Using the Candidates Test to Determine Absolute (Global) Extrema: AP Calculus Study Guide
Introduction
Howdy, math wizards! 🧙♂️ Get ready to dive into the magical world of absolute (or global) extrema. This guide will teach you how to find the highest highs and lowest lows of a function using the legendary Candidates Test. So grab your calculus wand and let’s uncover those elusive maximums and minimums!
The Candidates Test: Your Calculus Detective Kit
To determine the absolute maximum and minimum of a function over a closed interval, we dust off our detective hats and use the Candidates Test. Absolute extrema are the highest (maximum) and lowest (minimum) values of a function across its entire domain. Unlike finding a local extremum (which is like finding the best pizza place in your neighborhood), finding absolute extrema is like uncovering the best pizza place in the entire city! 🍕
So, how do we embark on this quest?
The Quest for Absolute Extrema
Here are the steps for the Candidates Test, offered to you straight from the calculus playbook:
 Find the Critical Points: These are the points where the first derivative (f') of the function equals zero or is undefined. Think of these as the potential contenders for our extrema title. 🥇
 Evaluate the Critical Points: Plug these points back into the original function to get the corresponding yvalues. These are the contestant scores, so to speak.
 Evaluate the Endpoints: Don't forget the endpoints of your interval – they too are vying for the title of absolute extrema.
 Compare the Values: Now, it’s the climax! Compare all the yvalues found. The highest score is our absolute maximum, and the lowest score is our absolute minimum. 🎉
The Candidates Test is a superhero in scenarios where functions are complex or tricky to handle with traditional algebraic methods. Consider it your utility belt full of gadgets when battling tough calculus problems.
Candidates Test Walkthrough: A Live Demonstration
Imagine we have a function h(x) = x³ + 3x² + 16, and we need to find the absolute maximum on the interval [4, 0]. Let’s go Sherlock Holmes on this!
Step 1: Find Critical Points We start by taking the derivative of h(x): [ h'(x) = 3x^2 + 6x ]
Set the derivative to zero to find the critical points: [ 0 = 3x^2 + 6x ] [ 0 = 3x(x + 2) ]
So, we have critical points at x = 2 and x = 0. These are the candidates, folks!
Step 2: Evaluate Critical Points Plug the critical points back into the original function: [ h(2) = (2)^3 + 3(2)^2 + 16 = 8 + 12 + 16 = 20 ] [ h(0) = 0^3 + 3(0)^2 + 16 = 16 ]
Step 3: Evaluate Endpoints Now, look at the endpoints: [ h(4) = (4)^3 + 3(4)^2 + 16 = 64 + 48 + 16 = 0 ] (Since h(0) was already evaluated, no need to do it again – thank you, math efficiency!)
Step 4: Compare All Values We’ve got these values: [ h(2) = 20 ] [ h(0) = 16 ] [ h(4) = 0 ]
Clearly, h(2) = 20 is the absolute maximum value over the interval [4, 0]. 🎈
Candidates Test Practice Problems
Ready to prove your calculus prowess? Here are a couple of practice problems for you!
Candidates Test Problem 1:
Let ( f(x) = x^4 + 2x^2 ). Find the absolute maximum on the interval [0, 2].
Candidates Test Problem 2:
Let ( g(x) = x^5 + 5x ). Find the absolute minimum on the interval [2, 1].
Solutions to Practice Problems:
Problem 1 Solution:

Find Critical Points: [ f'(x) = 4x^3 + 4x ] [ 0 = 4x(x^2  1) ] [ 0 = 4x(x  1)(x + 1) ] So, critical points are x = 1, 0, 1. Only x = 0, 1 lie within [0, 2].

Evaluate Critical Points: [ f(0) = 0 ] [ f(1) = 1 ]

Evaluate End Points: [ f(2) = 8 ]

Compare Values: [ f(0) = 0, ] [ f(1) = 1, ] [ f(2) = 8 ]
Absolute maximum is ( f(1) = 1 ).
Problem 2 Solution:

Find Critical Points: [ g'(x) = 5x^4 + 5 ] [ 0 = 5(x^4  1) ] [ 0 = 5(x^2 + 1)(x  1)(x + 1) ] Critical points are x = ±1 which lie within the interval [2, 1].

Evaluate Critical Points: [ g(1) = 4 ] [ g(1) = 4 ]

Evaluate End Points: [ g(2) = 22 ]

Compare Values: [ g(1) = 4, ] [ g(1) = 4, ] [ g(2) = 22 ]
Absolute minimum is ( g(1) = 4 ).
Conclusion
Congratulations, you’ve mastered the Candidates Test for finding absolute extrema! 🎓 Use these steps as part of your math toolkit, and you'll be ready to tackle those AP Calculus questions like a boss. Remember, math is a journey full of peaks and valleys, and now you know exactly how to find them! Happy calculusing! 🏔📉
Key Terms to Review:
 Absolute Extrema: The highest and lowest values a function can reach over a given interval.
 Candidates Test: A method to determine function behavior at critical points and endpoints.
 First Derivative Test: Identifies where a function is increasing or decreasing, and finds local extrema points.
 Points of Inflection: Spots on a graph where the curve changes concavity.
 Second Derivative Test: Determines whether critical points are local maxima, minima, or neither by checking concavity.
Alright math geniuses, you’ve got this! 💪🔢