fog & gof - Algebra-1

1. Fundamental Concepts

Definition: The composition of two functions \(f\) and \(g\) is a new function where the output of one function becomes the input of another. Specifically, \(f \circ g\) (read as "f composed with g") means applying \(g\) first and then \(f\), while \(g \circ f\) means applying \(f\) first and then \(g\).
Notation: \(f \circ g(x) = f(g(x))\) and \(g \circ f(x) = g(f(x))\)
Order Matters: In general, \(f \circ g \neq g \circ f\). The order in which functions are composed can lead to different results.

2. Key Concepts

Basic Rule: \(f \circ g(x) = f(g(x))\)

Degree Preservation: The domain of \(f \circ g\) is all \(x\) in the domain of \(g\) such that \(g(x)\) is in the domain of \(f\).

Application: Used in various fields including physics, engineering, and computer science for modeling sequential processes.

3. Examples

Example 1 (Basic)

Problem: Given \(f(x) = x^2 + 1\) and \(g(x) = 2x - 3\), find \(f \circ g(x)\).

Step-by-Step Solution:

Substitute \(g(x)\) into \(f(x)\): \(f(g(x)) = (2x - 3)^2 + 1\)
Simplify: \(f(g(x)) = 4x^2 - 12x + 9 + 1 = 4x^2 - 12x + 10\)

Validation: Substitute \(x = 1\): Original: \(f(2 \cdot 1 - 3) = (2 \cdot 1 - 3)^2 + 1 = (-1)^2 + 1 = 2\); Simplified: \(4 \cdot 1^2 - 12 \cdot 1 + 10 = 4 - 12 + 10 = 2\) ✓

Example 2 (Intermediate)

Problem: Given \(f(x) = \sqrt{x}\) and \(g(x) = x^2 + 1\), find \(g \circ f(x)\).

Step-by-Step Solution:

Substitute \(f(x)\) into \(g(x)\): \(g(f(x)) = (\sqrt{x})^2 + 1\)
Simplify: \(g(f(x)) = x + 1\)

Validation: Substitute \(x = 4\): Original: \(g(\sqrt{4}) = (\sqrt{4})^2 + 1 = 4 + 1 = 5\); Simplified: \(4 + 1 = 5\) ✓

4. Problem-Solving Techniques

Visual Strategy: Use diagrams to represent the flow of function compositions.
Error-Proofing: Always check the domains of the functions involved to ensure they are compatible.
Concept Reinforcement: Practice with a variety of functions to understand how different types of functions interact under composition.