Extreme Value Theorem, Global Versus Local Extrema, and Critical Points - AP Calculus AB Study Guide

The Core Idea: Extreme Value Theorem, Global Versus Local Extrema, and Critical Points

This topic addresses a fundamental question in calculus: how to guarantee the existence of, and then locate, the absolute highest and lowest points of a function over a specific, closed interval. The central pillar is the Extreme Value Theorem (EVT), which provides the conditions under which a function is guaranteed to have an absolute maximum and an absolute minimum value. The theorem states that if a function is continuous over a closed interval (e.g., $[a,b]$), then these extreme values must exist.

Once we know these extreme values exist, the next step is to find them. The key insight is that these absolute (or global) extrema can only occur at a few specific locations: either at the endpoints of the interval or at special points within the interval called critical points. A critical point is a point in the function's domain where its derivative is either zero or undefined. By identifying all critical points within the interval and also considering the endpoints, we create a complete list of "candidates" for the locations of the absolute extrema. Evaluating the original function at each of these candidate points allows us to determine the absolute maximum and minimum values by simply comparing the results.

Key Definitions and Theorems

The Extreme Value Theorem (EVT)

If a function $f(x)$ is continuous on a closed interval$[a,b]$, then $f(x)$ is guaranteed to attain both an absolute maximum value and an absolute minimum value on that interval.

• Condition 1: Continuity. The function must be continuous everywhere on $[a,b]$. There can be no jumps, holes, or vertical asymptotes within the interval.

• Condition 2: Closed Interval. The interval must be closed, meaning it includes its endpoints (e.g., $[1,5]$, not $(1,5)$).

Definition of a Critical Point

A critical point of a function $f(x)$ is a point $x=c$ in the domain of $f(x)$ where either:

1. $f^{\prime }(c)=0$ (the tangent line is horizontal)

2. $f^{\prime }(c)$ is undefined (the tangent line is vertical, or there is a corner or cusp)

Understanding Global vs. Local Extrema

It is crucial to distinguish between global (absolute) and local (relative) extrema.

• A global maximum/minimum on an interval is the absolute highest/lowest $y$-value the function achieves across the entire interval. The Extreme Value Theorem guarantees the existence of these on a closed interval.

• A local maximum/minimum is a point that is higher/lower than all the nearby points. A function can have many local maxima and minima.

The core procedure in this topic, often called the Candidates Test, is designed to find the global extrema on a closed interval. The candidates for the global extrema are the endpoints of the interval and all critical points inside the interval. A local extremum might also be a global extremum, but we must test all candidates to be certain.

For example, a function may have a local maximum at $x=2$, but the function's value at the endpoint $x=5$ might be even higher, making $f(5)$ the global maximum.

Core Concepts & Rules

• EVT Guarantee: A function must be continuous on a closed interval $[a,b]$` for the Extreme Value Theorem to apply. If these conditions are met, an absolute maximum and an absolute minimum are guaranteed to exist.

• Location of Extrema: The absolute extrema of a continuous function on a closed interval $[a,b]$ can only occur at the endpoints ($x=a$ and $x=b$) or at critical points located in the open interval $(a,b)$.

• Finding Critical Points: To find all critical points for a function $f(x)$, you must find its derivative, $f^{\prime }(x)$, and then identify all $x$-values in the domain of $f(x)$ for which $f^{\prime }(x)=0$ or $f^{\prime }(x)$ is undefined.

• The Candidates Test: The procedure for finding absolute extrema on a closed interval $[a,b]$ is as follows:

  1. Confirm the function $f(x)$ is continuous on $[a,b]$.

  2. Find the derivative, $f^{\prime }(x)$.

  3. Find all critical points by solving $f^{\prime }(x)=0$ and finding where $f^{\prime }(x)$ is undefined.

  4. Make a list of candidates: the endpoints $a$ and $b$, and any critical points that lie within the interval $(a,b)$.

  5. Evaluate the original function $f(x)$ at each candidate point.

  6. The largest value from step 5 is the absolute maximum value, and the smallest value is the absolute minimum value.

Step-by-Step Example 1: Basic Application

Problem: Find the absolute maximum and minimum values of the function $f(x)=x^3-6x^2+5$ on the closed interval $[-1,5]$.

