Problem 87 - College Algebra

Question

Find the composition when \(f(x)=x^{2}+2\) for all \(x \geq 0\) and \(g(x)=\sqrt{x-2}\). $$ (f \circ g)(6) ;(g \circ f)(6) $$

Step-by-Step Solution

Answer

\((f \circ g)(6) = 6\), \((g \circ f)(6) = 6\).

Step 1: Understand the Composition of Functions

To solve for compositions of functions \((f \circ g)(x)\) and \((g \circ f)(x)\), recall that \((f \circ g)(x) = f(g(x))\) and \((g \circ f)(x) = g(f(x))\). This means to first apply the function \(g\) and then \(f\) for \(f \circ g\), and vice versa for \(g \circ f\).

Step 2: Compute \(g(6)\)

Given \(g(x) = \sqrt{x - 2}\), calculate \(g(6)\).\[ g(6) = \sqrt{6 - 2} = \sqrt{4} = 2 \]

Step 3: Compute \((f \circ g)(6)\)

Now substitute \(g(6)\) into \(f\) to get \((f \circ g)(6)\).\[ (f \circ g)(6) = f(g(6)) = f(2) \]Since \(f(x) = x^2 + 2\), we have:\[ f(2) = 2^2 + 2 = 4 + 2 = 6 \]

Step 4: Compute \(f(6)\)

Now calculate \(f(6)\) using \(f(x) = x^2 + 2\).\[ f(6) = 6^2 + 2 = 36 + 2 = 38 \]

Step 5: Compute \((g \circ f)(6)\)

Use the result from Step 4 to substitute in \(g(x)\).\[ (g \circ f)(6) = g(f(6)) = g(38) \]Since \(g(x) = \sqrt{x - 2}\), we substitute to find:\[ g(38) = \sqrt{38 - 2} = \sqrt{36} = 6 \]

Key Concepts

Understanding Composition of Functions

Function composition is an essential concept in mathematics that combines two functions into one. When we talk about the composition of functions, we often use the notation \((f \circ g)(x)\), which means plugging the output of one function \(g(x)\) into another function \(f(x)\). This can also be considered as applying one function to the results of another function. For example:

• \((f \circ g)(x) = f(g(x))\)
• \((g \circ f)(x) = g(f(x))\)

So, in the given exercise, if we apply \(f\) after \(g\), or \(f \circ g\), we find \(g(6)\) first and use that result in \(f\). Conversely, when we apply \(g\) after \(f\), or \(g \circ f\), we do the opposite.This method is helpful for solving problems where direct evaluation might be complicated, simplifying computations by dealing with function layers in steps.

Basics of Algebraic Functions

Algebraic functions are functions built from basic operations like addition, subtraction, multiplication, division, and taking roots. The functions in the given problem are both algebraic. Specifically, they involve squaring and root-taking operations, which are very common in algebra. Given functions:

• \(f(x) = x^2 + 2\), which involves squaring \(x\) and then adding 2.
• \(g(x) = \sqrt{x - 2}\), which involves subtracting 2 from \(x\) and then taking the square root.

Understanding these straightforward operations helps in evaluating and composing algebraic functions efficiently. These functions are part of what we often work with in algebra, enabling us to solve equations, graph curves, and model real-world situations.

Steps for Function Evaluation

Function evaluation is a process where you substitute a specific value into a function and compute the result. Let's walk through each step of evaluating a composite function using the exercise:1. **Evaluate the Inner Function**: For \((f \circ g)(x)\), start by evaluating \(g(6) = \sqrt{6 - 2} = \sqrt{4} = 2\). Once we find \(g(6)\), we move to the outer function.2. **Substitute into the Outer Function**: Next, substitute the result into \(f\), meaning compute \(f(2)\). Using \(f(x) = x^2 + 2\), we find \(f(2) = 4 + 2 = 6\).3. **Evaluate for \((g \circ f)(x)\):** - Start with \(f(6)\): Substitute 6 into \(f\) to get \(f(6) = 38\). - Then, compute \(g(38) = \sqrt{38 - 2} = \sqrt{36} = 6\).By following these steps sequentially, you can efficiently evaluate compositions and understand the interaction between the functions. Evaluations like this help to break down more complicated problems into manageable parts by leveraging both individual function results and composition.