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

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

The fundamental goal of this topic is to find the absolute highest and lowest points (the absolute maximum and minimum values) of a function on a closed, continuous interval. The Extreme Value Theorem guarantees that for any function that is continuous over a closed interval $[a,b]$, an absolute maximum and an absolute minimum value must exist.

The "Candidates Test" provides a systematic procedure to find these values. It is based on the critical concept that these absolute extrema can only occur at two types of locations: the endpoints of the interval ($a$ and $b$) or at critical points within the interval (places where the derivative is zero or undefined). By identifying all these potential locations (the "candidates") and comparing the function's output values at each one, we can definitively determine the absolute highest and lowest values on the entire interval.

Key Theorems and Definitions

The Extreme Value Theorem (EVT)

The Extreme Value Theorem is the foundational guarantee for this process. It states:

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

Critical Point

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

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

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

Absolute extrema on a closed interval can only occur at critical points or at the endpoints of the interval.

Understanding the Conditions

The conditions of the Extreme Value Theorem—that the function must be continuous and the interval must be closed—are essential. If either condition is not met, the existence of an absolute maximum or minimum is not guaranteed.

• Continuity: Consider a function with a removable discontinuity (a hole) or an infinite discontinuity (a vertical asymptote) within an interval. The function may approach a value without ever reaching it, or it may increase or decrease without bound. In such cases, an absolute extremum might not exist.

• Closed Interval: Consider the function $f(x)=x$ on the open interval $(0,5)$. The function values get arbitrarily close to 5 (as $x$ approaches 5) and arbitrarily close to 0 (as $x$ approaches 0), but the function never actually reaches these values. Therefore, it has no absolute maximum or minimum on this open interval. The closed interval $[0,5]$ is necessary to provide the endpoints $x=0$ and $x=5$ as candidates where the extrema can occur.

Core Concepts & Rules

• Guaranteed Existence: The Extreme Value Theorem guarantees that an absolute maximum and an absolute minimum exist for any continuous function on a closed interval.

• Location of Extrema: Absolute extrema on a closed interval $[a,b]$ can only occur at the endpoints ($x=a$ and $x=b$) or at critical points inside the interval $(a,b)$.

• The Candidates Test Procedure: This is the method for finding absolute extrema.

  1. Verify that the function is continuous on the closed interval.

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

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

  4. Discard any critical points that are not within the interval $[a,b]$.

  5. Make a list of candidates, which includes the endpoints ($a$ and $b$) and the valid critical points from the previous step.

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

  7. The largest value from this list is the absolute maximum value, and the smallest value is the absolute minimum value.

Step-by-Step Example 1: Basic Application

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

Step 1: Check Conditions

The function $f(x)$ is a polynomial, which is continuous everywhere. Therefore, it is continuous on the closed interval $[-3,5]$. The Extreme Value Theorem applies.

Step 2: Find the Derivative

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

Step 3: Find Critical Points

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

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

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

The critical points are $x=0$ and $x=4$.

The derivative is a polynomial, so it is never undefined.

Step 4: Identify Candidates

The candidates for absolute extrema are the endpoints and the critical points within the interval.

• Endpoint: $x=-3$

• Endpoint: $x=5$

• Critical Point: $x=0$ (which is in $[-3,5]$)

• Critical Point: $x=4$ (which is in $[-3,5]$)

Our list of candidates is $\{-3,0,4,5\}$.

Step 5: Evaluate the Function at Each Candidate

We test each candidate by plugging it into the original function, $f(x)=x^3-6x^2+5$.

• $f(-3)=(-3{)}^3-6(-3{)}^2+5=-27-6(9)+5=-27-54+5=-76$

• $f(0)=(0{)}^3-6(0{)}^2+5=0-0+5=5$

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

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

Step 6: State the Conclusion

Compare the output values: $\{-76,5,-27,-20\}$.

• The largest value is 5.

• The smallest value is -76.

On the interval $[-3,5]$, the absolute maximum value of $f(x)$ is 5, which occurs at $x=0$. The absolute minimum value is -76, which occurs at $x=-3$.

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

Find the absolute minimum value of $g(t)=2t+\cos (t)$ on the interval $[0,\pi ]$.

Step 1: Check Conditions

The function $g(t)$ is the sum of a polynomial ($2t$) and a trigonometric function ($\cos (t)$), both of which are continuous everywhere. Therefore, $g(t)$ is continuous on the closed interval $[0,\pi ]$. The Extreme Value Theorem applies.

