Problem 88 - College Algebra

Question

For the following exercises, find the composition when \(f(x)=x^{2}+2\) for all \(x \geq 0\) and \(g(x)=\sqrt{x-2}\). $$(g \circ f)(a) ;(f \circ g)(a)$$

Step-by-Step Solution

Answer

Both compositions (g ∘ f)(a) and (f ∘ g)(a) simplify to a.

Step 1: Understand the Composition Notation (g ∘ f)(x)

In the composition \((g \circ f)(x)\), the function \(f(x)\) is applied first, and then the result is used as the input for \(g(x)\). This can be written as \(g(f(x))\). We need to substitute \(f(x) = x^2 + 2\) into \(g(x)\).

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

Given \(f(x) = x^2 + 2\), substitute into \(g(x)\) which is \(g(y) = \sqrt{y-2}\). Therefore, substitute \(y\) for \(f(x)\): \(g(f(x)) = \sqrt{(x^2 + 2) - 2} = \sqrt{x^2}\).

Step 3: Simplify g(f(x))

Simplifying \(\sqrt{x^2}\) gives \(x\) since we assume \(x\) is non-negative based on \(x \geq 0\). Thus, \((g \circ f)(x) = g(f(x)) = x\).

Step 4: Understand the Composition Notation (f ∘ g)(x)

In the composition \((f \circ g)(x)\), the function \(g(x)\) is applied first, and then the result is used as the input for \(f(x)\). This can be written as \(f(g(x))\). We need to substitute \(g(x) = \sqrt{x-2}\) into \(f(x)\).

Step 5: Substitute g(x) into f(x)

Given \(g(x) = \sqrt{x-2}\), substitute into \(f(x)\) which is \(f(y) = y^2 + 2\). Therefore, substitute \(y\) for \(g(x)\): \(f(g(x)) = (\sqrt{x-2})^2 + 2\).

Step 6: Simplify f(g(x))

Simplifying \((\sqrt{x-2})^2\) gives \(x-2\). Therefore, \(f(g(x)) = (x - 2) + 2 = x\). Thus, \((f \circ g)(x) = f(g(x)) = x\).

Key Concepts

Composition of Functions

Composition of functions is a mathematical operation where one function's output becomes the input of another. It's like a sequence of actions. First, you perform the operation of one function, and then you use that result in another function. Let's break it down with an example from our exercise:
Imagine you have two functions:

• Function \( f(x) = x^2 + 2 \)
• Function \( g(x) = \sqrt{x-2} \)

If you want to find the composition \((g \circ f)(x)\), you first compute \(f(x)\) and use its result as an input into \(g(x)\). Mathematically, this looks like \(g(f(x))\). Specifically, for our example:

• First, calculate \(f(x)\), which results in \(x^2 + 2\).
• Insert the result into \(g(x)\): \(g(f(x)) = \sqrt{x^2}\).

Thus, \((g \circ f)(x) = x\). Remember, function composition is about chaining operations efficiently.

Domain and Range

Understanding the domain and range of composed functions is vital. The domain represents all possible input values, while the range denotes possible output values.
In the exercise, for the composition function \( (g\circ f)(x) \), we start by considering the domains of \(f\) and \(g\):

• Function \(f(x) = x^2 + 2\) has a domain \(x \geq 0\).
• Function \(g(x) = \sqrt{x - 2}\) requires \(x \geq 2\) to ensure the square root is defined.

The composition \((g\circ f)(x) = x\) implies both \(g(f(x)) = \sqrt{x^2}\), valid for all \(x \geq 0\), as squaring and taking a square root work cleanly over non-negative values. Consequently, the domain of \((g\circ f)(x)\) is \(x \geq 0\). For \((f\circ g)(x) = x\), ensure the result of \(g(x)\) fits in \(g\) and \(f\)'s domains for smooth transitions and valid calculations.

Function Operations

Function operations include addition, subtraction, multiplication, division, and composition. Each technique allows functions to combine, yielding varied results. We'll focus here on composition as per our exercise.
In composing functions, each function operation affects the others:

• Addition: \((f + g)(x) = f(x) + g(x)\)
• Subtraction: \((f - g)(x) = f(x) - g(x)\)
• Multiplication: \((f \cdot g)(x) = f(x) \cdot g(x)\)
• Division: \((f / g)(x) = \frac{f(x)}{g(x)}\), provided \(g(x) eq 0\).
• Composition: \((g \circ f)(x) = g(f(x))\).

In our example, composition does more than add or multiply; it links the sequence of calculations. For instance, \((g \circ f)(x)\) streamlined results in operations \(f(x)\) in \(g(x)\) directly, offering an intersection of processes rather than mere arithmetic.