Problem 57 - Applied Calculus

Question

57-58. For each pair of functions \(f(x)\) and \(g(x)\), find and fully simplify a. \(f(g(x))\) and b. \(g(f(x))\) $$ f(x)=2 x-6 ; \quad g(x)=\frac{x}{2}+3 $$

Step-by-Step Solution

Answer

Both compositions \(f(g(x))\) and \(g(f(x))\) simplify to \(x\).

Step 1: Understand the Functions and Notation

You are given two functions, \(f(x) = 2x - 6\) and \(g(x) = \frac{x}{2} + 3\). The task is to find the compositions \(f(g(x))\) and \(g(f(x))\). Here, \(f(g(x))\) means plugging \(g(x)\) into \(f(x)\), and \(g(f(x))\) means plugging \(f(x)\) into \(g(x)\).

Step 2: Compute f(g(x))

Substitute \(g(x)\) into \(f(x)\). We have \(f(g(x)) = f\left(\frac{x}{2} + 3\right)\). Replace \(x\) in \(f(x) = 2x - 6\) with \(\frac{x}{2} + 3\) to get \(2\left(\frac{x}{2} + 3\right) - 6\).

Step 3: Simplify f(g(x))

Distribute the 2: \(2\left(\frac{x}{2} + 3\right) = 2\times\frac{x}{2} + 2\times3 = x + 6\). Now, subtract 6 from this expression: \((x + 6) - 6 = x\). Thus, \(f(g(x)) = x\).

Step 4: Compute g(f(x))

Substitute \(f(x)\) into \(g(x)\). We have \(g(f(x)) = g(2x - 6)\). Replace \(x\) in \(g(x) = \frac{x}{2} + 3\) with \((2x - 6)\) to get \(\frac{2x - 6}{2} + 3\).

Step 5: Simplify g(f(x))

Simplify \(\frac{2x - 6}{2}\) which is \(\frac{2x}{2} - \frac{6}{2} = x - 3\). Then, add 3 to \(x - 3\): \((x - 3) + 3 = x\). Therefore, \(g(f(x)) = x\).

Key Concepts

Mathematics Education

Learning how to compose functions is a vital part of mathematics education. Function composition is a concept that helps us understand how two or more functions can work together to form a new function. The process involves plugging one function into another to see what output it produces.
Here, the task was to compose two functions, which means to find both possible compositions of these functions:

• First, find \( f(g(x)) \), where you substitute the output of \( g(x) \) into \( f(x) \).
• Second, find \( g(f(x)) \), where the output of \( f(x) \) goes into \( g(x) \).

Understanding this operation is essential as it serves as a building block for higher learning in algebra and calculus, preparing students to tackle more complex mathematics with confidence. Function composition is not just a formal operation but a way to perceive one process being applied after another. This sequential action forms the cornerstone of more sophisticated concepts, such as inverse functions and transformations.

Algebra

Algebra is a cornerstone in the study of mathematics that deals with variables, expressions, and equations. The function composition exercise perfectly illustrates the application of algebraic techniques.
When solving for \( f(g(x)) = x \) and \( g(f(x)) = x \), one must substitute and simplify algebraic expressions, which helps in reinforcing the student’s ability to handle variables and constants. For instance, in \( f(g(x)) \), replacing \( g(x) = \frac{x}{2} + 3 \) into \( f(x) = 2x - 6 \) involves basic algebraic manipulations such as distribution and combining like terms:

• Distribute to manage coefficients: Apply the distributive property to expand expressions like \( 2\left(\frac{x}{2} + 3\right)\).
• Simplify to reach the conclusion: Similar terms are combined to simplify the expression to \( x \).

These steps improve fluency in algebra, making solving equations and simplifying expressions second nature, which are everyday skills in mathematics.

Calculus

Calculus relies heavily on understanding functions and their compositions. While the main focus of calculus might be limits, derivatives, and integrals, composing functions gives insight into how complex functions can behave.
Understanding function composition in calculus has several useful implications:

• **Chain Rule**: An essential technique for differentiating compositions of functions. Knowing how to compose functions in algebra helps when working with the chain rule in calculus, as it requires knowledge of how functions interact.
• **Limits and Continuity**: Composing functions allows us to discuss limits and continuity across more complicated function chains, crucial in studying behaviors of functions as they approach certain points or infinity.

By visualizing and understanding the relationships in composed functions, students develop a deeper understanding of how calculus can describe real-world phenomena, making this an invaluable concept in mathematics education.