Problem 21 - Intermediate Algebra

Question

Show that \((f \circ g)(x)=x\) and \((g \circ f)\) \((x)=x\) for each pair of functions. \(f(x)=4 x+2\) and \(g(x)=\frac{x-2}{4}\)

Step-by-Step Solution

Answer

Both compositions equal x, confirming they are inverses.

Step 1: Substitute and Simplify

First, we need to find \((f \circ g)(x)\), which means \(f(g(x))\). Substitute \(g(x) = \frac{x-2}{4}\) into \(f(x)\). Thus, we have:\(f(g(x)) = f\left(\frac{x-2}{4}\right) = 4\left(\frac{x-2}{4}\right) + 2\). Simplifying this, we get:\[4\left(\frac{x-2}{4}\right) + 2 = (x - 2) + 2 = x\].So, \((f \circ g)(x) = x\).

Step 2: Substitute and Simplify Again

Next, find \((g \circ f)(x)\), which means \(g(f(x))\). Substitute \(f(x) = 4x + 2\) into \(g(x)\). Thus, we have:\(g(f(x)) = g(4x + 2) = \frac{(4x + 2) - 2}{4}\). Simplifying this, we get:\[\frac{4x + 2 - 2}{4} = \frac{4x}{4} = x\].Therefore, \((g \circ f)(x) = x\).

Key Concepts

Inverse Functions

Inverse functions are a fascinating concept in mathematics. Essentially, an inverse function "reverses" the effect of the original function. If you have a function \( f \) and its inverse \( g \), applying \( f \) then \( g \) will return you to your original input, \( x \). This is expressed as \((f \circ g)(x) = x\) and \((g \circ f)(x) = x\). To show this relationship, it's crucial that both compositions of \( f \) and \( g \) simplify to the identity function.

• Each function has a specific reverse operation.
• The inverse of a function \( f \) is usually denoted as \( f^{-1} \).
• A function must be bijective (one-to-one and onto) to have an inverse.

For our functions, \( f(x) = 4x + 2 \) and \( g(x) = \frac{x-2}{4} \), substituting and simplifying both compositions prove they are indeed inverses of each other.

Algebraic Functions

Algebraic functions describe various types of expressions using algebraic operations such as addition, subtraction, multiplication, division, and taking roots. The given functions \( f(x) = 4x + 2 \) and \( g(x) = \frac{x-2}{4} \) are classic examples.

• Polynomial functions, like \( f(x) = 4x + 2 \), involve terms raised to whole number powers.
• Rational functions, such as \( g(x) = \frac{x-2}{4} \), are ratios of polynomial functions.
• Simplifying these functions often involves substituting algebraic expressions and observing patterns.

Understanding the underlying algebraic expressions can help identify the nature and behavior of the functions. This also facilitates the process of finding an inverse function, as done in our exercise.

Identity Function

The identity function is a fundamental concept in function composition. In mathematical terms, the identity function on an element \( x \) is \( x \) itself. It is often denoted as \( I(x) = x \). This function just "leaves things as they are," mapping every element to itself.

• In the context of function compositions, saying \((f \circ g)(x) = x\) implies the result is the identity function.
• For any function \( f \), composing with the identity function doesn’t alter the function: \( f \circ I = f \) and \( I \circ f = f \).
• This function is critical in defining the concept of inverse functions, as shown through our exercise.

Essentially, after composing a function with its inverse, if you arrive at the identity function, it confirms that you selected the correct inverse pair.