Extreme Value Theorem, Global Versus Local Extrema, and Critical Points - AP Calculus BC 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 find, the absolute maximum and minimum values of a function over a specific, closed interval. The central concept 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. The theorem states that if a function is continuous on a closed interval $[a,b]$, then it must attain both a highest and a lowest value somewhere in that interval.

To find these extreme values, we must identify all potential locations where they can occur. These locations are called "candidates." The key insight is that absolute extrema on a closed interval can only happen at two types of places: the endpoints of the interval ($a$ and $b$) or at "critical points" within the interval. A critical point is a point in the interior of the domain where the function's derivative is either zero or undefined. By identifying all critical points and endpoints, we can create a list of all possible candidates and then test them to find the absolute highest and lowest function values.

Key Definitions and Theorems

Definition: Critical Point

A critical point is a point $c$ in the interior of the domain of a function $f$ for which $f^{\prime }(c)=0$ or $f^{\prime }(c)$ is undefined.

• $f^{\prime }(c)=0$: This corresponds to a point where the tangent line to the graph of $f$ is horizontal.

• $f^{\prime }(c)$ is undefined: This corresponds to a point where the graph of $f$ may have a sharp corner (cusp) or a vertical tangent line.

Theorem: The Extreme Value Theorem (EVT)

If a function $f$ is continuous on a closed interval$[a,b]$, then $f$ has both an absolute minimum and an absolute maximum value on that interval.

Understanding the Conditions

The Extreme Value Theorem is powerful, but it only applies if its two conditions are met: the function must be continuous, and the interval must be closed. If either condition is violated, the function is not guaranteed to have an absolute maximum or minimum on the interval.

• Condition 1: Continuity. Consider the function $f(x)=\{\begin{array}{ll}x& \text{if }0\le x<1\\ 0& \text{if }x=1\end{array}$ on the closed interval $[0,1]$. This function is not continuous at $x=1$. It approaches a value of 1, but never reaches it. Therefore, it has no absolute maximum on $[0,1]$. The EVT does not apply.

• Condition 2: Closed Interval. Consider the function $f(x)=x^2$ on the open interval $(0,2)$. The function gets arbitrarily close to 0 as $x$ approaches 0, and arbitrarily close to 4 as $x$ approaches 2, but it never actually reaches these values because the endpoints are not included in the interval. Therefore, the function has no absolute minimum or maximum on $(0,2)$. The EVT does not apply.

The procedure for finding absolute extrema, often called the Candidates Test, relies on the fact that for a continuous function on a closed interval, the absolute extrema must occur at either a critical point or an endpoint.

Core Concepts & Rules

• Existence of Extrema: The Extreme Value Theorem guarantees that a continuous function on a closed interval $[a,b]$ will always have an absolute maximum and an absolute minimum value.

• Identifying Critical Points: A critical point of $f$ is an x-value, $c$, in the interior of the domain where $f^{\prime }(c)=0$ or $f^{\prime }(c)$ is undefined. These are potential locations for local and global extrema.

• The Candidates Test: To find the absolute (global) extrema of a continuous function $f$ on a closed interval $[a,b]$:

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

  2. Identify all critical points of $f$ that lie within the interval $(a,b)$.

  3. Create a list of candidates, which includes the endpoints $a$ and $b$, and all critical points found in Step 2.

  4. Evaluate the function $f(x)$ at each candidate point from the list.

  5. The largest value from Step 4 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 $f(x)=2x^3-3x^2-12x+1$ on the closed interval $[-2,3]$.

Step 1: Verify Conditions

The function $f(x)$ is a polynomial, so it is continuous everywhere. The interval $[-2,3]$ is a closed interval. Therefore, the Extreme Value Theorem applies, and we are guaranteed to find an absolute maximum and minimum.

Step 2: Find the Derivative

Find $f^{\prime }(x)$ to locate the critical points.

$$f^{\prime }(x)=6x^2-6x-12$$

Step 3: Find Critical Points

Set $f^{\prime }(x)=0$ and solve for $x$. The derivative is a polynomial, so it is never undefined.

$$6x^2-6x-12=0$$

$$6(x^2-x-2)=0$$

$$6(x-2)(x+1)=0$$

The critical points are $x=2$ and $x=-1$. Both of these points are within the interval $[-2,3]$.

Step 4: Create a Table of Candidates

The candidates for the location of the absolute extrema are the critical points ($-1,2$) and the endpoints ($-2,3$). Evaluate $f(x)$ at each of these values.

Step 5: State the Conclusion

By comparing the values in the table:

• The absolute maximum value is $8$, which occurs at $x=-1$.

• The absolute minimum value is $-19$, which occurs at $x=2$.

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

Problem: Find the absolute extrema of $g(x)=x(12-2x{)}^{1/2}$ on the interval $[0,6]$.

Step 1: Verify Conditions

The function involves a square root, so its domain is restricted to where $12-2x\ge 0$, which means $12\ge 2x$, or $x\le 6$. The function is continuous on its domain, $(-\infty ,6]$. The given interval $[0,6]$ is a closed interval within this domain, so $g(x)$ is continuous on $[0,6]$. The EVT applies.

