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Chapter 2: Problem 29

Find the functions \(f \circ g, g \circ f, f \circ f,\) and \(g \circ g\) and their domains. $$f(x)=2 x+3, \quad g(x)=4 x-1$$

Short Answer

Expert verified
\((f \circ g)(x) = 8x + 1; (g \circ f)(x) = 8x + 11; (f \circ f)(x) = 4x + 9; (g \circ g)(x) = 16x - 5\), all with domain \((-\infty, \infty)\).

Step by step solution

01

Understanding the Functions

First, let's understand the given functions:\[ f(x) = 2x + 3 \] and \[ g(x) = 4x - 1 \] These functions are linear, meaning they have no restrictions on their domains. Both have domain \( (-\infty, \infty) \).
02

Calculating \( f \circ g \)

The function \( f \circ g \) means applying \( g(x) \) first and then applying \( f \) to that result. We calculate it as follows:\[ f(g(x)) = f(4x - 1) = 2(4x - 1) + 3 \]Simplifying:\[ = 8x - 2 + 3 = 8x + 1 \]Hence, \( f \circ g = 8x + 1 \).
03

Calculating \( g \circ f \)

The function \( g \circ f \) applies \( f(x) \) first and then \( g \) to that result:\[ g(f(x)) = g(2x + 3) = 4(2x + 3) - 1 \]Simplifying:\[ = 8x + 12 - 1 = 8x + 11 \]Thus, \( g \circ f = 8x + 11 \).
04

Calculating \( f \circ f \)

The function \( f \circ f \) applies \( f(x) \) to itself:\[ f(f(x)) = f(2x + 3) = 2(2x + 3) + 3 \]Simplifying:\[ = 4x + 6 + 3 = 4x + 9 \]Therefore, \( f \circ f = 4x + 9 \).
05

Calculating \( g \circ g \)

The function \( g \circ g \) applies \( g(x) \) to itself:\[ g(g(x)) = g(4x - 1) = 4(4x - 1) - 1 \]Simplifying:\[ = 16x - 4 - 1 = 16x - 5 \]Hence, \( g \circ g = 16x - 5 \).
06

Determining the Domains

Because both \( f(x) \) and \( g(x) \) are linear functions, there are no restrictions on the composed functions' domains. Thus, the domains for \( f \circ g \), \( g \circ f \), \( f \circ f \), and \( g \circ g \) are all \( (-\infty, \infty) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Functions
Linear functions are among the simplest and most essential types of functions in mathematics. A linear function can be expressed in the form \( f(x) = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. This means that the graph of a linear function is a straight line. Here, the given functions are \( f(x) = 2x + 3 \) and \( g(x) = 4x - 1 \). Both functions are linear:
  • The slope of \( f(x) \) is 2, and it crosses the y-axis at 3.
  • The slope of \( g(x) \) is 4, and it crosses the y-axis at -1.
Linear functions are straightforward because they increase or decrease steadily. They provide useful models for many real-world situations where a constant rate of change is involved. For instance, speed, rates of payment, and the relationship between distance and time can often be modeled using linear functions. Moreover, their simplicity makes them useful for exploring more complex mathematical concepts, such as function composition.
Domain of Functions
The domain of a function is crucial because it defines all possible input values (or x-values) for which the function is defined. For both linear functions we have, their domain is all real numbers, expressed as \((-\infty, \infty)\). This means that you can plug any real number into \( f(x) = 2x + 3 \) or \( g(x) = 4x - 1 \) and receive a valid output, since linear functions have no restrictions like square roots or denominators that could cause problems. When working with compositions of functions, it's essential to consider the domain carefully. Though linear functions generally have unrestricted domains, their compositions might not be so straightforward if involving non-linear functions or real-world scenarios where certain values might be excluded. Here, because the functions involved and their compositions are all linear, their domains remain \((-\infty, \infty)\). However, always verify conditions in different context compositions.
Mathematical Functions
Mathematical functions describe a relationship where each input is related to exactly one output. This concept is fundamental in mathematics and is the foundation for understanding function composition. Function composition refers to the process of combining two functions to create a new function.In this exercise, the compositions \( f \circ g, g \circ f, f \circ f, \) and \( g \circ g \) were explored:
  • \( f \circ g(x) \) means you first apply \( g \), then \( f \).
  • \( g \circ f(x) \) means you first apply \( f \), then \( g \).
  • \( f \circ f(x) \) involves applying \( f \) to itself.
  • \( g \circ g(x) \) involves applying \( g \) to itself.
Each composition results in another linear function, which shows the robustness of linear functions under composition in maintaining their linear nature. Function composition is vital in understanding behavior and complex relationships in mathematical modeling and analysis.

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Most popular questions from this chapter

Determine whether the function \(f\) is even, odd, or neither. If \(f\) is even or odd, use symmetry to sketch its graph. $$f(x)=1-\sqrt[3]{x}$$

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If \(f(x)=\sqrt{2 x-x^{2}},\) graph the following functions in the viewing rectangle \([-5,5]\) by \([-4,4] .\) How is each graph related to the graph in part (a)? (a) \(y=f(x)\) (b) \(y=f(-x)\) (c) \(y=-f(-x)\) (d) \(y=f(-2 x)\) (e) \(y=f\left(-\frac{1}{2} x\right)\)

Changing Temperature Scales The temperature on a certain afternoon is modeled by the function $$C(t)=\frac{1}{2} t^{2}+2$$ where \(t\) represents hours after 12 noon \((0 \leq t \leq 6),\) and \(C\) is measured in "C. (a) What shifting and shrinking operations must be performed on the function \(y=t^{2}\) to obtain the function \(y=C(t) ?\) (b) Suppose you want to measure the temperature in "F instead. What transformation would you have to apply to the function \(y=C(t)\) to accomplish this? (Use the fact that the relationship between Celsius and Fahrenheit degrees is given by \(F=\frac{9}{5} C+32 .\) ) Write the new function \(y=F(t)\) that results from this transformation.
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