Problem 16 - College Algebra

Question

Given \(f(x)=\frac{x}{2+x}\) and \(g(x)=\frac{2 x}{1-x}:\) a. Find \(f(g(x))\) and \(g(f(x))\). b. What does the answer tell us about the relationship between \(f(x)\) and \(g(x) ?\)

Step-by-Step Solution

Answer

The functions \(f(x)\) and \(g(x)\) are inverses.

Step 1: Understand the Problem

We are given two functions, \(f(x)\) and \(g(x)\), and need to find their compositions: \(f(g(x))\) and \(g(f(x))\). Then we will analyze what these compositions tell us about the relationship between \(f(x)\) and \(g(x)\).

Step 2: Find f(g(x))

To find \(f(g(x))\), substitute \(g(x) = \frac{2x}{1-x}\) into \(f(x) = \frac{x}{2+x}\). This gives:\[f(g(x)) = \frac{\frac{2x}{1-x}}{2 + \frac{2x}{1-x}}\]Simplify the expression:\[f(g(x)) = \frac{2x}{1-x} \times \frac{1-x}{2(1-x) + 2x}\]Combine the terms in the denominator:\[f(g(x)) = \frac{2x}{1-x} \times \frac{1-x}{2 + 2x - 2x} = \frac{2x}{2(1-x) + 2x} = \frac{2x}{2}\]Finally, simplify the expression:\[f(g(x)) = x\]

Step 3: Find g(f(x))

To find \(g(f(x))\), substitute \(f(x) = \frac{x}{2+x}\) into \(g(x) = \frac{2x}{1-x}\). This gives:\[g(f(x)) = \frac{2 \cdot \frac{x}{2+x}}{1 - \frac{x}{2+x}}\]Simplify the expression by combining fractions:\[g(f(x)) = \frac{2x}{2+x} \times \frac{2+x}{2+x - x}\]Simplify the denominator:\[g(f(x)) = \frac{2x}{2+x} \times \frac{2+x}{2}\]Cancel out the \(2+x\) and simplify:\[g(f(x)) = \frac{2x}{2} = x\]

Step 4: Analyze the Results

The expressions \(f(g(x)) = x\) and \(g(f(x)) = x\) show that both function compositions return \(x\). This implies that \(f(x)\) and \(g(x)\) are inverse functions of each other.

Key Concepts

Inverse Functions

Inverse functions are an interesting concept in mathematics. They essentially "undo" each other. When you compose a function with its inverse, the result is the identity function, which simply returns the original input value. In other words, if you have a function \( f(x) \) and its inverse \( g(x) \), then \( f(g(x)) = x \) and \( g(f(x)) = x \). This property is evident in the exercise where composing the two given functions in either order yields \( x \).

Inverse functions are super useful because they allow us to solve equations that involve complex operations more easily. Knowing that functions are inverses can simplify problem-solving, like when reversing the order of transformations.

• The output of one becomes the input of the other.
• Inverse functions can often be graphed as mirror images over the line \( y = x \).

Understanding how to find and use inverse functions is essential in calculus and algebra, where transformations and solving equations are key operations.

Rational Functions

Rational functions are functions represented by the ratio of two polynomials. The general form is \( f(x) = \frac{p(x)}{q(x)} \), where both \( p(x) \) and \( q(x) \) are polynomials.

Rational functions can exhibit a wide range of behaviors, depending on the degrees of the numerator and the denominator. They may have asymptotes, which occur where the function is undefined. For example, the function \( f(x) = \frac{x}{2+x} \) from the exercise has a vertical asymptote at \( x = -2 \) because the denominator equals zero when \( x \) is \(-2\). The function \( g(x) = \frac{2x}{1-x} \) has a vertical asymptote at \( x = 1 \).

• Rational functions are discontinuous at the points where their denominators equal zero.
• They can have horizontal asymptotes, revealing end-behavior of the function as \( x \) approaches infinity.

Rational functions are crucial in modeling real-world situations, especially where rates and ratios are involved. Hence, understanding their properties and behavior can help in science and engineering fields.

Composition of Functions

Function composition is like a mathematical "chain reaction." You take the output from one function and use it as the input for another. It's a way to combine functions to create more complex expressions or solutions.

In the exercise, you found \( f(g(x)) \) and \( g(f(x)) \), demonstrating composition by substituting one function into another. When you compose functions, you're essentially stacking them, performing one function's operation and then immediately another's. Composition can reveal relationships between functions, like detecting inverse pairs as seen in the exercise results.

A key point is to always consider the domains and ranges. Not every function can be composed with another in every instance because the output of the first function must fall within the domain of the second function.

• Composition is denoted as \( (f \circ g)(x) = f(g(x)) \).
• It's crucial in calculus for dealing with changes of variables and transformations.

The power of composition lies in its ability to create new functions from existing ones, providing flexibility for problem-solving within various branches of mathematics.