Problem 37 - Algebra and Trigonometry

Question

\(29-40\) 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

- \( f(g(x)) = \frac{2x-1}{2x} \), Domain: \( x \neq 0 \) - \( g(f(x)) = \frac{x-1}{x+1} \), Domain: \( x \neq -1 \) - \( f(f(x)) = \frac{x}{2x+1} \), Domain: \( x \neq -\frac{1}{2} \) - \( g(g(x)) = 4x-3 \), Domain: all real numbers

Step 1: Determine f(g(x))

To find \( f \circ g \), we first substitute \( g(x) \) into \( f(x) \). Begin by replacing \( x \) in \( f(x) = \frac{x}{x+1} \) with \( g(x) = 2x - 1 \). This results in:\[ f(g(x)) = \frac{2x - 1}{(2x - 1) + 1} \]Simplify the expression:\[ f(g(x)) = \frac{2x - 1}{2x} \]

Step 2: Find the Domain of f(g(x))

The function \( f(g(x)) = \frac{2x - 1}{2x} \) is defined for all \( x \) such that the denominator is not zero. Set \( 2x eq 0 \), which gives \( x eq 0 \). Thus, the domain of \( f \circ g \) is all real numbers except \( x = 0 \).

Step 3: Determine g(f(x))

To find \( g \circ f \), substitute \( f(x) \) into \( g(x) \). Replace \( x \) in \( g(x) = 2x - 1 \) with \( f(x) = \frac{x}{x+1} \). This becomes:\[g(f(x)) = 2 \left(\frac{x}{x+1}\right) - 1 \]This simplifies to:\[g(f(x)) = \frac{2x}{x+1} - 1 = \frac{2x - (x+1)}{x+1} = \frac{x-1}{x+1} \]

Step 4: Find the Domain of g(f(x))

The function \( g(f(x)) = \frac{x-1}{x+1} \) is defined for all \( x \) such that the denominator is not zero. Solve \( x+1 eq 0 \), leading to \( x eq -1 \). Therefore, the domain of \( g \circ f \) is all real numbers except \( x = -1 \).

Step 5: Determine f(f(x))

For \( f \circ f \), substitute \( f(x) \) into itself. Replace \( x \) in \( f(x) = \frac{x}{x+1} \) with \( \frac{x}{x+1} \). The function becomes:\[f(f(x)) = \frac{\frac{x}{x+1}}{\frac{x}{x+1} + 1} \]Simplifying, this results in:\[f(f(x)) = \frac{x}{x + 2x + 1} = \frac{x}{2x + 1} \]

Step 6: Find the Domain of f(f(x))

The function \( f(f(x)) = \frac{x}{2x+1} \) is defined for all \( x \) where the denominator is not zero. Solve \( 2x+1 eq 0 \), giving \( x eq -\frac{1}{2} \). Thus, the domain of \( f \circ f \) is all real numbers except \( x = -\frac{1}{2} \).

Step 7: Determine g(g(x))

For \( g \circ g \), substitute \( g(x) \) into itself. Replace \( x \) in \( g(x) = 2x - 1 \) with \( 2x - 1 \). The function becomes:\[g(g(x)) = 2(2x - 1) - 1 = 4x - 2 - 1 = 4x - 3 \]

Step 8: Find the Domain of g(g(x))

The function \( g(g(x)) = 4x - 3 \) is a linear function, which is defined for all real numbers. Therefore, the domain of \( g \circ g \) is all real numbers.

Key Concepts

Understanding the Domain of a Function

The domain of a function represents all the possible input values (independent variables) for which the function is defined. In simpler terms, it refers to the set of all possible values that you can put into a function to get a valid output. To find the domain:

• Identify any restrictions on the variable, such as avoiding division by zero.
• Consider any square roots or even-numbered roots, which must have non-negative radicands.
• Look for values that result in undefined operations, like taking the logarithm of a negative number.

For the exercise provided, when finding the composition of functions like \(f(g(x))\) or \(g(f(x))\), the domain is determined by both the individual functions and their combinations. For instance, in the case of \(f(g(x)) = \frac{2x - 1}{2x}\), we avoid division by zero, so the domain excludes \(x = 0\). In \(g(f(x)) = \frac{x-1}{x+1}\), it excludes \(x = -1\). Think of the domain as the starting point for analyzing any function, as it tells you where the function "lives" on the number line.

Rational Functions Explained

Rational functions are fractions where both the numerator and the denominator are polynomials. They can express ratios of linear or higher-degree polynomials. The most common characteristic of rational functions is that they can have undefined points, notably where the denominator is zero. Here are some key points:

• To determine undefined points, set the denominator equal to zero and solve for the variable.
• These undefined points often result in vertical asymptotes on a graph of the function, indicating where the function grows infinitely large or small.
• Rational functions are usually continuous, except for at these undefined points.

In our exercise, the rational functions are evident in \(f(x) = \frac{x}{x+1}\) and \(f(g(x)) = \frac{2x-1}{2x}\). It's important to simplify these expressions and carefully analyze the denominators, as the domain is drastically impacted by where these expressions break down. Understanding and manipulating rational functions prepare you to handle more complex algebraic expressions effortlessly.

The Simplicity of Linear Functions

Linear functions are the simplest type of function. They create a straight line when graphed and have the general form \(y = mx + b\), where \(m\) represents the slope and \(b\) is the y-intercept. Some essential properties include:

• The domain of a linear function is all real numbers because there are no restrictions like division by zero.
• The slope \(m\) signifies the rate of change, showing how \(y\) changes with respect to \(x\).
• Linear functions display constant rates of change and are continuously increasing or decreasing.

In our exercise, \(g(x) = 2x - 1\) is a linear function. Its straightforward nature means its domain includes all real numbers, since there are no divisions or roots involved. Linear functions are foundational in algebra, and understanding them well is crucial as they build the basis for more complex algebraic concepts you will encounter.