Problem 83 - Elementary and 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

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

Step 1: Understand the Functions

We have two functions given. The first function is \(f(x) = x + 1\) and the second function is \(g(x) = 2x - 5\). Our goal is to show that the composition of these functions in different orders results in different outputs.

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

The notation \((f \circ g)(x)\) means \(f(g(x))\). This indicates we need to substitute \(g(x)\) into \(f(x)\). We know from step 1 that \(g(x) = 2x - 5\), so substitute this into \(f(x) = x + 1\):\[f(g(x)) = f(2x - 5) = (2x - 5) + 1 = 2x - 4.\]

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

The notation \((g \circ f)(x)\) means \(g(f(x))\). This indicates we need to substitute \(f(x)\) into \(g(x)\). We know from step 1 that \(f(x) = x + 1\), so substitute this into \(g(x) = 2x - 5\):\[g(f(x)) = g(x + 1) = 2(x + 1) - 5 = 2x + 2 - 5 = 2x - 3.\]

Step 4: Compare \((f \circ g)(x)\) and \((g \circ f)(x)\)

From Step 2, we found that \((f \circ g)(x) = 2x - 4\). From Step 3, we found that \((g \circ f)(x) = 2x - 3\). Now we compare the two:\[(f \circ g)(x) = 2x - 4 \quad \text{and} \quad (g \circ f)(x) = 2x - 3.\]Clearly, \(2x - 4 eq 2x - 3\).

Step 5: Conclude the Non-Equality

Since \((f \circ g)(x)\) and \((g \circ f)(x)\) result in different expressions, we can conclude that \((f \circ g)(x) eq (g \circ f)(x)\). This shows the non-commutative nature of function composition in this case.

Key Concepts

Commutativity in Mathematics

In mathematics, a property is said to be commutative if changing the order of the operands does not change the result. Simple examples include addition and multiplication of numbers:

• For addition: \( a + b = b + a \)
• For multiplication: \( a \times b = b \times a \)

However, function composition is generally not commutative. This means that for two functions, \(f(x)\) and \(g(x)\), the composition \((f \circ g)(x)\) is not necessarily equal to \((g \circ f)(x)\). This is because the output of one function becomes the input of the other, and switching their order can lead to different results. In our exercise,

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

were shown to yield different expressions and thus demonstrate the non-commutative nature of these functions.

Understanding Algebraic Functions

Algebraic functions consist of operations like addition, subtraction, multiplication, division, and exponentiation on variables and constants. These functions play a crucial role in problem-solving because they can be manipulated using algebraic rules. In our exercise, the functions given are:

• \(f(x) = x + 1\)
• \(g(x) = 2x - 5\)

These simple algebraic functions involve basic operations like addition and multiplication. Understanding these basics can help you easily perform more complex operations, such as composition of functions. When dealing with algebraic functions, it's essential to:

• Identify the operations involved, such as addition or multiplication
• Consider how two functions might interact when composed together

This knowledge will be useful when examining how they behave through function composition, which is demonstrated in the exercise.

Composition of Functions

Composition of functions is a process where the output of one function becomes the input of another. It's a way of combining two functions to form a new one. Notationally, it's written as \((f \circ g)(x)\), which means "\(f\) composed with \(g\)" or "\(f\) of \(g(x)\)." Understanding this can provide insight into complex mathematical relationships. Let's dive into the exercise as an example:

• \((f \circ g)(x) = f(g(x))\) results in \(f(2x - 5) = 2x - 4\)
• \((g \circ f)(x) = g(f(x))\) results in \(g(x + 1) = 2x - 3\)

When performing function compositions, the order is crucial as demonstrated. This means that the tasks the functions perform depend on their sequential order. This distinctly leads to the final expressions being different when \(f\) and \(g\) are composed in different ways—highlighting the importance of order in the composition of functions.