Problem 3 - Precalculus Mathematics for Calculus

Question

If the rule of the function \(f\) is "add one" and the rule of the function \(g\) is "multiply by 2 ," then the rule of \(f \circ g\) is "_______________" and the rule of \(g \circ f\) is "_____________"

Step-by-Step Solution

Answer

The rule of \(f \circ g\) is "multiply by 2, then add 1," and \(g \circ f\) is "add 1, then multiply by 2."

Step 1: Understanding Composite Functions

The problem involves two functions, \(f\) and \(g\), with specific rules. The function \(f(x)\) adds 1 to any input \(x\), so \(f(x) = x+1\). The function \(g(x)\) multiplies any input \(x\) by 2, so \(g(x) = 2x\). Our goal is to find the rules for the composite functions \(f \circ g\) and \(g \circ f\).

Step 2: Calculating \(f \circ g\)

First, we need to evaluate the composite function \(f \circ g\). This means applying \(g(x)\) first and then applying \(f\). So, \((f \circ g)(x) = f(g(x))\). Given \(g(x) = 2x\), we substitute it into \(f\): \(f(2x) = 2x + 1\). Therefore, the rule for \(f \circ g\) is "multiply by 2, then add 1."

Step 3: Calculating \(g \circ f\)

Next, we determine \(g \circ f\), which means applying \(f(x)\) first, then applying \(g\). So, \((g \circ f)(x) = g(f(x))\). From \(f(x) = x+1\), we substitute into \(g\): \(g(x+1) = 2(x+1) = 2x + 2\). Hence, the rule for \(g \circ f\) is "add 1, then multiply by 2."

Key Concepts

Function Composition

In mathematics, function composition is a fundamental concept that involves combining two functions to form a new function. The notation for composition is typically symbolized by the small circle "∘" between the two functions. Suppose we have two functions, \( f \) and \( g \). The composite function \( f \circ g \) means you apply \( g \) first and then apply \( f \) to the result. In other words, it is expressed as \( f(g(x)) \). Similarly, \( g \circ f \) implies applying \( f \) first, followed by \( g \), which is expressed as \( g(f(x)) \).

Function composition allows us to streamline a series of operations into a single expression.

• It provides a powerful way to create complex functions from simpler ones.
• The output of the inner function becomes the input for the outer function.
• It's important to perform the operations in the given order, as reversing them can yield different results.

Understanding this concept is crucial when dealing with algebraic functions and operations involving functions.

Algebraic Functions

Algebraic functions are functions that can be formed using algebraic expressions. These expressions involve variables and constants using operations like addition, subtraction, multiplication, division, and taking roots. The functions \( f(x) = x + 1 \) and \( g(x) = 2x \) are simple examples of algebraic functions.

These functions play a vital role in various mathematical processes because:

• They are typically easy to manipulate and combine—enabling formation of new functions.
• They lay the groundwork for exploring more complex mathematical concepts.
• The rules governing them are based on fundamental arithmetic operations.
• They're used to model real-world situations where quantities vary in a predictable manner.

In our exercise, we used algebraic functions to understand how different operations affect outputs, providing a solid base for understanding compositions like \( f \circ g \) and \( g \circ f \).

Operations on Functions

Operations on functions involve applying arithmetic operations like addition, subtraction, multiplication, or division to combine two or more functions. These operations are essential for exploring how different functions can work together to produce a specific output.

For instance, consider the two functions in our exercise:

• Adding \( f(x) = x + 1 \) and \( g(x) = 2x \) results in \((f + g)(x) = (x + 1) + 2x = 3x + 1\).
• Subtracting gives \((f - g)(x) = (x + 1) - 2x = -x + 1\).
• Multiplying results in \((f \cdot g)(x) = (x + 1) \, \cdot \, 2x = 2x^2 + 2x\).

Function composition itself is an operation that focuses specifically on chaining operations, making the result of one function the input for a second. This allows for intricate function behavior through seemingly simple rules. By understanding and utilizing these operations, students can solve complicated functional equations by breaking them into simple steps, just as we've done in the solution.