Problem 83 - Intermediate Algebra

Question

If \(f(x)=x+1\) and \(g(x)=2 x-5,\) show that \((f \circ g)(x) \neq(g \circ f)(x)\).

Step-by-Step Solution

Answer

The compositions are not equal, \((f \circ g)(x) = 2x-4\) and \((g \circ f)(x) = 2x-3\).

Step 1: Define the Composite Functions

To find the compositions \((f \circ g)(x)\) and \((g \circ f)(x)\), we need to understand what each composition represents. Specifically, \((f \circ g)(x)\) means applying \(g(x)\) first, then applying \(f(x)\); \((g \circ f)(x)\) means applying \(f(x)\) first, then applying \(g(x)\).

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

First, calculate \(g(x)=2x-5\). Then, plug this into \(f(x)\):\[f(g(x)) = f(2x-5) = (2x-5) + 1 = 2x-4.\] So, \((f \circ g)(x) = 2x - 4.\)

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

First, calculate \(f(x)=x+1\). Then, plug this into \(g(x)\):\[g(f(x)) = g(x+1)= 2(x+1) - 5 = 2x + 2 - 5 = 2x - 3.\] So, \((g \circ f)(x) = 2x - 3.\)

Step 4: Compare the Two Compositions

We now have \((f \circ g)(x) = 2x - 4\) and \((g \circ f)(x) = 2x - 3\). Clearly, \((f \circ g)(x) eq (g \circ f)(x)\) because \(2x - 4 eq 2x - 3\) for any value of \(x\). This shows that the two compositions are not equal, confirming that \((f \circ g)(x) eq (g \circ f)(x)\).

Key Concepts

Function Composition

Function composition is a vital concept in mathematics, especially in understanding how different functions interact. Simply put, composing two functions involves taking the output of one function and using it as the input for another. This operation can reveal important relationships in algebra and calculus. In the context of our original problem, we encounter compositions like \((f \circ g)(x)\) and \((g \circ f)(x)\).

To compose \((f \circ g)(x)\), we first evaluate \(g(x)\) and then substitute this value into \(f(x)\). Conversely, for \((g \circ f)(x)\), \(f(x)\) is evaluated first, followed by substituting into \(g(x)\). These operations highlight that the order of composition matters. In many scenarios, reversing the order of function composition results in different outcomes, which illustrates the non-commutative nature of function composition.

Intermediate Algebra

Intermediate algebra serves as a bridge between basic algebra and more advanced topics. It involves manipulation and understanding of functions, which includes operations like function composition. This level of algebra helps in solving equations and understanding the behaviour of functions, ensuring a more robust grasp of mathematical principles.

In our exercise, we help students appreciate this by demonstrating the non-equivalence of two composite functions \((f \circ g)(x)\) and \((g \circ f)(x)\). Intermediate algebra encourages students to

• Recognize function forms
• Apply algebraic operations correctly
• Understand the hierarchy of operations

When functions are combined, as shown in our example, it shows students the importance of order and precision, therefore building a foundation for more advanced topics.

Algebraic Operations

Algebraic operations form the backbone of problem-solving in mathematics. Understanding how to manipulate expressions and functions using operations such as addition, subtraction, multiplication, and function composition is essential. In our step-by-step solution, we performed several algebraic operations to compute the composite functions.

For the function \(f(x) = x + 1\) and \(g(x) = 2x - 5\), we:

• Calculated \(g(x)\), resulting in \(2x - 5\)
• Substituted \(g(x)\) into \(f(x)\) to find \((f \circ g)(x)\) as \(2x - 4\)
• Similarly, computed \(f(x)\), and then \(g(f(x))\) to form \((g \circ f)(x)\)

These steps reinforce how meticulous application of algebraic operations can derive new functions that reflect the characteristics of their composed parts. Operations like these underscore the importance of maintaining clarity and procedural accuracy when performing algebraic manipulations.