Ch. 2 - Functions and Graphs

Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook

Blitzer 8th Edition

Ch. 2 - Functions and Graphs

Problem 38a

Chapter 3, Problem 38a

In Exercises 27–38, evaluate each function at the given values of the independent variable and simplify. f(x) = |x+3|/|x + 3| a. f(5)

Verified step by step guidance

Step 1: Understand the function f(x) = |x + 3| / |x + 3|. This function involves absolute values in both the numerator and denominator. The absolute value of a number is its distance from zero, so it is always non-negative.

Step 2: Substitute the given value of x = 5 into the function. This means replacing x with 5 in both the numerator and denominator: f(5) = |5 + 3| / |5 + 3|.

Step 3: Simplify the expression inside the absolute value symbols. Calculate 5 + 3, which equals 8. So the function becomes f(5) = |8| / |8|.

Step 4: Evaluate the absolute values. Since 8 is already positive, |8| = 8. Therefore, the function simplifies to f(5) = 8 / 8.

Step 5: Simplify the fraction. Divide the numerator by the denominator: 8 / 8 simplifies to 1. Thus, the value of f(5) is 1.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Absolute Value Function

The absolute value function, denoted as |x|, represents the distance of a number x from zero on the number line, always yielding a non-negative result. For example, |5| equals 5, while |-5| also equals 5. This function is crucial in evaluating expressions that involve both positive and negative values.

Function Evaluation

Function evaluation involves substituting a specific value for the independent variable in a function to determine the corresponding output. In this case, evaluating f(5) means replacing x with 5 in the function f(x) = |x + 3| / |x + 3|, which simplifies the process of finding the function's value at that point.

Simplification of Rational Expressions

Simplification of rational expressions involves reducing fractions to their simplest form by canceling common factors in the numerator and denominator. In the given function, |x + 3| appears in both the numerator and denominator, which can lead to simplification, provided that the denominator is not zero, ensuring the expression remains valid.

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Related Practice

Textbook Question

In Exercises 27–38, evaluate each function at the given values of the independent variable and simplify. f(x) = |x+3|/|x + 3| b. f(-5)

Textbook Question

Find f/g and determine the domain for each function. f(x) = 3 − x², g(x) = x² + 2x − 15

Textbook Question

In Exercises 37–40, find the slope of the line passing through each pair of points or state that the slope is undefined. Then indicate whether the line through the points rises, falls, is horizontal, or is vertical. (-1, -2) and (-3, -4)

Textbook Question

Evaluate each function at the given values of the independent variable and simplify. f(x) = |x+3|/(x + 3) c. f(−9 - x)

Textbook Question

Graph the given functions, f and g, in the same rectangular coordinate system. Select integers for x, starting with -2 and ending with 2. Once you have obtained your graphs, describe how the graph of g is related to the graph of f. f(x) = x, g(x) = x + 3

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Textbook Question

Find $f+g$ , $f-g$ , $fg$ , and $\frac{f}{g}$ . Determine the domain for each function.

$f$\left$(x$\right$)=5-x^2$ , $g$\left$(x$\right$)=x^2+4x-12$