}{x-\pi }$ on $[\pi /2,3\pi /2]$

Solution:

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.

EVT applies only to option C.

Exam 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.

3. Critical Points

A critical point of a function $f(x)$ is defined as an interior point $c$ (meaning $c$ is inside the domain of $f$, not an endpoint) where either $f^{\prime }(c)=0$ or $f^{\prime }(c)$ does not exist. Critical points are the only possible locations for local extrema, by Fermat’s Theorem, which states that if $f(x)$ has a local extremum at an interior point $c$, then $c$ must be a critical point.

There are two key edge cases to remember:

1. Endpoints of an interval are never critical points, because they are not interior points. We still check endpoints as candidates for global extrema, but they do not count as critical points by AP CED definition.
2. A point where the original function $f(x)$ is undefined can never be a critical point, even if $f^{\prime }(x)$ is also undefined there. Critical points must lie in the domain of the original function.

Worked Example

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

Solution:

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

Exam tip: Always confirm that a point where the derivative is undefined is actually 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.

4. Local vs Global (Absolute) Extrema

Extrema are classified by the interval over which they are the maximum or minimum:

• Global (Absolute) Extremum: A global maximum on an interval $I$ is a value $f(c)$ such that $f(c)\ge f(x)$ for all $x$ in $I$. A global minimum is $f(c)\le f(x)$ for all $x$ in $I$. 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. Global extrema can occur at critical points (interior) or endpoints.
• Local (Relative) Extremum: A local maximum is a value $f(c)$ such that $f(c)\ge f(x)$ for all $x$ in some small open interval around $c$. A local minimum follows the same rule with the inequality reversed. Local extrema only occur at interior points (endpoints can never be local extrema, since you cannot have an open interval around them within the domain). A 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.

Solution:

1. Find critical points: $f^{\prime }(x)=3x^2-3=3(x-1)(x+1)$, so critical points at interior points $x=-1$ and $x=1$, both in the domain.
2. Classify local extrema via the first derivative test: $f^{\prime }(x)$ changes from positive to negative at $x=-1$, so $f(-1)=(-1{)}^3-3(-1)+1=3$ is a local maximum. $f^{\prime }(x)$ changes from negative to positive at $x=1$, so $f(1)=1-3+1=-1$ is a local minimum.
3. Evaluate all candidates for global extrema: evaluate $f$ at critical points and endpoints: $f(-3)=-17$, $f(-1)=3$, $f(1)=-1$, $f(2)=3$.
4. Compare values: The largest value is 3, and the smallest value is $-17$.

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 $-17$.

Exam 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.

5. Common Pitfalls (and how to avoid them)

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

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

Which of the following statements about $f(x)=\frac{x^2}{x-3}$ 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.

Worked Solution: First, check the conditions for EVT: the interval $[0,5]$ is closed, but $f(x)$ has a vertical asymptote and discontinuity at $x=3$, which lies inside the interval. This means $f(x)$ is not continuous on the entire interval, so EVT does not apply, eliminating options A and B. 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. The correct answer is C.

Question 2 (Free Response)

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.

Worked Solution: (a) Compute the derivative of the polynomial: $f^{\prime }(x)=6x^2-30x+24=6(x^2-5x+4)=6(x-1)(x-4)$. Setting $f^{\prime }(x)=0$ gives critical points at $x=1$ and $x=4$, both interior points on $(0,5)$ in the domain of $f$. There are no points where $f^{\prime }(x)$ is undefined. Critical points are $x=1$ and $x=4$.

(b) Use the first derivative test to classify: For $0<x<1$, $f^{\prime }(x)>0$ (f increasing); for $1<x<4$, $f^{\prime }(x)<0$ (f decreasing); for $4<x<5$, $f^{\prime }(x)>0$ (f increasing). $f^{\prime }$ changes from positive to negative at $x=1$, so $f(1)=2(1)-15(1)+24(1)+4=15$, a local maximum of 15 at $x=1$. $f^{\prime }$ changes from negative to positive at $x=4$, so $f(4)=2(64)-15(16)+24(4)+4=-12$, a local minimum of -12 at $x=4$.

(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: $f(0)=4$, $f(1)=15$, $f(4)=-12$, $f(5)=-1$. Comparing values, the largest value is 15 and the smallest is -12. Global maximum value = 15, global minimum value = -12.

Question 3 (Application / Real-World Style)

A small coffee shop estimates that their daily profit from selling $x$ lattes is given by $P(x)=-0.002x^3+0.4x^2+1.5x-200$, where $0\le x\le 150$, and $P(x)$ is measured in dollars. The shop can make at most 150 lattes per day. Find the global maximum daily profit the shop can earn, and the number of lattes that gives this maximum profit.

Worked Solution: $P(x)$ is a polynomial, so it is continuous on the closed interval $[0,150]$, so EVT applies. Find critical points: $P^{\prime }(x)=-0.006x^2+0.8x+1.5$. Set $P^{\prime }(x)=0$, and solve via quadratic formula to get the only positive root $x\approx 135$, which is inside the interval. Evaluate $P$ at all candidates: $P(0)=-200$, $P(135)\approx 3128$, $P(150)=2275$. The maximum value is approximately $\$3128$ at 135 lattes. In context: The coffee shop earns its maximum daily profit of roughly $\$3130$ when it sells approximately 135 lattes per day.

7. Quick Reference Cheatsheet

8. 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 the conditions for EVT, you cannot correctly solve global optimization problems, which make up a large portion of FRQ points on the AP exam. This topic also feeds into the broader study of curve sketching and applied problem-solving across the rest of the course, where you will regularly need to find maximum and minimum values to answer real-world questions.

• First Derivative Test
• Second Derivative Test
• Applied Optimization
• Curve Sketching