Problem 12 - Intermediate Algebra

Question

Complete each solution. Let \(f(x)=3 x-1\) and \(g(x)=2 x+3 .\) Find \(f \circ g\). $$ \begin{aligned} (f \circ g)(x) &=f(\square) \\ &=f(\square)\\\ &=3(\square)-1 \\ &=\square+\square-1 \\ (f \circ g)(x) &=6 x+8 \end{aligned} $$

Step-by-Step Solution

Answer

The composition \((f \circ g)(x)\) is equal to \(6x + 8\).

Step 1: Understand Function Composition

The expression \((f \circ g)(x)\) means \(f(g(x))\). This means you are substituting \(g(x)\) into every instance of \(x\) in the function \(f(x)\).

Step 2: Substitute g(x) Into f(x)

Start by substituting \(g(x) = 2x + 3\) into \(f(x) = 3x - 1\). So, \(f(g(x)) = f(2x + 3)\).

Step 3: Apply f to g(x)

Substitute \(2x + 3\) in place of \(x\) in the expression \(f(x) = 3x - 1\), so it becomes \(f(2x + 3) = 3(2x + 3) - 1\).

Step 4: Simplify the Expression

Distribute and simplify the expression: \(3(2x + 3) - 1 = 6x + 9 - 1\).

Step 5: Final Calculation

Simplify \(6x + 9 - 1\) to \(6x + 8\). Thus, \((f \circ g)(x) = 6x + 8\).

Key Concepts

Intermediate Algebra Essentials

Intermediate algebra acts as a bridge between basic algebra and more advanced topics in mathematics. It covers skills and concepts that are essential for solving practical problems and for understanding advanced concepts later on. In this exercise, finding the composite function \((f \circ g)(x)\) draws on these intermediate algebra skills. We use operations involving algebraic expressions, such as substitution and simplification.

• Substitution: This involves taking one expression and replacing variables with another expression, just as we replaced \(g(x)\) in \(f(x)\).
• Simplification: This step requires arithmetic skills to combine and streamline expressions, which in our exercise meant expanding and combining like terms.

Mastering these skills is crucial to achieving proficiency in algebra, paving the way for more complex mathematics.

Understanding Algebraic Functions

Algebraic functions are expressions that define relationships between numbers using variables, constants, and operations like addition and multiplication. For example, in this exercise, we have two functions: - \(f(x) = 3x - 1\)- \(g(x) = 2x + 3\)These expressions showcase how each function manipulates the input \(x\).

• Linear Formations: Both \(f(x)\) and \(g(x)\) are linear because their highest degree is 1. They form straight lines when graphed on a coordinate plane.
• Operations Involved: The operations include multiplication of \(x\) by a constant and adjustment by addition or subtraction of another fixed number.

Functions like these form the backbone of formula manipulation in algebra, enhancing problem-solving capability and reasoning skills.

The Magic of Composite Functions

Composite functions involve the operation of one function within another, denoted as \((f \circ g)(x)\) which means \(f(g(x))\). In our exercise, this means that we take the output of \(g(x)\) and use it as the input for \(f(x)\). Here’s how the magic happens:1. **Substitute:** We start with \(g(x) = 2x + 3\), and substitute it into \(f(x) = 3x - 1\). This substitution creates a nested scenario: \(f(2x + 3)\).2. **Layer of Simplicity:** By replacing \(x\) in \(f(x)\) with the entire expression \(2x + 3\), we layer the complexity, resulting in merging two function operations: \(3(2x + 3) - 1\).3. **Simplification: - Distribute the 3: \(6x + 9\) - Adjust by subtracting 1: \(6x + 8\)Composite functions enhance the understanding of how different functions interact. They're essential in modeling complex systems in calculus and beyond, enabling students to tackle real-world scenarios with interconnected operations.Take time to practice these steps yourself—you will see how each function transformation opens a new window into understanding mathematics.