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

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Extreme Value Theorem conditions and application, definition of critical points, Fermat’s Theorem, classification of local vs global (absolute) extrema, and the closed interval method for finding global extrema on closed intervals.

You should already know: How to compute derivatives of algebraic, transcendental, and composite functions. What it means for a function to be continuous on an interval. How to simplify and solve rational and polynomial equations.

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 BC 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 finding and classifying the highest and lowest values of a function on an interval, and it is the backbone of all optimization and curve sketching work in AP Calculus. It is part of Unit 5: Analytical Applications of Differentiation, which makes up 15-18% of the total AP Calculus BC exam score. Concepts from this topic appear in both multiple-choice (MCQ) and free-response (FRQ) sections: standalone MCQ questions test your understanding of theorem conditions and definitions, while FRQ questions embed these concepts as the first step in larger problems like optimization or graph analysis. Mastery of this topic is non-negotiable for every subsequent topic in the unit, as all other derivative-based analysis of extrema builds on these definitions and rules.

2. Critical Points

A critical point of a function $f$ is defined as an interior point $c$ (meaning $c$ is inside the open interval of the domain we are studying, not an endpoint) that meets one of two conditions: either $f^{\prime }(c)=0$, or $f^{\prime }(c)$ does not exist (but $c$ is still in the domain of the original function). This definition comes 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. In other words, all local extrema occur at critical points—this means when we search for extrema, we only need to check critical points (and endpoints for global extrema) to find them. A common misconception is that all critical points are extrema; this is not true: for example, $f(x)=x^3$ has a critical point at $x=0$ where $f^{\prime }(0)=0$, 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: $$f^{\prime }(x)=\frac{2}{3}{x}^{-1/3}(4-x)+x^{2/3}(-1)$$
2. Simplify the derivative by factoring out $\frac{1}{3}{x}^{-1/3}$: $$f^{\prime }(x)=\frac{2(4-x)-3x}{3x^{1/3}}=\frac{8-5x}{3x^{1/3}}$$
3. Locate points where $f^{\prime }(x)=0$ or $f^{\prime }(x)$ is undefined:
   • $f^{\prime }(x)=0$ when the numerator equals 0: $8-5x=0⟹x=\frac{8}{5}=1.6$, which is an interior point of $[-1,4]$ and in the domain of $f$.
   • $f^{\prime }(x)$ is undefined when the denominator equals 0: $x=0$, which is an interior point of $[-1,4]$ and in the domain of $f$.
4. The critical points on $[-1,4]$ are $x=0$ and $x=\frac{8}{5}$.

Exam 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 also undefined there, it cannot be a critical point.

3. 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 of a function on an interval, but it does not tell you how to find them. The formal statement is: If a function $f(x)$ is continuous on every point of a closed, bounded interval $[a,b]$, then $f$ must attain at least one global maximum and at least one global minimum on $[a,b]$.

Two conditions must both be satisfied for EVT to apply: (1) continuity over the entire interval (no discontinuities anywhere inside or at the endpoints), and (2) the interval is closed (includes both endpoints) and bounded (has finite length). If either condition fails, EVT does not guarantee that global extrema exist—they might exist by chance, but you cannot rely on the theorem to confirm they exist. For example, $f(x)=1/x$ on $(0,1)$ is continuous but the interval is open, so EVT does not apply, and indeed there is 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. First, find all discontinuities of $f(x)$: as a rational function, $f(x)$ is discontinuous where the denominator equals 0: $x^2-1=0⟹x=1,x=-1$.
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 $[2,4]$.

Exam 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.

4. Local vs Global (Absolute) Extrema

Extrema are classified as either local (relative) or global (absolute), based on what interval they are the maximum/minimum over:

• A global (absolute) maximum of $f$ on an interval $I$ is a value $f(c)$ such that $f(c)\ge f(x)$ for all $x$ in $I$.
• A global (absolute) minimum of $f$ on an interval $I$ is a value $f(c)$ such that $f(c)\le f(x)$ for all $x$ in $I$.
• A local (relative) maximum of $f$ is a value $f(c)$ such that $f(c)\ge f(x)$ for all $x$ in some small open interval around $c$ (only nearby points, not the entire interval).
• A local (relative) minimum follows the same logic, with $f(c)\le f(x)$ for nearby $x$.

Key distinctions the AP exam tests: (1) Global extrema can occur at critical points or endpoints, but local extrema only occur at interior critical points (endpoints cannot be local extrema, because you cannot have an open interval around the endpoint within your domain). (2) 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). (3) 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: $f(-1)=-3$, $f(0)=1$, $f(2)=-3$, $f(4)=17$.
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 $x=2$.