Step 2: Find the Derivative

Use the Product Rule: $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$. Let $u=x$ and $v=(12-2x{)}^{1/2}$.

$$g^{\prime }(x)=(1)(12-2x{)}^{1/2}+(x)\cdot \frac{1}{2}(12-2x{)}^{-1/2}\cdot (-2)$$

$$g^{\prime }(x)=\sqrt{12-2x}-\frac{x}{\sqrt{12-2x}}$$

Step 3: Find Critical Points

Critical points occur where $g^{\prime }(x)=0$ or $g^{\prime }(x)$ is undefined.

• Where is $g^{\prime }(x)$ undefined?

  The derivative is undefined when the denominator is zero: $\sqrt{12-2x}=0$, which means $12-2x=0$, so $x=6$. However, a critical point must be in the interior of the domain. Since $x=6$ is an endpoint of our interval, it is not considered a critical point, but it is still a candidate we must test.

• Where is $g^{\prime }(x)=0$?

  $$\sqrt{12-2x}-\frac{x}{\sqrt{12-2x}}=0$$

  To solve, find a common denominator:

  $$\frac{(\sqrt{12-2x})(\sqrt{12-2x})-x}{\sqrt{12-2x}}=0$$

  $$\frac{(12-2x)-x}{\sqrt{12-2x}}=0$$

  A fraction is zero only when its numerator is zero:

  $$12-3x=0$$

  $$x=4$$

  This critical point, $x=4$, is inside the interval $[0,6]$.

Step 4: Create a Table of Candidates

The candidates are the endpoints ($0,6$) and the critical point ($4$).

Step 5: State the Conclusion

By comparing the values in the table:

• The absolute maximum value is $8$, which occurs at $x=4$.

• The absolute minimum value is $0$, which occurs at $x=0$ and $x=6$.

Using Your Calculator

While the process of finding critical points is analytical, a graphing calculator can be used to verify results and speed up calculations. The justification on an AP Exam free-response question must still come from the calculus (finding the derivative and setting it to zero).

To find the extrema of $f(x)=2x^3-3x^2-12x+1$ on $[-2,3]$:

1. Graph to Visualize:

   • Enter Y_1 = 2X^3 - 3X^2 - 12X + 1.

   • Set the window to $Xmin=-2$, $Xmax=3$. Use $ZoomFit$ (often $ZOOM0$) to adjust the Y-axis. This gives a visual confirmation of where the max/min might be. You can visually see a local max near $x=-1$ and a local min near $x=2$.

2. Find Critical Points Numerically:

   • Enter Y_2 = \text{nDeriv}(Y_1, X, X) or use the d/dx template. This graphs the derivative.

   • Use the $CALC$ menu (2ndTRACE) and select $2:zero$. Find where $Y_2$ crosses the x-axis within your interval. You will find zeros at $x=-1$ and $x=2$.

3. Evaluate Candidates:

   • On the home screen, you can quickly evaluate the function at the candidates.

   • $Y_1(-2)$ [ENTER] -> -3` * $Y_1(-1) [ENTER] -> 8` * $Y_1(2) [ENTER] -> -19` * $Y_1(3) [ENTER] -> -8` * This allows you to quickly build your candidate table after finding the critical points analytically. ## AP Exam Quick Hit ### Common Question Types - **Standard Polynomial/Trigonometric Function:** "Find the absolute maximum value of $f(x) = x - 2\sin(x) on the interval $[0,2\pi ]$." This requires finding the derivative, setting it to zero, solving for $x$, and testing the critical points and endpoints.

• Function with a Cusp: "Determine the absolute minimum value of $h(x)=(x-1{)}^{2/3}+4$ on $[0,9]$." This tests whether you remember to check for critical points where the derivative is undefined (in this case, at $x=1$).

• Graphical or Tabular Data: "The derivative of a function $g$ is given by the graph of $g^{\prime }(x)$ above. If $g$ is continuous on $[0,5]$, at what value of $x$ does $g$ attain its absolute minimum?" You must identify critical points from the x-intercepts of the $g^{\prime }(x)$ graph and then use information about the area under the curve to compare the values at the critical points and endpoints.

Common Mistakes

• Forgetting to Test Endpoints: This is the most frequent error. Students find the critical points, test them, and state a conclusion without ever evaluating the function at the interval's endpoints.

• Ignoring Undefined Derivatives: Students often only set the derivative equal to zero and forget to check for x-values where the derivative is undefined (e.g., denominators of $f^{\prime }(x)$ equal to zero, or points that create zero inside an even root in the denominator).

• Mistaking the Location for the Value: The question might ask for the "absolute maximum value," which is the y-coordinate ($f(c)$), but the student provides the x-coordinate ($c$). Read the question carefully.

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

• Incorrectly Justifying the Answer: On an FRQ, simply stating the answer is not enough. You must show the analytical work (the derivative, the critical points) and often a table of candidates to justify your conclusion. A calculator-found answer without the supporting calculus will not earn full credit.