Problem 27 - Calculus with Concepts in Calculus

Question

Write \(h\) as the composite \(g \circ f\) of two functions \(f\) and \(g\) (neither of which is equal to \(h\) ). $$ h(x)=\sqrt{x-3} $$

Step-by-Step Solution

Answer

Choose \(f(x) = x-3\) and \(g(u) = \sqrt{u}\), then \(h(x) = (g \circ f)(x)\).

Step 1: Identify the Inner Function

The function \(h(x) = \sqrt{x-3}\) can be broken down into simpler parts. To do this, we first identify an inner function \(f(x)\) such that \(x-3\) is the expression inside the square root. So, we define \(f(x) = x-3\).

Step 2: Define the Outer Function

Now that we have \(f(x) = x-3\), we need a second function \(g\) that takes the output of \(f(x)\) and converts it into \(h(x)\). Notice that \(h(x) = \sqrt{f(x)}\), so we define the outer function as \(g(u) = \sqrt{u}\), where \(u\) represents the input from \(f(x)\).

Step 3: Verify the Composite Function

The composite function \((g \circ f)(x)\) means \(g(f(x))\). Substitute \(f(x) = x-3\) into \(g(u)\): \(g(f(x)) = g(x-3) = \sqrt{x-3}\). This matches \(h(x)\), confirming that \(h(x) = (g \circ f)(x)\).

Key Concepts

Inner Function

In function composition, the inner function is the initial function applied to the input variable. It acts as the base layer in the operation, performing the first step in a sequence of computations. In the context of our example with the function \(h(x) = \sqrt{x-3}\), the inner function \(f(x)\) is selected based on the expression inside the outer operation—in this instance, the expression inside the square root.

• Here, \(f(x) = x - 3\).
• This expression \(x - 3\) is the part inside the square root, and serves as the inner function because it is calculated first, before being further processed by the outer function.

Understanding the role of inner functions is crucial. They transform the initial input into a new form, which is then passed to the outer function for additional operations.

Outer Function

The outer function in function composition takes the result of the inner function as its input. It is the second layer in the sequence of operations, applying its rules to the output of the inner function. In our example of \(h(x) = \sqrt{x-3}\), the outer function \(g\) operates on the result of the inner function \(f(x) = x-3\).

• We defined the outer function as \(g(u) = \sqrt{u}\), taking \(u\) as the output from \(f(x)\).
• Once the inner operation \(f(x)\) has been completed, the outer function \(g\) takes that output (here, \(x-3\)) and applies its own operation, which in this case is to find the square root.

By identifying and applying the outer function correctly, you ensure that the composite function accurately replicates the original expression.

Composite Function

The composite function ties together the inner and outer functions, resulting in a new function that accomplishes the same task as a complex original function but in a stepwise manner. In our exercise, the goal was to express \(h(x) = \sqrt{x-3}\) in terms of two simpler functions, \(f\) and \(g\), where \(h\) serves as the composite function \(g \circ f\).

• The notation \((g \circ f)(x)\) stands for \(g(f(x))\), meaning you first apply \(f\) to \(x\), then apply \(g\) to the result.
• Substituting \(f(x) = x-3\) into \(g(u)\), we have \(g(f(x)) = g(x-3) = \sqrt{x-3}\), which perfectly reconstructs \(h(x)\).

Mastering the concept of composite functions allows you to simplify complex operations, break them down into manageable parts, and better understand the relationship between different mathematical processes.