Problem 30 - 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 \(h\) ). $$ h(x)=\left(x+\frac{1}{x}\right)^{5 / 2} $$

Step-by-Step Solution

Answer

Functions are: \( f(x) = x + \frac{1}{x} \), \( g(u) = u^{5/2} \).

Step 1: Understanding the Function Structure

The given function is \( h(x) = \left(x + \frac{1}{x}\right)^{5/2} \). We want to express this function as a composition of two functions \( g(f(x)) \).

Step 2: Identifying the Inner Function

Identify a possible inner function, \( f(x) \). Since \( h(x) \) involves the expression \( x + \frac{1}{x} \), set \( f(x) = x + \frac{1}{x} \).

Step 3: Determining the Outer Function

With \( f(x) = x + \frac{1}{x} \), we observe that \( h(x) = f(x)^{5/2} \). Therefore, the function \( g(u) \) must be such that \( g(u) = u^{5/2} \). This will make \( g(f(x)) = \left(x + \frac{1}{x}\right)^{5/2} \), which matches \( h(x) \).

Step 4: Verification

Compose the functions to check the result. If we let \( f(x) = x + \frac{1}{x} \) and \( g(u) = u^{5/2} \), then \( g(f(x)) = g\left(x + \frac{1}{x}\right) = \left(x + \frac{1}{x}\right)^{5/2} \), which is indeed \( h(x) \).

Key Concepts

Composite Functions

A composite function is essentially a combination where one function is applied to the result of another. Think of it like putting one machine's output directly into another machine. In mathematics, we use the notation \( g \circ f \) to denote the composition of two functions \( f(x) \) and \( g(x) \). This represents applying \( f(x) \) first, and then applying \( g(x) \) to the result.

Composite functions help us describe complex operations in a manageable way. Consider a scenario where measuring a final quantity requires multiple steps - each step can be represented as a function, and these can be composed into a single operation.

For example, in the exercise given, we create a composite function \( h(x) = (g \circ f)(x) = g(f(x)) \), which breaks down a complex expression into simpler, interpretable functions.

Outer Function

The outer function in a composite function setup is the function that is applied last. It operates on the result produced by the inner function. Think of it as the final layer in a process where each layer performs a different task.

In mathematical terms, if you have a composition \( g(f(x)) \), \( g \) is your outer function. This final function takes the output of \( f(x) \) and performs its operation, thereby giving us the complete result.

In our specific example, \( g(u) = u^{5/2} \) acts as the outer function. It takes an input \( u \), which is the result from the inner function \( f(x) = x + \frac{1}{x} \), and computes \( u^{5/2} \). Hence, \( g \) decides the final form of the expression. Being the last step, getting \( g \) correct ensures the composite function \( h(x) \) matches the original challenge.

Inner Function

An inner function is the function applied first within the composite function framework. It's like a preparatory step that transforms or adjusts the input before the outer function takes over.

In a composition \( g(f(x)) \), \( f(x) \) is the inner function. Its role is to take a direct input \( x \) and modify it in a specific way, setting the stage for the outer function.

For the exercise at hand, \( f(x) = x + \frac{1}{x} \) was identified as the inner function. This choice captures the core transformation of the input variable, simplifying \( h(x) = \left(x + \frac{1}{x}\right)^{5/2} \) into something more manageable for further manipulation by the outer function \( g(u) \). The idea is to break a tough problem into smaller, easier steps, and the inner function is crucial in beginning this breakdown effectively.