Problem 54 - Algebra and Trigonometry

Question

Express the function in the form \(f \circ g\) $$ H(x)=\sqrt{1+\sqrt{x}} $$

Step-by-Step Solution

Answer

\(H(x) = f(g(x))\) where \(f(x) = \sqrt{1 + x}\) and \(g(x) = \sqrt{x}\).

Step 1: Identify the Inner Function

To express \(H(x)\) as a composite function \(f(g(x))\), we need first to identify the inner function. Since the expression \(1 + \sqrt{x}\) is nested inside the square root, let's choose \(g(x) = \sqrt{x}\).

Step 2: Define the Intermediate Function

We now define an intermediate expression which will be used as an input for the outer function. Substitute \(\sqrt{x}\) in the part \(1 + \sqrt{x}\) to form \(u(x) = 1 + g(x) = 1 + \sqrt{x}\).

Step 3: Identify the Outer Function

Now, the expression \(\sqrt{1 + g(x)}\) is what we need to complete the function. This expression is \(f(u) = \sqrt{u}\). Hence, \(f(x) = \sqrt{x}\).

Step 4: Express as Composition

Express \(H(x)\) as a composition of the two functions: \(f(g(x)) = f(\sqrt{x}) = \sqrt{1 + \sqrt{x}}\). Thus, we have identified \(f(x) = \sqrt{x}\) and \(g(x) = \sqrt{x}\), resulting in \(H(x) = f(g(x))\).

Key Concepts

Inner Function

To understand the concept of an inner function in the context of composite functions, picture it as the base layer of a nested function. It's the first operation applied to the input, and its output becomes the input for the next function, known as the outer function. In the exercise, we dealt with the function \[H(x) = \sqrt{1+\sqrt{x}}\]Here, the expression inside the innermost square root, \(\sqrt{x}\), acts as our inner function, denoted as:

• \(g(x) = \sqrt{x}\)

The inner function simplifies the complexity a composite function might present at first glance. Recognizing the correct inner function is crucial as it helps in breaking down the function into manageable segments. By identifying \(\sqrt{x}\) as \(g(x)\), we can better handle the progression towards using function composition.

Outer Function

The outer function in a composite function is akin to the final layer of operations that takes in the results processed by the inner function. It's what happens after the initial simplification through the inner function. Continuing from where we identified our inner function \(g(x)\), let's determine our outer function for \[H(x) = \sqrt{1+\sqrt{x}}\]After applying \(g(x) = \sqrt{x}\), the resulting expression is \(1 + g(x)\). The operations don't stop here; this expression is further processed by another function, the square root in this case, making our outer function:

• \(f(u) = \sqrt{u}\)

Thus, the outer function takes the result from \(g(x) = \sqrt{x}\) and produces the final output of the composite function. Identifying this correctly allows for a clearer path to forming and understanding the entire function composition system.

Function Composition

Function composition is a fascinating and powerful concept in mathematics where you apply one function to the results of another. It's like combining two operations into a single, smooth process. Consider it as creating more sophisticated functions from simple ones. In the given exercise, we expressed \[H(x) = \sqrt{1 + \sqrt{x}}\]as a composite function, denoted by \(f(g(x))\). Here’s a brief breakdown:

• The inner function \(g(x) = \sqrt{x}\)
• The outer function \(f(u) = \sqrt{u}\)

Thus, the composition is expressed as \(f(g(x)) = \sqrt{1 + \sqrt{x}}\). This means that \(g(x)\) first processes the input \(x\), and the result is then passed into \(f\) to give the final output. This ordered processing, where one function follows another, underscores the elegance and utility of function composition, making complex transformations more approachable and easier to digest.