Using the Candidates Test to Determine Absolute (Global) Extrema - AP Calculus BC Study Guide

The Core Idea: Using the Candidates Test to Determine Absolute (Global) Extrema

The fundamental challenge in finding the absolute highest and lowest points of a function is that, over its entire domain, such points may not even exist. However, the Extreme Value Theorem (EVT) provides a powerful guarantee: if a function is continuous over a closed, bounded interval, it is guaranteed to achieve both an absolute maximum and an absolute minimum value within that interval. This theorem provides the "why" we can be certain a solution exists.

The next logical question is where these extreme values can possibly occur. For a continuous function on a closed interval, the absolute maximum and minimum values can only be located at two types of places: the endpoints of the interval or at any critical points that lie within the interval. The Candidates Test is the direct, procedural application of this knowledge. It is not a test for if an extremum exists, but rather a systematic method to find the guaranteed extrema by identifying all possible locations (the "candidates"), evaluating the function at each of these locations, and simply comparing the resulting values to find the largest and smallest.

Key Theorems and Definitions

The Extreme Value Theorem (EVT)

This theorem provides the foundational guarantee for this topic.

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

Conditions:

1. The function $f$ must be continuous on $[a,b]$.

2. The interval $[a,b]$ must be closed.

If either of these conditions is not met, the guarantee does not apply, and an absolute extremum may or may not exist.

Critical Point

A critical point of a function $f$ is a point $c$ in the interior of the domain of $f$ at which either $f^{\prime }(c)=0$ or $f^{\prime }(c)$ is undefined. These are the only interior points where a function can have a local extremum.

The Candidates Test

This is the procedure used to find absolute extrema of a continuous function on a closed interval. The "candidates" are the $x$-values where an absolute extremum could possibly occur.

The set of candidates for absolute extrema of a function $f$ on an interval $[a,b]$ consists of:

1. The endpoints of the interval: $x=a$ and $x=b$.

2. All critical points $x=c$ such that $a<c<b$.

Understanding the Conditions and the Process

The power of the Candidates Test lies in its simplicity, but that simplicity is built upon the strict conditions of the Extreme Value Theorem. It is crucial to understand why these conditions are necessary. A function that is not continuous might have a "hole" or a vertical asymptote where a maximum or minimum value would have been, preventing it from ever being reached. Similarly, a function on an open interval, like $(0,1)$, might approach a maximum or minimum value as $x$ gets closer and closer to an endpoint, but it never actually reaches it because the endpoints are not included in the interval.

The Candidates Test works by narrowing down an infinite number of points in an interval to a finite list of potential locations for extrema. Because we know from the EVT that an absolute maximum and minimum must exist (given the conditions are met), and we know they can only occur at endpoints or critical points, our task is reduced from an infinite search to a simple comparison. By evaluating the function at every candidate, we create a complete list of all possible absolute extreme values. The largest value on that list must be the absolute maximum, and the smallest must be the absolute minimum. This process is not just a calculation; it is a logical argument and serves as the required justification on the AP Exam for claims about absolute extrema on a closed interval.

Core Concepts & Rules

• The EVT Guarantee: A function $f$ that is continuous on a closed interval $[a,b]$ is guaranteed to have an absolute maximum and an absolute minimum value on that interval. Always verify these two conditions before proceeding.

• Candidate Locations: The absolute extrema of $f$ on $[a,b]$ can only occur at the endpoints ($x=a$, $x=b$) or at critical points ($f^{\prime }(x)=0$ or $f^{\prime }(x)$ is undefined) located inside the interval $(a,b)$.

• The Candidates Test Procedure: To find the absolute extrema of a continuous function $f$ on a closed interval $[a,b]$, follow these steps:

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

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

  3. Create a list of candidates, which includes the endpoints $a$ and $b$, along with any critical points that fall within the interval $(a,b)$. Discard any critical points outside the interval.

  4. Evaluate the original function $f(x)$ at every candidate $x$-value.

  5. Compare the output values: the largest is the absolute maximum value, and the smallest is the absolute minimum value.

Step-by-Step Example 1: Polynomial on a Closed Interval

Problem: Find the absolute maximum and minimum values of the function $f(x)=2x^3-3x^2-12x+1$ on the interval $[-2,3]$.