Exam 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.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Calling an endpoint of the interval a critical point. Why: 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 move: 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 move: Claiming EVT applies to a continuous function on an open interval, or a discontinuous function on a closed interval. Why: Students memorize "EVT gives extrema" but forget that both conditions are required to apply the theorem. Correct move: For any EVT application question, explicitly check "continuous on entire interval" and "closed bounded interval" one after the other before concluding.
• Wrong move: Only finding points where $f^{\prime }(x)=0$ when searching for critical points, ignoring points where $f^{\prime }(x)$ is undefined. Why: Most introductory examples use polynomials that are differentiable everywhere, so students forget the second half of the critical point definition. Correct move: After solving $f^{\prime }(x)=0$, always check where $f^{\prime }(x)$ is undefined, then confirm those points are interior and in the domain of $f$ to add them to your critical point list.
• Wrong move: Classifying a global extremum at an endpoint as a local extremum. Why: 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 move: 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 move: Assuming every critical point is automatically a local extremum. Why: 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 move: Always test a critical point with the first or second derivative test to confirm it is an extremum, do not skip this step.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

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? A) $f(x)=\frac{1}{x^2-1}$ B) $f(x)=x^3-3x+2$ C) $f(x)=e^{-x^2}$ D) $f(x)=\sqrt{x+2}$

Worked Solution: First, eliminate options that do not satisfy EVT conditions (continuous on the entire closed interval $[-2,2]$): Option A has discontinuities at $x=\pm 1$, which are inside the interval, so A is eliminated. Next, count critical points for the remaining options: Option D: $f^{\prime }(x)=\frac{1}{2\sqrt{x+2}}$, which is never zero, and only undefined at the endpoint $x=-2$, so it has 0 critical points, eliminate D. Option C: $f^{\prime }(x)=-2x{e}^{-x^2}$, which is zero only at $x=0$, so it has one critical point, eliminate C. Option B: $f(x)$ is a polynomial, continuous everywhere, so EVT applies, and $f^{\prime }(x)=3x^2-3=3(x^2-1)$, which is zero at $x=1$ and $x=-1$, giving exactly two critical points. The correct answer is B.

Question 2 (Free Response)

Let $f(x)=x\ln x$ defined for $x>0$, and consider the interval $[\frac{1}{2},4]$. (a) Find all critical points of $f(x)$ on $[\frac{1}{2},4]$. Justify your answer. (b) Does the Extreme Value Theorem guarantee the existence of a global maximum and global minimum of $f(x)$ on this interval? Justify your answer. (c) Identify all global and local extrema of $f(x)$ on this interval.

Worked Solution: (a) Use the product rule to compute the derivative: $f^{\prime }(x)=\ln x+x\cdot \frac{1}{x}=\ln x+1$. Set $f^{\prime }(x)=0$: $\ln x=-1⟹x=1/e\approx 0.368$. $f^{\prime }(x)$ is defined for all $x>0$, so the only candidate critical point is at $x=1/e$, which is less than $1/2$, so it is outside the interval $[\frac{1}{2},4]$. There are no critical points on the interior of this interval. (b) $f(x)=x\ln x$ is continuous for all $x>0$, so it is continuous over the entire closed bounded interval $[\frac{1}{2},4]$. Both conditions of the Extreme Value Theorem are satisfied, so EVT does guarantee the existence of a global maximum and global minimum. (c) Evaluate $f$ at the endpoints: $f(1/2)=\frac{1}{2}\ln \left(\frac{1}{2}\right)=-\frac{\ln 2}{2}\approx -0.347$, $f(4)=4\ln 4=8\ln 2\approx 2.773$. The smallest value is $-\frac{\ln 2}{2}$ at $x=1/2$ (endpoint), and the largest value is $8\ln 2$ at $x=4$ (endpoint). There are no interior critical points, so there are no local extrema. Final result: Global maximum = $8\ln 2$ at $x=4$, global minimum = $-\frac{\ln 2}{2}$ at $x=1/2$, no local extrema.

Question 3 (Application / Real-World Style)

A small business models its daily profit over an 8-hour shift as $P(t)=-t^3+9t^2-18t+25$, where $P(t)$ is profit in hundreds of dollars, and $0\le t\le 8$ is hours after the start of the shift. Find the global maximum and minimum profit over the shift, and classify all local extrema.

Worked Solution: $P(t)$ is a polynomial, so it is continuous on the closed interval $[0,8]$, so EVT applies. Compute the derivative to find critical points: $P^{\prime }(t)=-3t^2+18t-18=-3(t^2-6t+6)$. Set $P^{\prime }(t)=0$: $t=\frac{6\pm \sqrt{36-24}}{2}=3\pm \sqrt{3}\approx 1.268,4.732$, both interior points of $[0,8]$, derivative exists everywhere, so these are the only critical points. Evaluate $P(t)$ at all candidates: $P(0)=25$, $P(3-\sqrt{3})\approx 21.92$, $P(3+\sqrt{3})\approx 28.08$, $P(8)=-103$. Classification: $t=3-\sqrt{3}$ is an interior local minimum, $t=3+\sqrt{3}$ is an interior local maximum (also global maximum), $t=8$ is endpoint, global minimum. The global maximum profit is approximately $2808at4hours44minutesintotheshift,andtheglobalminimumprofitisapproximately-$10300 at the end of the 8-hour shift. Interpretation: The business peaks in profit near the middle of the shift, and ends the shift with a net loss if this model holds for the full 8 hours.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational first step for all further work in analytical differentiation applications, starting with the Mean Value Theorem and Rolle’s Theorem, which also rely on checking continuity conditions for a function on an interval, just like the Extreme Value Theorem. Immediately after this, 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 use the methods from this chapter to find the maximum or minimum value of a real-world function on an interval—without correctly identifying critical points and checking endpoint values, optimization problems will always have missing or incorrect solutions. This topic also supports analytic curve sketching, where locating extrema is the core step to drawing an accurate graph.

Mean Value Theorem and Rolle's Theorem First Derivative Test for Local Extrema Second Derivative Test and Curve Sketching Optimization of Functions