Problem 35 - College Algebra

Question

Find functions \(f(x)\) and \(g(x)\) so the given function can be expressed as \(h(x)=f(g(x))\). $$ h(x)=\sqrt{2 x+6} $$

Step-by-Step Solution

Answer

Let \(g(x)=2x+6\) and \(f(x)=\sqrt{x}\).

Step 1: Understand the Function h(x)

The given function is \(h(x) = \sqrt{2x + 6}\). Our goal is to express \(h(x)\) as the composition of two functions, \(f(x)\) and \(g(x)\), such that \(h(x) = f(g(x))\).

Step 2: Identify the Inner Function g(x)

The part inside the square root, \(2x + 6\), can be seen as the inner function. Let's choose \(g(x) = 2x + 6\). This simplifies the expression for substitution into the outer function.

Step 3: Determine the Outer Function f(x)

Given \(g(x) = 2x + 6\), the composition \(f(g(x))\) must equal \(\sqrt{2x + 6}\). Thus, the outer function \(f(x)\) should operate on the result from \(g(x)\), so we define \(f(x) = \sqrt{x}\).

Step 4: Verify the Solution

Substitute \(g(x)\) into \(f(x)\) to ensure it matches \(h(x)\): \(f(g(x)) = f(2x+6) = \sqrt{2x + 6}\). This matches the original \(h(x)\), confirming our solution.

Key Concepts

Inner Function

When dealing with function composition, an important starting point is identifying the **inner function**. Think of it as the core or the input portion of a larger expression. In this case, the function given is a square root, specifically, \( h(x) = \sqrt{2x + 6} \). Here, the expression inside the square root, \(2x + 6\), acts as the "inner" part of the whole procedure. It is the piece that we process first, before applying the subsequent operation.- To locate the inner function, observe which part of the expression is encapsulated or operated upon by another function. - Here, the expression \(2x + 6\) is under the square root, making it the natural choice for \( g(x) = 2x + 6 \).By selecting \( g(x) = 2x + 6 \), we prepare it to be passed as an input to another function, simplifying our approach to solve the problem using composition.

Outer Function

Once the inner function, \( g(x) = 2x + 6 \), is identified, the next step is recognizing the **outer function**. This is the one that "wraps around" the entire process, accepting the result from the inner function. Think of it as the final operation performed in the sequence.- In the expression \( h(x) = \sqrt{2x + 6} \), the square root symbol represents this outer action.- Hence, the outer function is \( f(x) = \sqrt{x} \), meaning it takes whatever \( g(x) \) evaluates to and processes it by taking the square root.By understanding \( f(x) = \sqrt{x} \) as the outer function, you simplify \( h(x) \) into the composition \( f(g(x)) = \sqrt{g(x)} \), highlighting how different functions combine to form a composite function.

Composite Function

The amalgamation of two or more functions results in what is known as a **composite function**. It’s essentially one function nested within another. In simpler terms, it's like executing a series of steps, where each step is separated into its function, paving the way for complex calculations through straightforward parts.- The composite function can be represented as \( h(x) = f(g(x)) \), where each component function plays a role.- For instance, with our example, since \( g(x) = 2x + 6 \) and \( f(x) = \sqrt{x} \), the composite function \( h(x) \) becomes \( f(g(x)) = \sqrt{2x + 6} \).This approach highlights how simplifying a function into its components can make complex problems more approachable, stringing together the operations defined by \( f(x) \) and \( g(x) \) seamlessly.