Problem 69 - A Graphical Approach to College Algebra

Question

For certain pairs of functions \(f\) and \(g,(f \circ g)(x)=x\) and \((g \circ f)(x)=x .\) Show that this is true for the pairs. $$f(x)=4 x+2, g(x)=\frac{1}{4}(x-2)$$

Step-by-Step Solution

Answer

The pairs \(f(x) = 4x + 2\) and \(g(x) = \frac{1}{4}(x-2)\) are inverses since both compositions equal \(x\).

Step 1: Understanding the composition of functions

We are given two functions, \( f(x) = 4x + 2 \) and \( g(x) = \frac{1}{4}(x-2) \). We are tasked with showing that when combined, their compositions \((f \circ g)(x)\) and \((g \circ f)(x)\) both equal \(x\).

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

The composition \((f \circ g)(x)\) means applying \(g(x)\) first and then applying \(f(x)\) to the result. Start by substituting \(g(x)\) into \(f(x)\):\[f(g(x)) = f\left(\frac{1}{4}(x-2)\right)\]Now, replace \(x\) in \(f(x) = 4x + 2\) with \(\frac{1}{4}(x-2)\):\[f\left(\frac{1}{4}(x-2)\right) = 4\cdot\frac{1}{4}(x-2) + 2\]Simplifying:\[(x-2) + 2 = x\]Therefore, \((f \circ g)(x) = x\).

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

The composition \((g \circ f)(x)\) means applying \(f(x)\) first and then applying \(g(x)\) to the result. Substitute \(f(x)\) into \(g(x)\):\[g(f(x)) = g(4x + 2)\]Replace \(x\) in \(g(x) = \frac{1}{4}(x-2)\) with \(4x + 2\):\[g(4x + 2) = \frac{1}{4}((4x + 2) - 2)\]Simplifying:\[\frac{1}{4}(4x) = x\]Therefore, \((g \circ f)(x) = x\).

Step 4: Conclusion

Since both \((f \circ g)(x) = x\) and \((g \circ f)(x) = x\), the functions \(f(x) = 4x + 2\) and \(g(x) = \frac{1}{4}(x-2)\) are indeed inverses of each other, confirming the statement is true for these pairs of functions.

Key Concepts

Function Composition

Function composition is a key concept in mathematics that allows us to combine two functions to form a new function. It is denoted by the symbol \((f \circ g)(x)\), which reads as "\(f\) composed with \(g\)". This means we apply the function \(g(x)\) first and then apply \(f(x)\) to the result of \(g(x)\). Here's how it works:

• Start with the input \(x\).
• Apply \(g\): Calculate \(g(x)\).
• Take the result from step 2 and substitute it into \(f\): \(f(g(x))\).
• This final output is the result of \((f \circ g)(x)\).

This powerful tool is used to simplify complex problems by understanding how outputs of one function become inputs of another. It’s especially important to check carefully in which order functions are applied, as reversing the order can lead to different results.

Algebraic Functions

Algebraic functions are functions that can be constructed using algebraic operations—such as addition, subtraction, multiplication, division, and taking roots—on the variable \(x\). Examples include linear functions, quadratic functions, and more complex polynomial functions.

• Linear Functions: The simplest algebraic functions, with general form \(f(x) = ax + b\), where \(a\) and \(b\) are constants.
• Operations: Algebraic functions follow the standard arithmetic operations, allowing compositions like the one we solved: Join \(f(x) = 4x + 2\) with \(g(x) = \frac{1}{4}(x-2)\).
• Solution Concept: By substituting one function into another, you effectively simplify the problem into solving for \(x\) via arithmetic simplifications.

Understanding how these properties work is essential to solving equations and finding solutions for such problems in mathematics. Mastery of these techniques also helps in tackling more complicated scenarios.

Identity Function

The identity function is a special function where the output is exactly the same as the input. It is generally denoted as \(I(x) = x\). When you compose a function with the identity function, the original function remains unchanged. Here's why this is important:

• Usage in Function Inverses: When two functions satisfy \((f \circ g)(x) = x\) and \((g \circ f)(x) = x\), we can say they are inverses of each other because they've essentially reduced back to the identity function.
• Importance in Problem Solving: Finding an identity through function composition, like in our exercise, means checking that combining two processes brings you back to the start, validating inverses.
• Mathematical Simplicity: The identity function’s simplicity is key to understanding more complex relationships and operations between functions.

Not only does this concept aid in verifying solutions regarding inverse functions, but it also provides an intuitive understanding of how function compositions can revert to simpler forms.