Problem 28 - College Algebra

Question

For the following exercises, find functions \(f(x)\) and \(g(x)\) so the given function can be expressed as \(h(x)=f(\mathrm{g}(x))\) $$h(x)=\frac{3}{x-5}$$

Step-by-Step Solution

Answer

Choose \( f(x) = \frac{3}{x} \) and \( g(x) = x-5 \).

Step 1: Identify the Inner Function

To express the function \( h(x) = \frac{3}{x-5} \) in the form \( h(x) = f(g(x)) \), we first need to identify a suitable inner function, \( g(x) \). Here, a noticeable operation inside the function is \( x-5 \). So, let's set \( g(x) = x-5 \).

Step 2: Express the Outer Function

Now that we have \( g(x) = x-5 \), we determine \( f(x) \) such that \( f(g(x)) = h(x) \). Since \( h(x) = \frac{3}{x-5} \) and \( g(x) = x-5 \), we can replace \( x-5 \) with \( g(x) \). This means we need a function \( f(x) \) whose input produces \( \frac{3}{x} \). Therefore, \( f(x) = \frac{3}{x} \).

Step 3: Verification

Verify that substituting \( g(x) \) into \( f(x) \) gives us \( h(x) \). Substitute \( g(x) = x-5 \) into \( f(x) = \frac{3}{x} \), which gives \( f(g(x)) = \frac{3}{x-5} \). This is indeed \( h(x) \), confirming that our choices of \( f(x) \) and \( g(x) \) are correct.

Key Concepts

Inner Function

The inner function is a crucial part of function decomposition. It is the part of the function that is first applied to the input. In our example, with the function \( h(x) = \frac{3}{x-5} \), we can identify the inner function as the transformation that occurs to the variable before any additional operations. Notice the expression \( x-5 \) within the denominator. This subtraction operation happens before any further transformations, making it the inner function. Hence, we define the inner function as \( g(x) = x-5 \).

• Defines the initial transformation applied to the variable \( x \)
• Forms the input for the outer function \( f(x) \)
• In our example: \( g(x) = x-5 \)

Recognizing the inner function simplifies how we analyze complex functions by breaking them down into more manageable parts. This method helps in understanding and solving problems involving compositions.

Outer Function

Once we have identified the inner function, the next step is understanding the outer function. The outer function \( f(x) \) is applied to the result of the inner function. In the context of \( h(x) = \frac{3}{x-5} \) and starting from our inner function \( g(x) = x-5 \), the outer function is responsible for transforming \( g(x) \) into \( h(x) \). Since the transformation \( \frac{3}{x-5} \) can be seen as performing \( \frac{3}{x} \) on \( x-5 \), we define it as the outer function \( f(x) = \frac{3}{x} \).

• Applies a further transformation on the inner function's output
• In our example: \( f(x) = \frac{3}{x} \)

The outer function encompasses the final operation that yields the expressed function \( h(x) = f(g(x)) \), providing a systematic breakdown of complex expressions.

Composition of Functions

The composition of functions is the process of combining two functions where the output of one function becomes the input of another. In simple terms, it's like stacking two transformations, one after the other. This concept is expressed with notation like \( f(g(x)) \) which reads as "\( f \) of \( g \) of \( x \)." Through function composition, each element is treated as part of a sequential process.

• Processes one function's output through another function
• Described as \( h(x) = f(g(x)) \)
• Ensures the entire operation is captured in a simple expression

In our example, when we combine \( f(x) = \frac{3}{x} \) with \( g(x) = x-5 \), we obtain \( h(x) = \frac{3}{x-5} \). Composition is an essential concept as it allows for handling multi-step processes within a single function, aiding in both simplified notation and streamlined computation.