Problem 28 - 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 \(h(x)=f(g(x))\). \(h(x)=\frac{3}{x-5}\)

Step-by-Step Solution

Answer

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

Step 1: Identify the Inner Function Component

First, we observe the given function \( h(x) = \frac{3}{x-5} \). Notice that within the fraction, there is a transformation applied to \( x \), specifically the term \( x-5 \). This suggests that a potential inner function \( g(x) \) could be represented by \( g(x) = x - 5 \).

Step 2: Define the Inner Function \(g(x)\)

Based on our observation from Step 1, define the inner function as \( g(x) = x - 5 \). This function represents a horizontal shift by 5 units to the right, which is a preliminary transformation in the composite function structure.

Step 3: Identify the Outer Function Component

Now, we look at how \( g(x) \) is incorporated into the given function \( h(x) \). The expression can be seen as \( \frac{3}{(x-5)} \), which implies that the outer function involves the operation of taking the reciprocal. Thus, define \( f(u) = \frac{3}{u} \).

Step 4: Verify the Decomposition

To ensure our functions are correctly identified, substitute \( g(x) = x - 5 \) into \( f(u) \). Compute \( f(g(x)) = f(x-5) = \frac{3}{x-5} \). This is exactly \( h(x) \), confirming that our decomposition is correct.

Key Concepts

Inner Function

To understand function composition, it helps to start with the **inner function**. This is the function that you apply first, innermost in order. In this case, we looked at the expression \( \frac{3}{x-5} \). Within the fraction, the transformation \( x-5 \) is a clue to the inner function. - Here, we defined the inner function as \( g(x) = x - 5 \). This suggests a transformation where 5 is subtracted from \( x \). Visually, this means we're shifting all values of \( x \) 5 units to the right before any other operations are applied. By identifying the core transformation within the function structure, you can systematically determine the role of the inner function in composing the whole expression.

Outer Function

Next, we explore the concept of the **outer function**. This is the operation applied to the result of the inner function. After determining \( g(x) = x-5 \), the next step is to see how this output is transformed back into the original function \( h(x) = \frac{3}{x-5} \). - Here, we identify the outer function as \( f(u) = \frac{3}{u} \). The substitution of \( u \) with \( g(x) \) leads to the composite function. In simpler terms, you are taking the result from \( g(x) \) and plugging it into an outer operation that involves taking the reciprocal of the result and multiplying it by 3. Seeing the outer function as a wrap-around process can demystify complex transformations and demonstrate the power of function layering.

Composite Function Decomposition

The magic happens when functions come together in **composite function decomposition**. This concept involves breaking down a complex function into simpler, smaller parts, **the inner and outer functions**, that work together to form the original function.Let's revisit how we decomposed \( h(x)=\frac{3}{x-5} \):- First, we identified the inner transformation \( g(x)=x-5 \).- Next, we put it into \( f(u)=\frac{3}{u} \) as the outer layer.By confirming that \( f(g(x))=\frac{3}{g(x)}=\frac{3}{x-5} \), we see that our choices accurately recreate the original function. Decomposing the function into steps allows us to understand complicated expressions better, where each step contributes to understanding transformation sequences and makes solving such expressions more manageable.