Problem 33 - 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 \(h\) ). $$ h(x)=\sqrt{\sqrt{x}-1} $$

Step-by-Step Solution

Answer

Choose \( f(x) = \sqrt{x} - 1 \) and \( g(u) = \sqrt{u} \).

Step 1: Recognize the Form of h(x)

We need to express the function \( h(x) = \sqrt{\sqrt{x} - 1} \) as a composition of two functions \( g(f(x)) \), where neither \( g \) nor \( f \) is the same as \( h \). This means identifying an inner function \( f(x) \) and an outer function \( g(u) \) where \( u = f(x) \).

Step 2: Choose the Inner Function f(x)

Since the function \( h(x) \) involves a nested expression inside the square root, a natural choice for \( f(x) \) is the expression inside the outer square root: \( f(x) = \sqrt{x} - 1 \). This simplifies \( h(x) \) into a form that can be made simpler with another function.

Step 3: Define the Outer Function g(u)

With \( f(x) = \sqrt{x} - 1 \), the next step is to define \( g(u) \) such that \( g(f(x)) = h(x) \). Since \( h(x) = \sqrt{\sqrt{x} - 1} \), then \( g(u) \) should take \( u = f(x) = \sqrt{x} - 1 \) and transform it to \( \sqrt{u} \). Thus, \( g(u) = \sqrt{u} \).

Step 4: Verify the Composite Function

To ensure correctness, check that \( g(f(x)) = g(\sqrt{x} - 1) = \sqrt{\sqrt{x} - 1} \), which indeed simplifies back to the original function \( h(x) = \sqrt{\sqrt{x} - 1} \). This confirms that the functions \( f(x) = \sqrt{x} - 1 \) and \( g(u) = \sqrt{u} \) work together as \( h(x) = g(f(x)) \).

Key Concepts

Function Composition

In mathematics, functions are like machines that take an input, process it, and give an output. Function composition is when one function is applied to the result of another function. Imagine you have function f that takes an input x and gives you an output f(x). Another function, g, can take the output from f as its input, giving the result g(f(x)).
This process effectively "composes" the two functions into a new, single function. It's like running one machine's output directly into another machine. To express this function composition, we use the notation \((g \circ f)(x)\), which is read as "g circle f of x," meaning we apply function f first and then apply g to the result.

Inner and Outer Functions

When dealing with composite functions, it's crucial to identify their structure as having an inner and an outer function. The inner function, usually denoted as f(x), processes the initial input. After f(x) completes its task, its output becomes the input for the outer function, denoted as g(u).
In our example \( h(x) = \sqrt{\sqrt{x} - 1} \), we determined that the inner function f(x) is \( \sqrt{x} - 1 \). This choice is natural as it involves simplifying the contents under the outer square root. Once you have identified it, you can define the outer function as g(u), where u represents the result from f. Here, g(u) = \sqrt{u}, which effectively wraps around f(x), completing the composition and forming the function h.

Square Roots

The square root is a fundamental mathematical operation that finds a number which, when multiplied by itself, equals the given number. In function composition, square roots can play a pivotal role as they often represent the outer function wrapping around an inner result.
Consider the example where h(x) has a square root outside: \( h(x) = \sqrt{\sqrt{x} - 1} \). The square root, combined with composition, helps to gradually unravel nested expressions, step by step, by first evaluating the inner content \( \sqrt{x} - 1 \) and then taking the square root of that result.

• It's essential to understand that each square root simplifies the expression further.
• When dealing with nested square roots, take your time to process each layer individually.

This methodical approach ensures that complex compositions are handled logically and accurately.