Problem 41 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

\( f \circ g(x) = \frac{2x-1}{2x} \), domain \( x \neq 0 \); \( g \circ f(x) = \frac{x-1}{x+1} \), domain \( x \neq -1 \); \( f \circ f(x) = \frac{x}{2x+1} \), domain \( x \neq -1, -\frac{1}{2} \); \( g \circ g(x) = 4x-3 \), domain is all real numbers.

Step 1: Understand Function Composition

Function composition means combining two functions where the output of one function becomes the input of the other. The notation \( f \circ g \) indicates that you substitute \( g(x) \) into \( f(x) \). Similarly, \( g \circ f(x) \) means substituting \( f(x) \) into \( g(x) \). The functions \( f \circ f \) and \( g \circ g \) involve substituting a function into itself.

Step 2: Calculate \(f \circ g(x)\)

To find \(f \circ g(x)\), substitute \(g(x) = 2x - 1\) into \(f(x) = \frac{x}{x+1}\): \[ f(g(x)) = f(2x - 1) = \frac{2x - 1}{(2x - 1) + 1} = \frac{2x - 1}{2x} \] The domain excludes any value making the denominator zero. Here, solve \(2x = 0\), giving \(x eq 0\). Thus, the domain is all real numbers except 0.

Step 3: Calculate \(g \circ f(x)\)

To find \(g \circ f(x)\), substitute \(f(x) = \frac{x}{x+1}\) into \(g(x) = 2x - 1\):\[ g(f(x)) = g\left(\frac{x}{x+1}\right) = 2\left(\frac{x}{x+1}\right) - 1 = \frac{2x}{x+1} - \frac{x+1}{x+1} = \frac{x - 1}{x+1} \]The domain of this composition excludes values where \(x+1 = 0\), so \(x eq -1\). Thus, the domain is all real numbers except -1.

Step 4: Calculate \(f \circ f(x)\)

To find \(f \circ f(x)\), substitute \(f(x) = \frac{x}{x+1}\) into itself:\[ f(f(x)) = f\left(\frac{x}{x+1}\right) = \frac{\frac{x}{x+1}}{\frac{x}{x+1} + 1} = \frac{\frac{x}{x+1}}{\frac{x+x+1}{x+1}} = \frac{x}{2x+1} \]The domain excludes any value where \(2x+1 = 0\), so \(x eq -\frac{1}{2}\), and \(x+1 = 0\) so \(x eq -1\). The domain is all real numbers except -1 and -1/2.

Step 5: Calculate \(g \circ g(x)\)

To find \(g \circ g(x)\), substitute \(g(x) = 2x - 1\) into itself:\[ g(g(x)) = g(2x - 1) = 2(2x - 1) - 1 = 4x - 2 - 1 = 4x - 3 \]There are no restrictions as this is a linear function. The domain is all real numbers.

Key Concepts

Function Domains

The domain of a function is the set of all possible inputs (or "x" values) that do not cause the function to break, meaning no division by zero or taking the square root of a negative number.
For example:

• In the function \( f(x) = \frac{x}{x+1} \), the domain excludes any value of \( x \) that makes the denominator zero. Solving \( x+1 = 0 \) gives \( x = -1 \), so \( x eq -1 \).
• For the linear function \( g(x) = 2x - 1 \), there are no restrictions since linear functions are defined for all real numbers.
When working with composite functions like \( f \circ g \), it's crucial to check both the domain of the inside function \( g(x) \) and the resulting composed function \( f(g(x)) \). You must exclude any x-values that would make any part of the composition undefined, leading to an accurate domain.
Understanding these rules helps ensure you determine the correct domains and avoid errors when solving problems.

Composite Functions

Composite functions involve taking one function and plugging it into another. We often write this using the notation \( f \circ g \). This represents applying \( g \) first and then applying \( f \) to the result. Understanding how to compose functions expands our ability to analyze relationships between varying quantities.
Here's how you might think about this:

• Start with \( f \circ g(x) \). You plug the output from \( g(x) \) into \( f(x) \). This could look like \( f(g(x)) = f(2x - 1) \).
• Next, with \( g \circ f(x) \), input \( f(x) \) into \( g(x) \): \( g(f(x)) = g\left(\frac{x}{x+1}\right) \).
• With \( f \circ f(x) \) and \( g \circ g(x) \), you're substituting the functions into themselves, which can sometimes lead to interesting simplifications or highlight special properties of the function.

Essentially, composite functions are about creating something new by building on existing functions, allowing a deeper understanding and more flexibility in solving complex mathematical problems.

Rational Functions

Rational functions are fractions where both the numerator and the denominator are polynomials. They are central to many areas of mathematics due to their complexity and the need to carefully manage their domains to avoid undefined expressions.
Consider the function \( f(x) = \frac{x}{x+1} \). This is a simple rational function where:

• The numerator is a linear polynomial \( x \).
• The denominator is a linear polynomial \( x + 1 \).

Rational functions like \( f(x) \) are defined for all real numbers except where the denominator is zero. Hence, understanding where these undefined points occur is crucial.
When dealing with rational composite functions, these undefined points can change. For example, calculating \( f \circ g(x) \) involves a rational expression: \( f(g(x)) = \frac{2x - 1}{2x}\). Here, simply solving for zero in the new denominator can reveal entirely different domain restrictions than those in the original function. Keeping these changes in mind helps solve composite rational function problems accurately and avoids missteps.