Step 1: Verify Conditions of the EVT

The function $f(x)$ is a polynomial, which is continuous everywhere. Therefore, it is continuous on the closed interval $[-2,3]$. The Extreme Value Theorem guarantees that an absolute maximum and an absolute minimum exist on this interval.

Step 2: Find Critical Points

First, find the derivative of $f(x)$.

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

Next, find where $f^{\prime }(x)=0$ or is undefined. Since $f^{\prime }(x)$ is a polynomial, 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 $x=-1$.

Step 3: Identify Candidates

The candidates for the absolute extrema are the endpoints and any critical points that lie within the interval $[-2,3]$.

• Endpoint: $x=-2$

• Endpoint: $x=3$

• Critical Point: $x=-1$ (which is in $[-2,3]$)

• Critical Point: $x=2$ (which is in $[-2,3]$)

Our list of candidates is $\{-2,-1,2,3\}$.

Step 4: Evaluate the Function at Each Candidate

Now, we evaluate the original function $f(x)$ at each candidate value.

• $f(-2)=2(-2{)}^3-3(-2{)}^2-12(-2)+1=2(-8)-3(4)+24+1=-16-12+24+1=-3$

• $f(-1)=2(-1{)}^3-3(-1{)}^2-12(-1)+1=2(-1)-3(1)+12+1=-2-3+12+1=8$

• $f(2)=2(2{)}^3-3(2{)}^2-12(2)+1=2(8)-3(4)-24+1=16-12-24+1=-19$

• $f(3)=2(3{)}^3-3(3{)}^2-12(3)+1=2(27)-3(9)-36+1=54-27-36+1=-8$

Step 5: State the Conclusion

By comparing the output values ($-3,8,-19,-8$), we can identify the absolute extrema.

• 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: Let $g(x)=x-2\arctan (x)$. Find the absolute minimum value of $g$ on the interval $[0,\sqrt{3}]$.

Step 1: Verify Conditions of the EVT

The function $g(x)$ is a combination of $x$ (continuous everywhere) and $\arctan (x)$ (continuous everywhere). Therefore, $g(x)$ is continuous on the closed interval $[0,\sqrt{3}]$. The EVT applies.

Step 2: Find Critical Points

Find the derivative of $g(x)$. Recall that $\frac{d}{dx}(\arctan (x))=\frac{1}{1+x^2}$.

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

The derivative $g^{\prime }(x)$ is defined for all real numbers since the denominator $1+x^2$ is never zero. So, we only need to find where $g^{\prime }(x)=0$.

$$1-\frac{2}{1+x^2}=0$$

$$1=\frac{2}{1+x^2}$$

$$1+x^2=2$$

$$x^2=1$$

$$x=\pm 1$$

Step 3: Identify Candidates

The candidates are the endpoints and the critical points inside the interval $[0,\sqrt{3}]$.

• Endpoint: $x=0$

• Endpoint: $x=\sqrt{3}$

• Critical Point: $x=1$ (which is in $[0,\sqrt{3}]$)

• Critical Point: $x=-1$ (which is not in $[0,\sqrt{3}]$, so we discard it)

Our list of candidates is $\{0,1,\sqrt{3}\}$.

Step 4: Evaluate the Function at Each Candidate

Evaluate $g(x)=x-2\arctan (x)$ at each candidate.

• $g(0)=0-2\arctan (0)=0-2(0)=0$

• $g(1)=1-2\arctan (1)=1-2(\frac{\pi }{4})=1-\frac{\pi }{2}$

• $g(\sqrt{3})=\sqrt{3}-2\arctan (\sqrt{3})=\sqrt{3}-2(\frac{\pi }{3})=\sqrt{3}-\frac{2\pi }{3}$

Step 5: State the Conclusion

We must compare the values $0$, $1-\frac{\pi }{2}$, and $\sqrt{3}-\frac{2\pi }{3}$.

Using approximations: $\pi \approx 3.14$ and $\sqrt{3}\approx 1.73$.

• $g(0)=0$

• $g(1)\approx 1-\frac{3.14}{2}=1-1.57=-0.57$

• $g(\sqrt{3})\approx 1.73-\frac{2(3.14)}{3}=1.73-\frac{6.28}{3}\approx 1.73-2.09=-0.36$

Comparing the values, the smallest is $1-\frac{\pi }{2}$.

