Problem 27 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Let \(f(x)=\frac{x}{x+3}\) and \(g(x)=\frac{3 x}{1-x}\). (a) Find \(f(g(2))\) and \(g(f(2))\). (b) Find \(f(g(x))\) and \(g(f(x))\). (c) What does part (b) suggest about the relationship between \(f\) and \(g\) ?

Step-by-Step Solution

Answer

a) \(f(g(2)) = 2\), \(g(f(2)) = 2\), b) \(f(g(x)) = 1\), \(g(f(x)) = 1\), c) Functions \(f\) and \(g\) are inverses of each other.

Step 1: Evaluate \(f(g(2))\)

To find \(f(g(2))\), first evaluate \(g(2)\). Substitute \(x = 2\) into \(g(x)\) to get \(g(2) = \frac{3*2}{1-2} = -6\). Then substitute \(x = -6\) into \(f(x)\) to get \(f(g(2)) = \frac{-6}{-6+3} = 2\).

Step 2: Evaluate \(g(f(2))\)

To find \(g(f(2))\), first evaluate \(f(2)\). Substitute \(x = 2\) into \(f(x)\) to get \(f(2) = \frac{2}{2+3} = 0.4\). Then substitute \(x = 0.4\) into \(g(x)\) to get \(g(f(2)) = \frac{3*0.4}{1-0.4} = 2\).

Step 3: Evaluate \(f(g(x))\)

To find \(f(g(x))\), substitute \(g(x)\) into \(f(x)\) to get \(f(g(x)) = f(\frac{3x}{1-x}) = \frac{\frac{3x}{1-x}}{\frac{3x}{1-x}+3} = 1\).

Step 4: Evaluate \(g(f(x))\)

To find \(g(f(x))\), substitute \(f(x)\) into \(g(x)\) to get \(g(f(x)) = g(\frac{x}{x+3}) = \frac{3*\frac{x}{x+3}}{1-\frac{x}{x+3}} = 1\).

Step 5: Analyze the Relationship Between \(f\) and \(g\)

From the results in step 3 and step 4, we can see that both \(f(g(x))\) and \(g(f(x))\) provide a constant result of 1. This suggests that \(f\) and \(g\) are inverse functions of each other, meaning that one function undoes the operation of the other.

Key Concepts

Function Composition

Function composition is a process of applying one function to the results of another function. This process helps us understand how different functions interact with each other. When we compose two functions, say \(f(x)\) and \(g(x)\), it means that we first apply \(g(x)\) and then apply \(f(x)\) to the result, written as \(f(g(x))\).
For example, in the problem, we first found \(g(2)\), which was calculated as \(-6\). We then used this result as an input for \(f(x)\), and determined that \(f(g(2)) = 2\).

• This demonstrates how composition works by performing multiple function evaluations in sequence.
• Function composition is an important tool in calculus and algebra because it allows us to explore potential relationships and combinations between different functions.

It also provides insights into how nested functions can simplify or complicate expressions. Understanding the result of \(f(g(x))\), as the problem illustrates, can highlight special properties like the potential for functions to be inverses.

Rational Functions

Rational functions are a type of function expressed as the ratio of two polynomials. A common example is \(f(x) = \frac{x}{x+3}\). Rational functions can behave in complex ways due to division by expressions that may include variables.
When interpreting rational functions like in this exercise, it's crucial to consider the domain. For instance, the function \(f(x) = \frac{x}{x+3}\) is undefined when the denominator is zero, such as \(x = -3\).
Rational functions often lead to interesting results when composed, as different function structures can lead to cancellation or special relationships.

• They often create vertical asymptotes where the denominator equals zero, and zeros occur where the numerator equals zero.
• Understanding how different rational functions compose with one another can help identify special relationships, such as inverses in this case.

This is exactly what the exercise demonstrates: composing two specific rational functions results in a constant, suggesting a specific inverse relationship.

Algebraic Manipulation

Algebraic manipulation involves using algebraic techniques to simplify, expand, or rearrange expressions to solve equations or reveal relationships. In the given exercise, algebraic manipulation was used to determine \(f(g(x))\) and \(g(f(x))\).
This involved substituting one rational function into another, leading to simplified expressions. For instance, \(f(g(x))\) was calculated as: \[ f(g(x)) = \frac{\frac{3x}{1-x}}{\frac{3x}{1-x} + 3} = 1 \]
Through this evaluation, we apply manipulation techniques including division of rational expressions and simplification. Algebraic manipulation skills are essential in solving these kinds of problems because they allow us to transform complex expressions into simpler forms that are easier to understand and evaluate.

• Effective manipulation can expose solutions that are not immediately evident, such as constant functions.
• Simplifying expressions can also reveal underlying function properties or relationships like periodic behavior or inverses.

Developing these techniques helps reduce potential errors in computation and lays the groundwork for solving more advanced algebraic problems.