Problem 39 - 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 \(h(x)=f(g(x))\). \(h(x)=\frac{1}{(x-2)^{3}}\)

Step-by-Step Solution

Answer

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

Step 1: Identify the Inner Function

To express the given function \( h(x) \) as a composite function \( f(g(x)) \), we need to identify an inner function. Inspecting \( h(x) = \frac{1}{(x-2)^{3}} \), the expression \( (x-2) \) can be chosen as the inner function since it's a simple operation applied to \( x \). Let \( g(x) = x-2 \).

Step 2: Define the Outer Function

Once we have the inner function \( g(x) = x-2 \), the next step involves figuring out the outer function \( f(x) \). Substituting \( x-2 \) into the place of \( x \) in \( h(x) \), we get \( \frac{1}{(g(x))^3} \). Thus the outer function is \( f(u) = \frac{1}{u^3} \), where \( u = g(x) \).

Step 3: Combine Functions

Finally, combine the functions as \( f(g(x)) \). Since we have \( g(x) = x-2 \) and \( f(u) = \frac{1}{u^3} \), we can write \( h(x) = f(g(x)) = f(x-2) = \frac{1}{(x-2)^3} \), which confirms our decomposition is correct.

Key Concepts

Inner Function

In the context of composite functions, the inner function is the function that is applied to the variable first. When dealing with composite functions, it is important to identify the inner function correctly as it sets the groundwork for understanding the entire decomposition process.

For the given exercise where we have the function \( h(x) = \frac{1}{(x-2)^{3}} \), the inner function is the expression that simplifies the operation within the function. By looking at \( h(x) \), we can see that the manipulation inside the cube is \( x-2 \).

This tells us that the inner function is \( g(x) = x - 2 \).

Understanding the role of the inner function helps us to break down complex functions into simpler parts, making it easier to manage and understand subsequent operations. The selection of the inner function often relates to recognizing a repeated or nested pattern within the original function that can be isolated for simplification.

Outer Function

The outer function in a composite function is applied after the inner function. It acts upon the result of the inner function, and together they form the complete composite function.

Once we have determined that the inner function is \( g(x) = x - 2 \), we need to identify what happens after this transformation. In our example, we have \( h(x) = \frac{1}{(x-2)^{3}} \) which can be rewritten in terms of the inner function as \( \frac{1}{(g(x))^3} \).

This indicates that the outer function is \( f(u) = \frac{1}{u^3} \), where \( u \) represents the output from the inner function \( g(x) \).

By separating the outer function, we can see how it transforms the intermediate results of the inner function to complete the process, making it a crucial part of understanding the structure of composite functions.

Function Decomposition

Function decomposition is the process of breaking down a composite function into two or more simpler functions. This is usually done to make manipulation and analysis of the original function more straightforward.

In our example, we decomposed \( h(x) = \frac{1}{(x-2)^{3}} \) into two parts: an inner function \( g(x) = x-2 \) and an outer function \( f(u) = \frac{1}{u^3} \).

To complete function decomposition, we carefully analyze the structure of \( h(x) \) to find components that can stand independently. Then, confirm that they can be recombined to recreate the original function by substituting the inner function into the outer one:

• Start with recognizing patterns or transformations within \( h(x) \).
• Isolate these into distinct functions such as \( g(x) \) and \( f(u) \).
• Reconstruct the original function to ensure accuracy.

Function decomposition aids in deepening understanding, simplifying calculations, and solving complex problems by reducing them to fundamental operations that are easier to evaluate.