Step 2: Find the Derivative

$$g^{\prime }(t)=2-\sin (t)$$

Step 3: Find Critical Points

Set $g^{\prime }(t)=0$.

$$2-\sin (t)=0$$

$$\sin (t)=2$$

There are no values of $t$ for which $\sin (t)=2$, since the range of $\sin (t)$ is $[-1,1]$. Therefore, there are no critical points where the derivative is zero. The derivative is never undefined.

Step 4: Identify Candidates

Since there are no critical points within the interval, the only candidates for the absolute extrema are the endpoints.

• Endpoint: $t=0$

• Endpoint: $t=\pi$

Our list of candidates is $\{0,\pi \}$.

Step 5: Evaluate the Function at Each Candidate

We test each candidate by plugging it into the original function, $g(t)=2t+\cos (t)$.

• $g(0)=2(0)+\cos (0)=0+1=1$

• $g(\pi )=2(\pi )+\cos (\pi )=2\pi -1\approx 2(3.14)-1=5.28$

Step 6: State the Conclusion

Compare the output values: $\{1,2\pi -1\}$.

• The smallest value is 1.

On the interval $[0,\pi ]$, the absolute minimum value of $g(t)$ is 1.

Using Your Calculator

The Candidates Test is an analytical method, and you must show the steps of finding the derivative and identifying the candidates. A calculator is best used to assist with complex calculations or to verify your work.

1. Finding Difficult Critical Points:

If solving $f^{\prime }(x)=0‘isalgebraicallydifficult,youcanfindthecriticalpointsgraphically.\ast Enterthederivativeinto‘Y1‘.\ast Graph‘Y1‘overthegiveninterval$[a, b]. * Use the $CALC menu (2nd + TRACE) and select 2:zero.

• Use the cursor to set a "Left Bound" and "Right Bound" around an x-intercept and press ENTER for "Guess". The calculator will find the x-value of the critical point. Repeat for all x-intercepts in the interval.

2. Evaluating Candidates:

To quickly and accurately evaluate $f(x)$ at your list of candidates ($a,b,c_1,c_2,...$):

• Enter the original function $f(x)intoY1`.

• Go to the home screen.

• Access Y1 by pressing VARS -> Y-VARS -> 1:Function... -> 1:Y1.

• Evaluate at each candidate by typing Y1(a), Y1(b), Y1(c1)$, etc., and pressing ENTER` each time. This avoids manual calculation errors.

Example: For $f(x) = x^3 - 6x^2 + 5$ on [-3, 5]`, you would enter the function in `Y1` and then compute `Y1(-3)`, `Y1(0)`, `Y1(4)`, and `Y1(5)` on the home screen. ## AP Exam Quick Hit ### Common Question Types - **Standard Polynomial/Trigonometric Function:** "Find the absolute maximum and minimum values of $f(x) = x^4 - 8x^2 on the interval $[-1,3]$." This is a direct, by-the-book application of the Candidates Test.

• Particle Motion (Position Given): "The position of a particle is given by $s(t)=t^3-12t+3$ for $t\ge 0$. Find the particle's maximum distance from the origin on the interval $[0,3]$." This requires finding the absolute maximum and absolute minimum of $s(t)$ on $[0,3]$ and then determining which of these values has the larger magnitude.

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

Common Mistakes

• Forgetting to Test the Endpoints: This is the most frequent error. Students correctly find all the critical points but forget to include the interval's endpoints ($a$ and $b$) in their list of candidates.

• Plugging Candidates into $f^{\prime }(x)$ instead of $f(x)$: After finding the x-values of the candidates, students mistakenly substitute them into the derivative, $f^{\prime }(x)$, to compare values. You must always plug candidates into the original function, $f(x)$, to find the potential maximum/minimum values.

• Stopping at Critical Points: Students find the x-values of the critical points and stop, incorrectly assuming these are the answers. You must evaluate the function at these points and compare them to the values at the endpoints.

• Confusing Local and Absolute Extrema: Students may try to use the First Derivative Test to classify a critical point as a local max or min. While useful for other problems, this is not sufficient for finding absolute extrema on a closed interval. The Candidates Test is the required, definitive method.

• Answering with the Wrong Value: Pay close attention to whether the question asks for the maximum/minimum value (the y-value) or the x-value where it occurs. Providing the x-coordinate when the y-value is requested is a common mistake.