Step 1: Verify Conditions for EVT

The function $f(x)$ is a polynomial, which is continuous everywhere. Therefore, it is continuous on the closed interval $[-1,5]$. The EVT guarantees that an absolute maximum and minimum exist.

Step 2: Find the Derivative

Using the power rule, we find the derivative of $f(x)$.

$$f^{\prime }(x)=3x^2-12x$$

Step 3: Find Critical Points

First, we set $f^{\prime }(x)=0$ to find where the tangent is horizontal.

$$3x^2-12x=0$$

$$3x(x-4)=0$$

This gives us potential critical points at $x=0$ and $x=4$.

Next, we check where $f^{\prime }(x)$ is undefined. Since $f^{\prime }(x)$ is a polynomial, it is defined for all real numbers. There are no critical points from an undefined derivative.

Step 4: Identify Candidates

Our candidates for the location of the absolute extrema are the endpoints and the critical points that fall within the interval $[-1,5]$.

• Endpoint: $x=-1$

• Critical Point: $x=0$ (which is in $[-1,5]$)

• Critical Point: $x=4$ (which is in $[-1,5]$)

• Endpoint: $x=5$

Our list of candidates is $\{-1,0,4,5\}$.

Step 5: Evaluate the Function at Each Candidate

We plug each candidate value back into the original function, $f(x)=x^3-6x^2+5$.

• $f(-1)=(-1{)}^3-6(-1{)}^2+5=-1-6+5=-2$

• $f(0)=(0{)}^3-6(0{)}^2+5=0-0+5=5$

• $f(4)=(4{)}^3-6(4{)}^2+5=64-6(16)+5=64-96+5=-27$

• $f(5)=(5{)}^3-6(5{)}^2+5=125-6(25)+5=125-150+5=-20$

Step 6: State the Conclusion

By comparing the $y$-values, we can identify the absolute extrema.

• The highest value is 5, which occurs at $x=0$.

• The lowest value is -27, which occurs at $x=4$.

Answer: On the interval $[-1,5]$, the absolute maximum value of $f(x)$ is 5, and the absolute minimum value is -27.

Step-by-Step Example 2: Exam-Style Application

Problem: Find the absolute minimum value of $g(x)=(x-2{)}^{2/3}$ on the interval $[1,10]$.

Step 1: Verify Conditions for EVT

The function $g(x)=\sqrt[3]{(x-2{)}^2}$ is a composition of a polynomial and a cube root function. Cube root functions are continuous for all real numbers, so $g(x)$ is continuous everywhere, including on the closed interval $[1,10]$. The EVT applies.

Step 2: Find the Derivative

Using the chain rule:

$$g^{\prime }(x)=\frac{2}{3}(x-2{)}^{-1/3}\cdot \frac{d}{dx}(x-2)$$

$$g^{\prime }(x)=\frac{2}{3}(x-2{)}^{-1/3}\cdot 1=\frac{2}{3\sqrt[3]{x-2}}$$

Step 3: Find Critical Points

First, we set $g^{\prime }(x)=0$. A fraction is zero only when its numerator is zero.

$$2=0$$

This equation has no solution. Therefore, there are no critical points where the derivative is zero.

Next, we check where $g^{\prime }(x)$ is undefined. The derivative is undefined when the denominator is zero.

$$3\sqrt[3]{x-2}=0$$

$$\sqrt[3]{x-2}=0$$

$$x-2=0$$

$$x=2$$

The derivative is undefined at $x=2$. Since $x=2$ is in the domain of the original function $g(x)$ ($g(2)=0$), it is a critical point. This corresponds to a cusp on the graph of $g(x)$.

Step 4: Identify Candidates

Our candidates are the endpoints and the critical point within the interval $[1,10]$.

• Endpoint: $x=1$

• Critical Point: $x=2$ (which is in $[1,10]$)

• Endpoint: $x=10$

Our list of candidates is $\{1,2,10\}$.

Step 5: Evaluate the Function at Each Candidate

We plug each candidate value back into the original function, $g(x)=(x-2{)}^{2/3}$.

• $g(1)=(1-2{)}^{2/3}=(-1{)}^{2/3}=(\sqrt[3]{-1}{)}^2=(-1{)}^2=1$

• $g(2)=(2-2{)}^{2/3}=(0{)}^{2/3}=0$

• $g(10)=(10-2{)}^{2/3}=(8{)}^{2/3}=(\sqrt[3]{8}{)}^2=(2{)}^2=4$

Step 6: State the Conclusion

Comparing the values \{1, 0, 4\}\, the smallest value is 0.

