Ch. 2 - Functions and Graphs

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

Blitzer 8th Edition

Ch. 2 - Functions and Graphs

Problem 69

Chapter 3, Problem 69

Find a. (fog) (x) b. the domain of f o g. f(x) = x/(x+1), g(x) = 4/x

Verified step by step guidance

First, recall that the composition of functions (fog)(x) means f(g(x)). So, we need to substitute g(x) into the function f.

Given f(x) = \(\frac{x}{x+1}\) and g(x) = \(\frac{4}{x}\), substitute g(x) into f(x) to get (fog)(x) = f\(\left\)(\(\frac{4}{x}\)\(\right\)) = \(\frac{\frac{4}{x}\)}{\(\frac{4}{x}\) + 1}.

Simplify the expression for (fog)(x) by combining the terms in the denominator: \(\frac{4}{x}\) + 1 = \(\frac{4}{x}\) + \(\frac{x}{x}\) = \(\frac{4 + x}{x}\). Then rewrite (fog)(x) as \(\frac{\frac{4}{x}\)}{\(\frac{4 + x}{x}\)}.

To simplify the complex fraction, multiply numerator and denominator by x to eliminate the denominators inside the fraction: (fog)(x) = \(\frac{4}{4 + x}\).

For the domain of (fog), consider the domain restrictions from both f and g. First, g(x) = \(\frac{4}{x}\) requires x \(\neq\) 0. Next, f(g(x)) requires the denominator of f(g(x)) to be nonzero, so 4 + x \(\neq\) 0, which means x \(\neq\) -4. Therefore, the domain of (fog) is all real numbers except x = 0 and x = -4.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

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

Function Composition

Function composition involves applying one function to the result of another, denoted as (f ∘ g)(x) = f(g(x)). It requires substituting the entire expression of g(x) into f(x), which helps in combining two functions into a single expression.

Domain of a Function

The domain of a function is the set of all input values (x) for which the function is defined. When composing functions, the domain of (f ∘ g) includes all x-values in the domain of g such that g(x) is in the domain of f.

Restrictions from Denominators

Functions with variables in denominators have restrictions where the denominator cannot be zero. Identifying these values is crucial to determine the domain, especially when composing functions, to avoid division by zero.

Recommended video:

Guided course

02:58

Rationalizing Denominators

Related Practice

Textbook Question

Graph both equations in the same rectangular coordinate system and find all points of intersection. Then show that these ordered pairs satisfy the equations. x² + y² = 16, x-y = 4

views

Textbook Question

Use intercepts to graph each equation. 6x-9y-18 = 0

Textbook Question

In Exercises 65–70, use the graph of f to find each indicated function value. f(-3)

views

Textbook Question

Begin by graphing the square root function, f(x) = √x. Then use transformations of this graph to graph the given function. g(x) = √x + 1

Textbook Question

Graph both equations in the same rectangular coordinate system and find all points of intersection. Then show that these ordered pairs satisfy the equations. (x − 2)²+(y+3)² = 4, y = x - 3

views

Textbook Question

In Exercises 67–69, begin by graphing the absolute value function, f(x) = |x|. Then use transformations of this graph to graph the given function. r(x) = (1/2) |x + 2|