Problem 39 - College Algebra

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

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

Step 1: Understand the Composition Concept

We need to express the function \( h(x) = \frac{1}{(x-2)^{3}} \) as a composition of two functions, \( h(x) = f(g(x)) \). This means we must define two functions \( f(x) \) and \( g(x) \) such that when the function \( g(x) \) is plugged into \( f(x) \), it produces \( h(x) \).

Step 2: Identify Inner Function \( g(x) \)

Notice that \( (x-2) \) appears as part of an exponentiation expression within \( h(x) \). It's often useful to assign this kind of expression within a fraction to the inner function. We set \( g(x) = x-2 \). This choice simplifies understanding the inner transformations affecting \( h(x) \).

Step 3: Determine Outer Function \( f(x) \)

Now that we have \( g(x) = x - 2 \), the function \( h(x) \) can be rewritten using \( g(x) \) as \( h(x) = \frac{1}{g(x)^3} \). Thus, the outer function must transform \( g(x) \) to match \( h(x) \), leading to \( f(x) = \frac{1}{x^3} \).

Step 4: Verify the Composition

Substitute \( g(x) = x-2 \) into \( f(x) = \frac{1}{x^3} \) to ensure it recreates \( h(x) \). By substituting, \( f(g(x)) = f(x-2) = \frac{1}{(x-2)^3} \), which matches \( h(x) = \frac{1}{(x-2)^3} \). The decomposition \( f(x) \) and \( g(x) \) confirmed accurately form \( h(x) \).

Key Concepts

Decomposition of Functions

When dealing with function composition, the goal is to break a complex function into simpler parts. This process is known as decomposition. For example, we have the function \( h(x) = \frac{1}{(x-2)^3} \). Decomposition allows us to express \( h(x) \) in terms of two separate functions, often called \( f(x) \) and \( g(x) \). This helps simplify the function and understand its structure more intuitively.

• First, identify segments of the function that can be separated, like a fraction or exponent.
• Assign those segments to individual functions.
• Verify by composing the functions again to ensure they form the original function.

Through decomposition, complex expressions become more manageable, making it easier to solve problems or apply further mathematical techniques.

Inner and Outer Functions

In the context of function composition, identifying the inner and outer functions is crucial. The inner function is the one you first apply, working inside-out. Consider \( h(x) = \frac{1}{(x-2)^3} \). Here, the term \( x-2 \) is a natural choice for the inner function. Hence, we define \( g(x) = x-2 \). The outer function is the one that acts on the result of the inner function.

• Inner Function \( g(x) = x-2 \)
• Outer Function \( f(x) = \frac{1}{x^3} \)

In this example, the outer function \( f(x) \) transforms \( g(x) \) to reflect the original function \( h(x) \). This compositional mapping \( f(g(x)) \) reproduces the structure of \( h(x) \) through a sequence of operations. It's much like following a series of instructions, where you first modify \( x \) by subtracting 2, then apply an exponentiation and inverse function.

Exponentiation in Functions

Exponentiation is a mathematical operation that involves raising numbers to powers. In functions, exponentiation can shape the behavior and rate of change. In \( h(x) = \frac{1}{(x-2)^3} \), notice how the expression involves making something to the third power, but then also inversely through division.

• Part of the expression \( (x-2)^3 \) is managed by the inner function.
• The cube operation indicates a rapid change in values as \((x-2)\) varies.

Exponentiation contributes to non-linear transformations within functions, adding complexity and variety to their graphs. Here, \( (x-2)^3 \) impacts the steepness of \( h(x) \), making it crucial to general function behavior analysis.