Problem 81 - Precalculus Essentials

Question

Express the given function \(h\) as a composition of two functions \(f\) and \(g\) so that \(h(x)=(f \circ g)(x)\). $$h(x)=\frac{1}{2 x-3}$$

Step-by-Step Solution

Answer

The functions \(f\) and \(g\) that make up the composition of the function \(h(x)\) are \(g(x) = 2x-3\) and \(f(u) = \frac{1}{u}\). Replacing \(u\) with \(g(x)\) in \(f\), we get \(h(x) = \frac{1}{2x -3}\), which is the given function.

Step 1: Identify Inner Function

The first step is to identify the inner function. This function is usually a part of the original function that gets calculated first. Looking at \(h(x)=\frac{1}{2 x-3}\), this involves an operation on \(x\) before the final result gets calculated which is \(2x-3\). Therefore, let \(g(x) = 2x-3\), which is recognized as the inner function.

Step 2: Identify Outer Function

The outer function is derived from replacing the complete variable part from the original function with another variable. In this case, replacing the \(2x-3\) portion (which we have defined as \(g(x)\)) with a new variable, \(u\), would make it \(h(x)=\frac{1}{u}\). So, let's define the outer function as \(f(u) = \frac{1}{u}\). In this case, \(u\) is a placeholder for \(g(x)\).

Step 3: Compose the function

We have \(g(x) = 2x-3\) and \(f(u) = \frac{1}{u}\), which, in the composition form, can be expressed as \(h(x) = (f \circ g)(x)\) or \(h(x) = f(g(x))\). Replacing \(u\) in function \(f\) with \(g(x)\), we have \(h(x) = f(g(x)) = \frac{1}{2x - 3}\). This is the required solution.

Key Concepts

Inner Function

The inner function is a critical part in understanding function composition. In the example of expressing the function \( h(x) = \frac{1}{2x-3} \) as a composition, the inner function does some preliminary work before the outer function acts. Essentially, it takes the initial input (\( x \)) and processes it to form an intermediary result.

In this case, the inner function is \( g(x) = 2x - 3 \). It's the mathematical expression that changes \( x \) first — a simple linear transformation involving multiplication by 2, followed by subtracting 3. Thus, the output of \( g(x) \) becomes input for the next function. This layering allows us to simplify and break down complex functions into simpler parts, making analysis or calculation easier.

When identifying an inner function, look for a segment of the function that represents preliminary processing of the input.

Outer Function

The outer function completes the rest of the work in the composition. After the inner function has transformed the initial input, the outer function transforms the result from the inner function further. It typically applies the main operation required to reach the final output.

In our example, once the inner function transforms \( x \) to \( 2x - 3 \), the outer function \( f(u) = \frac{1}{u} \) takes over. Here, \( u \) is effectively a placeholder substituting \( g(x) \). Thus, the outer function computes the reciprocal of what the inner function outputs, completing the final transformation needed for \( h(x) = \frac{1}{2x-3} \).

By isolating the outer function, it becomes straightforward to focus on how the processed input is utilized, aiding in better understanding or modification of the expression.

Function Decomposition

Function decomposition is splitting a complicated function into a series of simpler functions. This stepwise breakdown helps in tackling complex algebraic problems by outlining them as smaller, manageable parts.

In the context of our function \( h(x) = \frac{1}{2x-3} \), decomposition involves identifying \( g(x) = 2x - 3 \) as the inner function and \( f(u) = \frac{1}{u} \) as the outer function. By doing so, we presented \( h(x) \) as a composition, \( (f \circ g)(x) = f(g(x)) \).

This approach serves several purposes:

• It makes complex functions easier to analyze and manipulate.
• It's useful for calculus applications, like chain rule differentiation, where knowing individual function components are essential.
• It simplifies complex computations by breaking them into sequential stages.

Function decomposition is a fundamental technique in math that underscores the importance of looking at problems in stages.