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Chapter 1: Problem 43

Find formulas for \(f \circ g\) and \(g \circ f,\) and state the domains of the functions. $$f(x)=\frac{1+x}{1-x}, g(x)=\frac{x}{1-x}$$

Short Answer

Expert verified
\(f \circ g(x) = \frac{1}{1-2x}, x \neq 1, \frac{1}{2}\); \(g \circ f(x) = -1, x \neq 1\).

Step by step solution

01

Understanding Composition

When asked to find the composition of two functions, recall that the notation \((f \circ g)(x)\) means \(f(g(x))\). Similarly, \((g \circ f)(x)\) means \(g(f(x))\). The composition is found by substituting one function into the other.
02

Finding \(f \circ g\)

To find \(f \circ g\), substitute \(g(x) = \frac{x}{1-x}\) into \(f(x) = \frac{1+x}{1-x}\). Thus, \(f(g(x)) = f\left(\frac{x}{1-x}\right) = \frac{1 + \frac{x}{1-x}}{1 - \frac{x}{1-x}}\). Simplify this expression to get \(f(g(x)) = \frac{1}{1-2x}\).
03

Determining the Domain of \(f \circ g\)

The domain of \(f(g(x)) = \frac{1}{1-2x}\) requires both \(g(x)\) to exist and \(f(g(x))\) to be defined. \(g(x) = \frac{x}{1-x}\) is undefined for \(x=1\) and \(f\left(\frac{x}{1-x}\right) = \frac{1}{1-2x}\) is undefined for \(x = \frac{1}{2}\). This results in the domain \(x \in \mathbb{R}, x eq 1, x eq \frac{1}{2}\).
04

Finding \(g \circ f\)

For \(g \circ f\), substitute \(f(x) = \frac{1+x}{1-x}\) into \(g(x) = \frac{x}{1-x}\). So \(g(f(x)) = g\left(\frac{1+x}{1-x}\right) = \frac{\frac{1+x}{1-x}}{1 - \frac{1+x}{1-x}}\). Simplify this to get \(g(f(x)) = -1\).
05

Determining the Domain of \(g \circ f\)

For \(g(f(x)) = -1\), check when \(f(x)\) is undefined: \(f(x) = \frac{1+x}{1-x}\) is undefined for \(x = 1\). Thus, the domain of \(g(f(x)) = -1\) is \(x \in \mathbb{R}, x eq 1\).

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Key Concepts

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

Domain of a Function
The domain of a function refers to the set of all possible inputs (or x-values) that the function can accept without resulting in any undefined expressions. For any given function, it's crucial to identify scenarios where calculations might go wrong, like division by zero or taking the square root of a negative number.
In the exercise we have two compositions, and their domains are influenced by the nature of the functions involved:
  • For the composition \(f \circ g(x)\), \(g(x) = \frac{x}{1-x}\) becomes undefined when \(x = 1\), as it causes division by zero.
  • Additionally, \(f(g(x)) = \frac{1}{1-2x}\) becomes undefined when \(1 - 2x = 0\), giving \(x = \frac{1}{2}\).
  • Thus, the domain for \(f \circ g\) is all real numbers except \(x = 1\) and \(x = \frac{1}{2}\).
  • On the flip side, for \(g \circ f(x)\), the input \(f(x) = \frac{1+x}{1-x}\) cannot equal zero, which happens at \((x = 1)\).
  • Therefore, the domain of \(g \circ f\) excludes \(x = 1\).
Take caution to spot these problematic x-values when determining the domain.
Rational Functions
Rational functions are expressions that represent the ratio or fraction of two polynomial functions. A simple example is \( rac{p(x)}{q(x)}\), where both \(p(x)\) and \(q(x)\) are polynomials. These kinds of functions often involve division, which inherently implies potential for undefined conditions, primarily where the denominator is zero.
In the exercise, both functions \(f(x)\) and \(g(x)\) are rational functions:
  • \(f(x) = \frac{1+x}{1-x}\) involves division by \(1-x\), making it undefined at \(x=1\).
  • \(g(x) = \frac{x}{1-x}\) also shares a similar form and undefined point for the same reason.
The intersections where these functions combine can sometimes create new conditions for undefined values, as seen when they are composed. Always scrutinize the denominator to avoid division by zero, as these can impact the function's domain.
Composition of Functions
The composition of functions involves creating a third function by applying one function to the results of another. In simpler terms, it answers the question, "What happens if I put one function inside another?" The notation used is \((f \circ g)(x)\), which means \(f(g(x))\).
Here's how to compose functions effectively:
  • First, substitute the entire function \(g(x)\) into \(f(x)\), replacing every instance of \(x\) in \(f\) with \(g(x)\).
  • For \(f \circ g(x)\), this involves turning \(f(x) = \frac{1+x}{1-x}\) into \(f\left(\frac{x}{1-x}\right)\).
  • Simplifying carefully gives \(f(g(x)) = \frac{1}{1-2x}\).
  • Repeating the process in reverse for \(g \circ f(x)\), we substitute \(f(x)\) into \(g(x)\), simplifying gives \(-1\).
Understanding how to compose functions involves attention to detail and clear substitution practices. It highlights how the inner function affects the outer and impacts the resulting domain.

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Most popular questions from this chapter

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