Problem 40 - College Algebra

Question

Find functions \(f(x)\) and \(g(x)\) so the given function can be expressed as \(h(x)=f(g(x))\). $$ h(x)=\left(\frac{1}{2 x-3}\right)^{2} $$

Step-by-Step Solution

Answer

The functions are \(f(x) = x^2\) and \(g(x) = \frac{1}{2x-3}\).

Step 1: Recognize the Form

The expression for \(h(x)\) is \(h(x)=\left(\frac{1}{2 x-3}\right)^{2}\). This suggests a function of another function, i.e., \(f(g(x))\). Look for expressions that can be identified as \(g(x)\) and \(f(x)\).

Step 2: Identify the Inner Function

The expression inside the parentheses can be considered as the inner function. So, let \(g(x) = \frac{1}{2x-3}\). This simplifies the problem, so now seek a function \(f(x)\) that could operate on \(g(x)\) to yield \(h(x)\).

Step 3: Determine the Outer Function

Given \(h(x)=\left(\frac{1}{2 x-3}\right)^{2}\), and knowing \(g(x) = \frac{1}{2x-3}\), we need \(f(x) = x^2\) since squaring \(g(x)\) results in \(h(x)\). Thus, \(f(x) = x^2\).

Step 4: Verify the Composition

Compose \(f(g(x))\) to verify the function. Substituting \(g(x)\) into \(f(x)\), we get:\[ f(g(x)) = (g(x))^2 = \left(\frac{1}{2x-3}\right)^2 = h(x) \]This matches the original function \(h(x)\), confirming our choices of \(f\) and \(g\).

Key Concepts

Inner Function

In the context of function composition, the **inner function** is the function that gets evaluated first. It acts as the input to another function, which is the outer function. Imagine that every time you have a composite function like \( h(x) = f(g(x)) \), there's a two-step journey happening. The inner function, denoted here by \( g(x) \), is what starts this journey.

In the given exercise, the challenge is to identify \( g(x) \) from \( h(x) = \left(\frac{1}{2x-3}\right)^2 \). Here, the expression \( \frac{1}{2x-3} \) inside the parentheses is chosen as the inner function. Thus, we set:

\[ g(x) = \frac{1}{2x-3} \]

This decision simplifies our process because it allows us to focus on transforming \( g(x) \) into \( h(x) \) using another function, the outer function. Recognizing the inner function is about identifying which part of the composite expression gets processed first.

Outer Function

The **outer function** comes into play after the inner function has been processed. It takes the result of the inner function and further transforms it to get the final output of the composition. In our scenario, after determining the inner function \( g(x) = \frac{1}{2x-3} \), the next step is to identify a suitable outer function \( f(x) \) that, in combination with \( g(x) \), recreates \( h(x) \).

We observe here that \( h(x) = \left(\frac{1}{2x-3}\right)^2 \) indicates that the entire result of \( g(x) \) is taken to the power of 2. Therefore, the outer function is simply squaring the outcome of \( g(x) \). So, we choose:

\[ f(x) = x^2 \]

By composing these, where \( f(g(x)) = (g(x))^2 = \left(\frac{1}{2x-3}\right)^2 = h(x) \), we verify that our selection of \( f(x) \) correctly represents the transformation required to achieve \( h(x) \). The outer function is crucial as it completes the second phase of transformation in the function composition process.

Function Decomposition

Function decomposition is like solving a puzzle, where you break down a complex function into simpler components. It involves identifying the inner and outer functions so that the original function can be expressed as a composition of these simpler functions. This makes complex problems more approachable by dealing with one function at a time.

In this exercise, function decomposition is applied to express \( h(x) = \left(\frac{1}{2x-3}\right)^2 \) in the form of \( f(g(x)) \). The steps followed are a great example of decomposition:

• Identify the expression that should be isolated as the inner function \( g(x) \).
• Determine what operation is performed on \( g(x) \) to produce \( h(x) \), which becomes \( f(x) \).
• Verify that combining \( f(x) \) and \( g(x) \) recreates the original function \( h(x) \).

Through decomposition, complex functions are broken down, making them easier to analyze or compute. This method is widely used in mathematics to simplify functions and solve equations more efficiently. By leveraging this approach, problems that initially seem daunting become manageable.