Problem 63 - College Algebra

Question

Find a. \((f \circ g)(x)\) b. \((g \circ f)(x)\) c. \((f \circ g)(2)\) d. \((g \circ f)(2)\) $$ f(x)-2 x-3, g(x)-\frac{x+3}{2} $$

Step-by-Step Solution

Answer

In a. we find that \(f \circ g(x) = x\), in b. \(g \circ f(x) = x\), in c. \(f \circ g(2) = 2\), and in d. \(g \circ f(2) = 2\). The interesting fact is that both composite functions are identity functions; i.e., they return the original value.

Step 1: Compute \(f \circ g(x)\)

To compute \(f \circ g(x)\), we replace \(x\) in \(f(x)\) with \(g(x)\). This gives us \(f \circ g(x) = f(g(x)) = f\left(\frac{x+3}{2}\right)= 2*\left(\frac{x+3}{2}\right)-3= x+3-3 = x\). So, \(f \circ g(x) = x\).

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

To compute \(g \circ f(x)\), we replace \(x\) in \(g(x)\) with \(f(x)\). This gives us \(g \circ f(x) = g(f(x))=g(2x-3)=\frac{2x-3+3}{2}=x\). So, \(g \circ f(x) = x\).

Step 3: Compute \(f \circ g(2)\)

Substitute \(x = 2\) into the composite function \(f \circ g(x)\) to yield \(f \circ g(2) = 2\).

Step 4: Compute \(g \circ f(2)\)

Substitute \(x = 2\) into the composite function \(g \circ f(x)\) to yield \(g \circ f(2) = 2\).

Key Concepts

Composite Function

Composite functions are like a recipe where you use one function as an ingredient for another function. You first apply one function and then, using its output, you apply another function. This process is denoted by the symbol \(f \circ g(x)\), read as "\(f\) composed with \(g\)". Here, \(f \circ g(x)\) means you substitute \(g(x)\) into \(f(x)\).
For example, in our exercise,

• \(f(x) = 2x - 3\)
• \(g(x) = \frac{x + 3}{2}\)

If we want to find \((f \circ g)(x)\), we first find \(g(x)\) and then plug it into \(f(x)\). This allows us to simplify complex function operations by breaking them down into more manageable steps.
Understanding composite functions helps build a strong foundation in more advanced algebra topics later on.

Function Operations

Function operations involve basic math operations like addition, subtraction, multiplication, and division performed on functions. However, they also include some more specific operations, such as finding composite functions. Each operation has rules to simplify or solve the functions involved. In this context, it’s important to understand how each function behaves individually before combining them.
When we talk about combining \(f(x)\) and \(g(x)\), our exercise demonstrates it through operations such as:

• \(f \circ g(x)\), where \(g(x)\) is substituted into \(f(x)\)
• \(g \circ f(x)\), where \(f(x)\) is substituted into \(g(x)\)

Sometimes operations are straightforward, like

• When \(f(x)\) adds \(g(x)\), the focus is on combining results.
• Other times, operations involve capturing an essence of one function and impacting another as seen in composition.

This shows that beyond working on numbers, functions can interact and transform through diverse operations.

Composition of Functions

Composition of functions helps in layering two functions to act as one. This technique is essential when dealing with complex mathematical problems, allowing the transformation of one function by another.To perform a composition of functions, you follow these steps:

• Identify the two functions you want to compose (e.g., \(f(x)\) and \(g(x)\)).
• Decide which function to apply first. For \((f \circ g)(x)\), apply \(g(x)\) first, then apply \(f(x)\) to the result.
• Substitute the output of the first function into the second.
• Identify the resulting expression or number.

In our specific example,

• \((f \circ g)(x) = x\)
• \((g \circ f)(x) = x\)

This means each function undoes the work of the other. They are inverses in this context.
This layered approach often simplifies problem-solving in algebra and calculus, enabling focus on intricate operations without being overwhelmed.