Problem 20 - Intermediate Algebra

Question

Show that \((f \circ g)(x)=x\) and \((g \circ f)\) \((x)=x\) for each pair of functions. \(f(x)=-2 x\) and \(g(x)=-\frac{1}{2} x\)

Step-by-Step Solution

Answer

Yes, both composite functions \\(f \circ g(x)\\) and \\(g \circ f(x)\\) equal \\ x\\ for the given functions.

Step 1: Understand Composite Function

A composite function \(f \circ g\)(x) implies applying function \ g(x)\ first and then applying function \ f(x)\ to the result. Similarly, \(g \circ f\)(x) means applying \ f(x)\ first and then \ g(x)\ to the result.

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

Substitute \(g(x) = -\frac{1}{2}x\) into \(f(x)\): \(f(g(x)) = f\left(-\frac{1}{2}x\right) = -2\left(-\frac{1}{2}x\right)\). Simplifying, we have: \(-2\times-\frac{1}{2}x = x\). Thus, \(f \circ g(x) = x\).

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

Substitute \(f(x) = -2x\) into \(g(x)\): \(g(f(x)) = g(-2x) = -\frac{1}{2}(-2x)\). Simplifying, we have: \(-\frac{1}{2} \times -2x = x\). Therefore, \(g \circ f(x) = x\).

Step 4: Conclude the Solution

We have calculated both compositions: - \(f \circ g(x) = x\)- \(g \circ f(x) = x\) Thus, the two given functions are indeed inverses of each other as the compositions yield the identity function.

Key Concepts

Composite Function

When people hear about composite functions, they might feel a bit overwhelmed, but there’s really nothing to worry about! A composite function is all about doing one thing after another. In simple terms, it means using two functions in sequence: you feed the output of the first function as input into the second one. Let’s break it down.- If you have two functions, say \( f(x) \) and \( g(x) \), the composite function \((f \circ g)(x)\) means you will first calculate \( g(x) \) and then use the resulting value to find \( f(g(x)) \).- For example, in the given exercise, \( g(x) = -\frac{1}{2}x \). We first use this function to transform \( x \) before we apply the function \( f(x) = -2x \).Why is it significant?- Composite functions help explore how different functions interact with each other.- By understanding the resulting value from a composite function, you can study changes in data and unexpected results in calculations.Look at the composite function as a mini-process chain. Each block represents a step that must occur in a specific order.

Identity Function

Now, let's talk about the identity function. The identity function stands out because it essentially does nothing to what it’s given: it returns the value that it receives as input. In mathematical terms:- The identity function is represented usually by \( I(x) = x \).So why is the identity function important?- It acts as a neutral element in function compositions. When you combine functions in such a way that the final result is an identity function, the functions you started with are likely inverses.- This is exactly what happened in the given exercise. When the composite functions \( (f \circ g)(x) \) and \( (g \circ f)(x) \) both equate to \( x \), we identify them as forming an identity function.Ah, the identity function! It's like standing at a mirror. Whatever goes in, comes out the same. It reaffirms that certain functions are capable of perfectly undoing each other when combined appropriately.

Function Composition

Function composition might remind us of layering, like putting one sandwich ingredient on top of the other. It’s essentially combining two functions. When you compose functions, you're stacking them to see how one function affects another.- Think about \( (f \circ g)(x) \) and \( (g \circ f)(x) \): the difference here lies in which function you apply first.- A proper understanding can aid in identifying inverse relationships between functions.The composition is dynamic:- By evaluating each function in order, the total effect may change drastically depending on the sequence.Consider these steps:1. Apply one function to obtain a result.2. Plug that result into the next function.The beauty of function composition, similar to the functions in the exercise, reflects in its ability to reveal inverse pairs which lead to the output returning to its original value, so each initial input is exactly as it was before any transformations.