Problem 43 - Algebra and Trigonometry

Question

Find the functions \(f \circ g, g \circ f, f \circ f,\) and \(g \circ g\) and their domains. $$ f(x)=\frac{x}{x+1}, \quad g(x)=\frac{1}{x} $$

Step-by-Step Solution

Answer

The compositions are \(f \circ g(x) = \frac{1}{1+x}\) (\(x \neq 0, -1\)), \(g \circ f(x) = \frac{x+1}{x}\) (\(x \neq 0, -1\)), \(f \circ f(x) = \frac{x}{2x+1}\) (\(x \neq -1, -\frac{1}{2}\)), and \(g \circ g(x) = x\) (\(x \neq 0\)).

Step 1: Find the function f∘g

To find the function \(f \circ g\), substitute \(g(x)\) into \(f(x)\). This gives \(f(g(x)) = f\left(\frac{1}{x}\right) = \frac{\frac{1}{x}}{\frac{1}{x}+1} = \frac{1/x}{(1+x)/x} = \frac{1}{1+x}\).

Step 2: Determine the domain of f∘g

The domain of \(f \circ g\) requires that \(x eq 0\) for \(g(x)\) to be defined and \(1+x eq 0\) (or \(x eq -1\)) for \(f(g(x))\) to be defined. Therefore, the domain of \(f \circ g(x)\) is all real numbers except \(x = 0\) and \(x = -1\).

Step 3: Find the function g∘f

To find the function \(g \circ f\), substitute \(f(x)\) into \(g(x)\). This gives \(g(f(x)) = g\left(\frac{x}{x+1}\right) = \frac{1}{\frac{x}{x+1}} = \frac{x+1}{x}\).

Step 4: Determine the domain of g∘f

The domain of \(g \circ f\) requires that \(x+1 eq 0\) (or \(x eq -1\)) for \(f(x)\) to be defined and \(x eq 0\) for \(g(f(x))\) to be defined. Thus, the domain of \(g \circ f(x)\) is all real numbers except \(x = 0\) and \(x = -1\).

Step 5: Find the function f∘f

To find \(f \circ f\), substitute \(f(x)\) into itself: \(f(f(x)) = f\left(\frac{x}{x+1}\right) = \frac{\frac{x}{x+1}}{\frac{x}{x+1}+1} = \frac{x/(x+1)}{(2x+1)/(x+1)} = \frac{x}{2x+1}\).

Step 6: Determine the domain of f∘f

The domain of \(f \circ f\) requires \(x+1 eq 0\) (or \(x eq -1\)) for \(f(x)\) and \(2x+1 eq 0\) (or \(x eq -\frac{1}{2}\)) for \(f(f(x))\) to be defined. Thus, the domain is all real numbers except \(x = -1\) and \(x = -\frac{1}{2}\).

Step 7: Find the function g∘g

For \(g \circ g\), substitute \(g(x)\) into itself: \(g(g(x)) = g\left(\frac{1}{x}\right) = \frac{1}{\frac{1}{x}} = x\).

Step 8: Determine the domain of g∘g

The domain of \(g \circ g\) requires \(x eq 0\), so the domain of \(g \circ g(x)\) is all real numbers except \(x = 0\).

Key Concepts

Domain of a Function

Understanding the domain of a function is crucial when dealing with any type of function, especially when composing them. The domain refers to all the possible input values (usually represented by 'x') that a function can accept without causing any undefined behavior, such as division by zero or taking the square root of a negative number.

When we find the composite functions like \(f \circ g\) or \(g \circ f\), we need to consider the restrictions from not just one, but both of the functions involved. For instance, if one of the functions in composition is a rational function, there will be restrictions where the denominator is zero.

In the provided exercise, we noted that certain values like \(x = 0\), \(x = -1\), or \(x = -\frac{1}{2}\) are excluded from the domains of the functions \(f\), \(g\), or their composites. These restrictions come from ensuring no division by zero occurs in any part of the function, maintaining its validity and continuity. It's always important to remember to check every step for potential issues like these when determining the domain of composite functions.

Rational Functions

A rational function is a type of function that is expressed as the ratio of two polynomials, such as \(f(x) = \frac{x}{x+1}\) or \(g(x) = \frac{1}{x}\). These functions are interesting because they can behave in unexpected ways around certain points, particularly where the denominator is zero.

Rational functions often have holes or asymptotes. A hole occurs when a certain input makes the numerator and denominator zero at the same time, while a vertical asymptote occurs when the denominator approaches zero, causing the function value to go towards infinity.

In the exercise, \(f(x)\) and \(g(x)\) both illustrate this by limiting inputs where the denominators become zero. For \(f(x)\), the input \(x = -1\) leads to a zero denominator; similarly, for \(g(x)\), \(x=0\) results in a division by zero. These characteristics are essential to understand not only for their individual functions but also in the context of composite functions, where these restrictions continue to play an instrumental role in determining the overall domain.

Composite Functions

Composite functions involve creating a new function by combining two functions such as \(f\) and \(g\). Mathematically, this is represented as \(f \circ g\) or \(g \circ f\), which fundamentally means substituting one function into another.

The process of finding a composite function \(f \circ g\), for example, involves inputting \(g(x)\) into \(f(x)\). So, if \(g(x) = \frac{1}{x}\), then \(f(g(x)) = f\left(\frac{1}{x}\right)\). This composition requires careful attention to see how values that work in one function might not be valid in the overall composite, as they could cause division by zero or other undefined scenarios in the composed function.

Moreover, when creating composite functions, remember that the domain of the composition \(f \circ g\) is tied to both domains of \(f\) and \(g\). You must find values that work for both to ensure the output remains valid across all components of the composite function. This is why when given real-world exercises like this, one should carefully outline the domain restrictions step-by-step to ensure no details are missed. By mastering these careful evaluations, composite functions can be a powerful tool in mathematics.