• The absolute minimum value is $1-\frac{\pi }{2}$, which occurs at $x=1$.

Using Your Calculator

The Candidates Test is an analytical procedure, and your written justification must show the full process. A calculator is a powerful tool for assistance and verification, not for the primary solution.

1. Finding Critical Points:

• Graph the derivative, $Y_1=f^{\prime }(x)$.

• Use the 2nd -> $CALC$ -> 2:zero function to find the $x$-intercepts of the derivative's graph. These are the critical points where $f^{\prime }(x)=0$.

• Visually inspect the graph of $f^{\prime }(x)$ for any discontinuities, which would indicate a critical point where $f^{\prime }(x)$ is undefined.

2. Verifying Conditions:

• Graph the original function, $Y_1=f(x)$, on the given interval $[a,b]$.

• Visually inspect the graph to confirm it is continuous (no jumps, holes, or vertical asymptotes).

3. Evaluating Candidates:

• This is the most common and effective use of the calculator for this topic.

• After you have analytically determined your list of candidates ($a,b,c_1,c_2,...$), you can quickly evaluate $f(x)$ at these points on the home screen.

• For example, to evaluate $f(2)$ for $f(x)=2x^3-3x^2-12x+1$, you can enter $2(2{)}^3-3(2{)}^2-12(2)+1$.

• Alternatively, enter the function into $Y_1$ and use the table (2nd -> GRAPH) or evaluate directly from the home screen by pressing VARS -> Y-VARS -> 1:Function... -> 1:Y1 and then entering (candidate_value). For example: Y1(2). This is highly efficient for evaluating multiple candidates.

AP Exam Quick Hit

Common Question Types

• Standard Analytical Problem (MCQ or FRQ): Given a function $f(x)$ and a closed interval $[a,b]$, find the absolute maximum or minimum value. You must show the full Candidates Test for justification on an FRQ.

  • Example: "Find the absolute maximum value of $f(x)=x\sqrt{4-x^2}$ on $[0,2]$."

• Problem from a Derivative Graph (FRQ): You are given the graph of $f^{\prime }(x)$ and an initial condition (e.g., $f(0)=5$). You are asked to find the absolute minimum value of $f(x)$ on a given closed interval.

  • Example: "The graph of $f^{\prime }(x)$ is shown on $[-3,4]$. If $f(0)=1$, what is the absolute minimum value of $f(x)$ on $[-3,4]$?"

  • Solution approach: The candidates are the endpoints ($-3,4$) and the $x$-intercepts of the $f^{\prime }(x)$ graph. You must use the Fundamental Theorem of Calculus and areas under the $f^{\prime }(x)$ curve to find the value of $f(x)$ at each candidate.

• Problem from a Derivative Table (FRQ): You are given a table of values for $f^{\prime }(x)$ and are told $f^{\prime }(x)$ is continuous. You may be asked to find the absolute extrema of $f(x)$.

  • Example: "A table of values for $g^{\prime }(x)$ is given. If $g(1)=3$, find the absolute maximum value of $g(x)$ on $[1,5]$."

  • Solution approach: The candidates are the endpoints and any values of $x$ in the table where $g^{\prime }(x)=0$. You would use methods like a Riemann sum or Trapezoidal Rule to approximate the change in $g(x)$ to find values at the other candidates.

Common Mistakes

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

• Considering Critical Points Outside the Interval: A function may have many critical points, but only those that lie within the specified interval are candidates for the absolute extrema on that interval. Always check if your critical points are in $(a,b)$.

• Evaluating the Derivative Instead of the Function: After creating the list of candidates, students sometimes mistakenly plug these $x$-values into $f^{\prime }(x)$ instead of the original function $f(x)$. Remember, you are looking for the highest/lowest function value, not the highest/lowest slope.

• Stopping at the $x$-value: The question often asks for the absolute maximum/minimum value (the $y$-coordinate), not the location where it occurs (the $x$-coordinate). Make sure to answer the question that is asked.

• Incorrect Justification: On an FRQ, simply stating that a point is a maximum because $f^{\prime }(x)$ changes from positive to negative is a justification for a local maximum. To justify an absolute maximum on a closed interval, you must use the Candidates Test and show the evaluation of the function at all candidates.