Book cover for College Algebra

College Algebra

Michael Sullivan

ISBN #9780321979476

10th Edition

5,609 Questions

Group icon
320,854 Students Helped

Homework Questions

Right arrow
Summary
Learning Objectives
Key Concepts
Example Problems
Explanations
Common Mistakes

Summary

This section focuses on functions by establishing that a function is a relation in which every input is associated with exactly one output. Key concepts include identifying the domain and range, performing operations on functions, and evaluating functions using the difference quotient. Understanding these fundamentals sets the stage for more complex topics in calculus and applied mathematics.

Learning Objectives

1

Determine whether a given relation represents a function using graphs, ordered pairs, and mappings.

2

Identify and compute the domain and range of a relation or function defined by an equation.

3

Evaluate a function at a specific input value and compute the difference quotient.

4

Perform operations on functions including sum, difference, product, and quotient, and determine the resulting domain.

Key Concepts

CONCEPT

DEFINITION

Relation

A correspondence between two sets. It pairs elements from one set (inputs) with elements in another set (outputs).

Function

A special type of relation in which each element in the domain is associated with exactly one element in the range.

Domain

The set of all possible input values of a function or relation.

Range

The set of all outputs or images corresponding to the inputs in the domain.

Mapping

A method of representing a relation by drawing arrows from each element in the domain to its corresponding element in the range.

Difference Quotient

An expression of the form [f(x+h) - f(x)]/h (with h ≠ 0) used to define the average rate of change and, in calculus, the derivative of a function.

Explicit vs. Implicit Function

An explicit function is one in which y is expressed directly in terms of x (e.g., y = f(x)), while an implicit function is defined by an equation involving both x and y.

Example Problems

Example 1

The inequality $-1<x<3$ can be written in interval notation as _____ $(\mathrm{pp} \cdot 120-121)$

Example 2

If $x=-2,$ the value of the expression $3 x^{2}-5 x+\frac{1}{x}$ is _____ $(\mathrm{pp} \cdot 20-23)$

Example 3

The domain of the variable in the expression $\frac{x-3}{x+4}$ is _____ $(\mathrm{pp} \cdot 20-23)$

Example 4

Solve the inequality: $3-2 x>5 .$ Graph the solution set. (pp. $123-126)$

Example 5

To rationalize the denominator of $\frac{3}{\sqrt{5}-2},$ multiply the numerator and denominator by _____ $(p .75)$
Scroll left
Scroll right

Step-by-Step Explanations

QUESTION

Given a set of ordered pairs or a mapping diagram, how can you decide if the relation is a function?

STEP-BY-STEP ANSWER:

Step 1: List all the input (first) elements from the ordered pairs or the domain indicated in the mapping.
Step 2: Check to see if any input is paired with more than one output.
Step 3: If each input corresponds to exactly one output, then the relation is a function.
Final Answer: The relation is a function if no input is assigned multiple outputs.

Determining Whether a Relation is a Function

QUESTION

Why does the equation x² + y² = 1 not define y as a function of x?

STEP-BY-STEP ANSWER:

Step 1: Solve the equation for y to see the relationship between x and y.
Step 2: Rearranging, we get y² = 1 - x².
Step 3: Taking the square root of both sides gives y = ±√(1 - x²).
Step 4: Notice that for a given x (for example, x = 0), there are two possible values of y (y = 1 and y = -1).
Final Answer: Because one input x produces two outputs, the equation does not represent a function.

Determining if an Equation Defines a Function

QUESTION

How do you compute the difference quotient for f(x) = 2x² - 3x?

STEP-BY-STEP ANSWER:

Step 1: Write the expressions for f(x + h) and f(x), where f(x) = 2x² - 3x.
Step 2: Compute f(x + h) = 2(x + h)² - 3(x + h) = 2(x² + 2xh + h²) - 3x - 3h.
Step 3: Expand and simplify: f(x + h) = 2x² + 4xh + 2h² - 3x - 3h.
Step 4: Subtract f(x) from f(x + h): (f(x+h) - f(x)) = (2x² + 4xh + 2h² - 3x - 3h) - (2x² - 3x) = 4xh + 2h² - 3h.
Step 5: Factor out h: h(4x + 2h - 3).
Step 6: Divide the entire expression by h (h≠0) to obtain the difference quotient: 4x + 2h - 3.
Final Answer: The difference quotient is (f(x+h) - f(x))/h = 4x + 2h - 3.

Computing the Difference Quotient

Scroll left
Scroll right

Common Mistakes

  • Assuming that if two different inputs produce the same output, the relation is not a function (it is acceptable for different inputs to yield the same output).
  • Failing to exclude values from the domain that would make an expression undefined (e.g., division by zero or invalid radical expressions).
  • Confusing function notation f(x) with multiplication of variables (e.g., misinterpreting f(x) as f multiplied by x).
  • Not properly simplifying the difference quotient, leading to incorrect cancellation of the factor h.