Problem 40 - Contemporary Precalculus

Question

Write the given function as the composite of two functions, neither of which is the identity function, as in Examples 6 and 7 . (There may be more than one way to do this.) $$g(x)=\sqrt{x+3}-\sqrt[3]{x+3}$$

Step-by-Step Solution

Answer

Question: Decompose the given function, \(g(x) = \sqrt{x+3}-\sqrt[3]{x+3}\), into the composite of two functions. Answer: The given function can be decomposed into the composite of two functions: \(f(x) = \sqrt{x}-\sqrt[3]{x}\) and \(h(x) = x+3\), such that \(g(x) = f(h(x))\).

Step 1: Define h(x) as the square root and cube root function combined

Let us define the function \(h(x)\) as: $$h(x) = x + 3$$ Our goal here is to transform h(x) so that it always takes a positive value while also ensuring that it consistently recurs when the given function is graphed.

Step 2: Define f(x) based on h(x)

Since we want to find the function for \(f(x)\) based on h(x) such that \(g(x) = f(h(x))\), we have: $$g(x) = f(h(x)) = f(x + 3)$$ To find \(f(x)\), we rewrite the given function \(g(x)\) in terms of \(h(x)\): $$g(x) = \sqrt{x+3}-\sqrt[3]{x+3} = \sqrt{h(x)}-\sqrt[3]{h(x)}$$ Now, we can define the function \(f(x)\) as: $$f(x) = \sqrt{x} - \sqrt[3]{x}$$ Thus, the composite of the two functions is: $$g(x) = f(h(x)) = \sqrt{x+3}-\sqrt[3]{x+3}$$ Now, we have successfully decomposed the given function into the composite of two functions, \(f(x)\) and \(h(x)\), neither of which is the identity function.

Key Concepts

Function Decomposition

Function decomposition is essentially about breaking down a complex function into simpler, more manageable parts. This process helps in understanding and solving advanced precalculus functions by dissecting them into their fundamental components. In our given exercise, we see this decomposition in action. By breaking the function \( g(x) = \sqrt{x+3} - \sqrt[3]{x+3} \) into two simpler functions, \( f(x) \) and \( h(x) \), we simplify our problem.

• We first define \( h(x) \) as \( x + 3 \), a simple linear function.
• This allows us to express \( g(x) \) as a composition of functions, \( g(x) = f(h(x)) \).

By rewriting \( g(x) \) as \( f(x+3) \), we can easily manipulate the function for various solutions and discussions. Function decomposition serves as a foundational tool in higher mathematics by facilitating easier manipulation and analysis.

Precalculus

Precalculus serves as a bridge to understanding calculus and other higher-level math concepts. It establishes groundwork through topics like functions, trigonometry, and algebra, which are essential before tackling calculus. Precalculus involves examining various types of functions, such as polynomial, rational, exponential, and logarithmic functions.

• Function analysis focuses on understanding how different functions operate and interconnect, allowing students to approach complex problems logically.
• Precalculus provides students with the tools to employ function composition and decomposition effectively, as seen in the exercise example \( g(x) \).

Students gain critical problem-solving skills that are vital in calculus, physics, and engineering through their work with functions in precalculus. Understanding these concepts makes it easier to visualize and solve complex equations by breaking them down into simpler components.

Function Composition

Function composition is combining two or more functions to create a new one. It forms a crucial part of the toolkit in precalculus used for solving intricate problems. When working with composed functions, you need to understand how each function affects the input and output of another.
In the exercise, \( g(x) = f(h(x)) \), where:

• \( h(x) = x + 3 \) handles input modification by shifting it positively.
• \( f(x) = \sqrt{x} - \sqrt[3]{x} \) manages the modification of the function's output after processing through \( h(x) \).

Function composition allows these two simpler functions to work in tandem, forming the complex function \( g(x) \). It also foreshadows operations in calculus, like differentiation and integration, where multiple layers of functions interact closely. Mastering function composition ensures a firm grasp of advanced mathematical concepts.