Answer: The absolute minimum value of $g(x)$ on the interval $[1,10]$ is 0.

Using Your Calculator

For problems involving the Extreme Value Theorem, the calculator is primarily a tool for verification and evaluation, not for finding the analytical justification required on the AP Exam. You must still show the process of finding the derivative and identifying critical points by hand.

1. Visualizing the Function and Estimating Extrema:

• Press Y= and enter the function, for example, $Y_1=X^3-6X^2+5$.

• Press WINDOW and set $Xmin$, $Xmax$, $Ymin$, and $Ymax$ to match the interval and expected range. For Example 1, set $Xmin=-1$ and $Xmax=5$.

• Press GRAPH to see the function's behavior on the interval. This can give you a good visual idea of where the max and min might be.

2. Evaluating the Function at Candidates:

• Once you have your list of candidates (e.g., $\{-1,0,4,5\}$), you can use the calculator to find the $y$-values quickly.

• From the home screen, you can type `Y_1(-1), $Y_1(0), etc. (To get Y_1`, press `VARS`, go to the `Y-VARS` menu, select `1:Function...`, and then `1:Y1`). - Alternatively, use the `TABLE` feature. Press `2nd` then `TBLSET`. Set `Indpnt:` to `Ask`. Then press `2nd` then `TABLE` and enter your candidate $x-values one by one.

3. Finding Critical Points for Complex Functions:

• If $f^{\prime }(x)=0$ is difficult to solve by hand, you can use the calculator.

• In Y=, set `Y_2 = nDeriv(Y_1, X, X)to graph the derivative of your function in $Y_1.

• Use the CALC menu (2ndTRACE) and select 2:zero to find the $x$-intercepts of the derivative graph ($Y_2$). These are your critical points where $f^{\prime }(x)=0$.

Important Note: On a Free Response Question, you must write down $f^{\prime }(x)=0$ and report the solution values. Simply stating that you used a calculator to find the zeros of the derivative is not sufficient justification.

AP Exam Quick Hit

Common Question Types

• Standard "Candidates Test" Problem: "Find the absolute maximum and minimum values of $f(x)=x-2\cos (x)$ on the interval $[0,2\pi ]$." This requires finding the derivative, setting it to zero, solving for $x$, and testing the endpoints and critical points.

• Justifying with the EVT: "Let $f$ be a function defined by $f(x)=...$. On the closed interval $[a,b]$, must there be a value $c$ for which $f(c)$ is the absolute minimum value? Justify your answer." The answer would involve confirming that $f(x)$ is continuous on $[a,b]$ and then invoking the Extreme Value Theorem.

• Finding Critical Points from a Graph of $f^{\prime }(x)$: Given the graph of a derivative $f^{\prime }(x)$, you might be asked to identify the $x$-values of the critical points of $f(x)$. These occur wherever the graph of $f^{\prime }(x)$ touches or crosses the $x$-axis ($f^{\prime }(x)=0$) or has a discontinuity (where $f^{\prime }(x)$ is undefined).

Common Mistakes

• Forgetting to Check the Endpoints: This is the most frequent error. Students correctly find the critical points but forget to include the interval's endpoints in their list of candidates. The absolute extremum on a closed interval very often occurs at an endpoint.

• Plugging Candidates into $f^{\prime }(x)$ instead of $f(x)$: After finding the $x$-values of the critical points, students mistakenly plug them into the derivative ($f^{\prime }(x)$) to find the "value." The question asks for the maximum/minimum value of the function, which is the $y$-value from the original function $f(x)$.

• Stopping at Critical Points: Students find the critical points and assume they are the locations of the extrema without comparing them to the values at the endpoints.

• Ignoring Critical Points Where $f^{\prime }(x)$ is Undefined: Many students only solve $f^{\prime }(x)=0$ and forget to look for $x$-values where the derivative is undefined (e.g., denominators of $f^{\prime }(x)$ equal to zero).

• Considering Critical Points Outside the Interval: A function may have several critical points, but only those that lie within the specified interval are candidates for the absolute extrema on that interval.