Problem 55 - Precalculus with Limits

Question

In Exercises 53-60, find two functions \(f\) and \(g\) such that \((f \circ g)(x) = h(x)\). (There are many correct answers.) \(h(x) = \sqrt[3]{x^2-4}\)

Step-by-Step Solution

Answer

The two functions that can make up \(h(x) = \sqrt[3]{x^2-4}\) could be \(f(x) = \sqrt[3]{x}\) and \(g(x) = x^2 - 4\).

Step 1: Identify the Inner Function

Firstly, we identify the inner function, which is the function that is within the cube root. In this case, we choose \(g(x) = x^2 - 4\). Here, \(x^2 - 4\) is our \(g(x)\), which is the function inside the cube root.

Step 2: Identify the Outer Function

Secondly, we identify the outer function, which in this case is the cube root. Hence, we can take \(f(x) = \sqrt[3] {x}\).

Step 3: Check the Composition

Finally, to validate our selected functions \(f\) and \(g\), we compose the functions \(f(g(x))\). Substituting \(g(x)\) into \( f(x)\), we get \(f(g(x)) = f(x^2 - 4) = \sqrt[3] {x^2 - 4}\). And this matches with our provided function \(h(x)= \sqrt[3]{x^{2}-4}\).

Key Concepts

Inner Function

The inner function is a fundamental concept when dealing with function composition. Imagine function composition like layers of an onion, where the inner layer, or the core, represents our inner function. This inner function is encapsulated and transformed by another, outer function.

In our example, we have a function expressed as \( h(x) = \sqrt[3]{x^2-4} \). Here, the expression \( x^2 - 4 \) is considered our inner function, which we denote as \( g(x) = x^2 - 4 \).

• It represents the part of the expression that is found inside another operation, such as a root or an exponent.
• Identifying the inner function helps to simplify and manage complex operations, making it easier to perform calculations involving the composite function.

By designating the inner function, we establish a base that is then modified by the outer operation, setting the stage for the concept of the outer function and composition.

Outer Function

The outer function builds upon the result of the inner function, applying another operation to complete the function composition process. It acts like the outside layer of the onion, enveloping and processing the core operation.

In the function \( h(x) = \sqrt[3]{x^2-4} \), after identifying \( x^2 - 4 \) as the inner function, we recognize the cube root \( \sqrt[3]{x} \) as our outer function. We denote this as \( f(x) = \sqrt[3]{x} \).

• The outer function applies an operation to the result of the inner function.
• It helps in constructing a new function by transforming the output of the inner function into a final desired form.

This transformation by the outer function is crucial as it completes the overall function composition, bridging the inner function with the final output.

Composite Functions

Composite functions represent the harmonious blend of the inner and outer functions. Combining these functions systematically, we achieve a new expression or function, where one function affects or operates on the results of another.

Consider our original function \( h(x) = \sqrt[3]{x^2-4} \). In this case, the composite function \((f \circ g)(x)\) is formed. This notation represents the function composition where \( g(x) = x^2 - 4 \) is computed first, and then \( f(x) = \sqrt[3]{x} \) is applied to the output of \( g(x) \).

• The idea is to observe how one function directly depends on the output of another.
• Composite functions are extremely useful in mathematics as they simplify complex function structures, enabling easier manipulation and calculation.
• The validation step involves ensuring that the composition \( f(g(x)) \) recreates our original \( h(x) = \sqrt[3]{x^2-4} \), confirming the correct interaction between \( f \) and \( g \).

Understanding the concept of composite functions allows one to break down, simplify, and analyze complex expressions effectively.