Problem 40 - College Algebra with Corequisite Support

Question

For the following exercises, 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

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

Step 1: Identify the Structure of the Function h(x)

The given function is \( h(x) = \left(\frac{1}{2x-3}\right)^2 \). We need to express this as a composition \( h(x) = f(g(x)) \). Recognize that the structure involves a fraction inside a power.

Step 2: Choose the Inner Function g(x)

For the inner function \( g(x) \), we can identify the inside fraction, i.e., \( g(x) = \frac{1}{2x-3} \). This is the simpler operation inside the parentheses.

Step 3: Choose the Outer Function f(x)

The outer function \( f(x) \) should take the result of \( g(x) \), and raise it to the power of 2. Therefore, choose \( f(x) = x^2 \). This represents the squaring function applied to the result of \( g(x) \).

Step 4: Verify the Composition

Compose the functions to ensure correctness: \( h(x) = f(g(x)) = f\left(\frac{1}{2x-3}\right) = \left(\frac{1}{2x-3}\right)^2 \). This matches the original function \( h(x) \).

Key Concepts

Algebraic Functions

Algebraic functions are equations made up of variables and constants, connected by arithmetic operations such as addition, subtraction, multiplication, division, and taking roots or powers. They play a vital role in mathematics, especially in calculus and algebra. Understanding algebraic functions is crucial because it allows us to model real-world situations and solve problems.

• An algebraic function can be simple like a linear equation, or complex like a rational or polynomial equation.
• In the exercise given, the function \( h(x) = \left(\frac{1}{2x-3}\right)^2 \) is an example of an algebraic function as it involves division and exponentiation.

Recognizing the form of the function helps you determine the operations involved and simplifies further manipulation.

Inner Function

In function composition, the inner function is the first function applied to the input. It feeds its output into the next function, called the outer function. Identifying the correct inner function is crucial for simplifying complex equations into manageable pieces.Here, the inner function \( g(x) = \frac{1}{2x-3} \) identifies the core of the given function.

• The choice of \( g(x) \) is straightforward because the expression \( \frac{1}{2x-3} \) is inside the parentheses in the original function \( h(x) \).
• This expression simplifies the more complex original function by breaking it into a simpler operation – a linear transformation.

Using the inner function helps lessen the computational complexity in the problem-solving process.

Outer Function

The outer function takes the result of the inner function and applies another operation to it. It completes the function composition by transforming the output of the inner function. In this exercise, \( f(x) = x^2 \) is the chosen outer function.

• It represents the final operation performed on \( g(x) \), raising it to the power of 2.
• This produces the squared expression seen in the original function \( h(x) = \left(\frac{1}{2x-3}\right)^2 \).

Understanding the role of the outer function is essential because it finalizes the composition and delivers the original function's structure.

Function Decomposition

Function decomposition is the process of breaking down a complex function into simpler parts, typically using inner and outer functions. This method proves beneficial in tackling complicated expressions or when finding derivatives and integrals.Through decomposition, you can express the original function as a combination of smaller, more manageable functions. In our exercise, the function \( h(x) = \left(\frac{1}{2x-3}\right)^2 \) is decomposed into:

• an inner function \( g(x) = \frac{1}{2x-3} \), and
• an outer function \( f(x) = x^2 \).

Each part performs a specific role, with \( g(x) \) representing the initial transformation and \( f(x) \) finalizing the expression. This breakdown makes it easier to understand, analyze, and manipulate the function.