Book cover for Biocalculus Calculus for the Life Sciences

Biocalculus Calculus for the Life Sciences

James Stewart

ISBN #9781133109631

1st Edition

2,565 Questions

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211,110 Students Helped

Homework Questions

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Summary
Learning Objectives
Key Concepts
Example Problems
Explanations
Common Mistakes

Summary

This section focuses on using derivatives to analyze and optimize functions by identifying extreme values. It introduces the definitions of absolute and local extrema, demonstrates how critical numbers are used in such analysis, and explains key theorems: the Extreme Value Theorem and Fermat’s Theorem. In addition, the Closed Interval Method is presented as a systematic approach for evaluating extrema on closed intervals, with real-world applications ranging from medical diagnostics to population dynamics.

Learning Objectives

1

Describe and distinguish between absolute (global) and local maximum and minimum values of a function.

2

Explain and apply the Extreme Value Theorem and Fermat’s Theorem in identifying extreme values.

3

Identify and compute critical numbers by finding where the derivative is zero or undefined.

4

Apply the Closed Interval Method to determine absolute extreme values on a closed interval.

5

Analyze real-world problems in biology and medicine using optimization techniques from differential calculus.

Key Concepts

CONCEPT

DEFINITION

Absolute Maximum/Minimum

An absolute maximum (or minimum) of a function f on a domain D is a value f(c) such that for all x in D, f(c) is greater (or less) than every other value of f(x). These are also known as global extrema.

Local Maximum/Minimum

A local maximum (or minimum) occurs at a point c if f(c) is the greatest (or least) value in some open interval containing c, even if it is not the overall extreme value for the entire domain.

Critical Number

A critical number of a function is any number c in its domain for which either f′(c) = 0 or f′(c) does not exist.

Extreme Value Theorem

This theorem states that if a function f is continuous on a closed interval [a, b], then f attains both an absolute maximum and an absolute minimum on that interval.

Fermat’s Theorem

Fermat’s Theorem declares that if f has a local maximum or minimum at c, and if the derivative f′(c) exists, then f′(c) must equal 0. However, the converse is not necessarily true.

Closed Interval Method

A systematic process to find absolute extrema of a continuous function on a closed interval by evaluating the function at its critical numbers within the interval and at the endpoints, and then comparing these values.

Example Problems

Example 1

Explain the difference between an absolute minimum and a local minimum.

Example 2

Suppose $f$ is a continuous function defined on a closed interval $[a, b] .$ (a) What theorem guarantees the existence of an absolute maximum value and an absolute minimum value for $f ?$ (b) What steps would you take to find those maximum and minimum values?

Example 3

$3-4$ For each of the numbers $a, b, c, d, r,$ and $s,$ state whether the function whose graph is shown has an absolute maximum or minimum, a local maximum or minimum, or neither a maximum nor a minimum.

Example 4

3-4 For each of the numbers a, b, c, d, r, and s, state whether the function whose graph is shown has an absolute maximum or minimum, a local maximum or minimum, or neither a maximum nor a minimum.

Example 5

Electrocardiogram A cardiologist looking at the rhythm strip shown might suspect right atrial hypertrophy because of the relatively tall peaked wave at $P$ (compare with Figure 6 ). State the local and absolute maximum and minimum values of the electric potential function $f(t)$ .
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Step-by-Step Explanations

QUESTION

Find the critical numbers of the function f(x) = x³ - 3x.

STEP-BY-STEP ANSWER:

Step 1: Differentiate f(x) to obtain f′(x). For f(x) = x³ - 3x, f′(x) = 3x² - 3.
Step 2: Set the derivative equal to zero: 3x² - 3 = 0.
Step 3: Divide both sides by 3 to get x² - 1 = 0.
Step 4: Solve for x: x² = 1, so x = 1 or x = -1.
Final Answer: The critical numbers are x = -1 and x = 1.

Critical Numbers

QUESTION

Find the absolute maximum and minimum values of f(x) = x² - 4x + 3 on the closed interval [0, 5].

STEP-BY-STEP ANSWER:

Step 1: Differentiate f(x) to obtain f′(x). f(x) = x² - 4x + 3 leads to f′(x) = 2x - 4.
Step 2: Set the derivative equal to zero: 2x - 4 = 0, which gives x = 2. Verify that x = 2 lies in [0, 5].
Step 3: Evaluate f(x) at the critical number and endpoints:
Step 4: Compare the computed values.
Final Answer: The absolute minimum value is -1 at x = 2 and the absolute maximum value is 8 at x = 5.

Closed Interval Method

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Common Mistakes

  • Assuming that every point where the derivative is zero is a local maximum or minimum, while some such points may be inflection points (e.g., f(x) = x³ at x = 0).
  • Neglecting to evaluate the function at the endpoints when using the Closed Interval Method, which can lead to missing the absolute maximum or minimum.
  • Confusing local extreme values with absolute extreme values, especially in functions that have multiple extreme points.
  • Overlooking cases where the derivative does not exist, yet the function still attains a local or absolute extreme value.