Question

Find \(f\) and \(g\) so that \(p=f \circ g\). (a) \(p(x)=\frac{2}{\left(x^{2}+x+1\right)^{3}}\) (b) \(p(x)=\frac{1}{x^{3}+3 x}\)

Short answer

(a) \( f(u) = \frac{2}{u^3} \), \( g(x) = x^2 + x + 1 \); (b) \( f(u) = \frac{1}{u} \), \( g(x) = x^3 + 3x \).

Step-by-step solution

Understanding Composite Functions

To decompose a function into two functions, say \( f(x) \) and \( g(x) \), such that \( p(x) = f(g(x)) \), we need to recognize the composition of functions. Here, \( g(x) \) is first applied and then \( f \) is applied to that result.

Decompose Function (a)

For \( p(x) = \frac{2}{(x^2 + x + 1)^3} \), identify the part of the function to be \( g(x) \). Let's set \( g(x) = x^2 + x + 1 \) and \( f(u) = \frac{2}{u^3} \), where \( u = g(x) \). Verify that \( f(g(x)) = \frac{2}{((x^2 + x + 1))^3} = p(x) \), matching the original function.

Decompose Function (b)

For \( p(x) = \frac{1}{x^3 + 3x} \), let's set \( g(x) = x^3 + 3x \) and \( f(u) = \frac{1}{u} \), where \( u = g(x) \). Verify that \( f(g(x)) = \frac{1}{x^3 + 3x} = p(x) \), ensuring we have the correct decomposition.

Key concepts

Function Decomposition

Function decomposition is like peeling an onion. You start with the outer layer (the full function) and peel away to identify its inner parts. Imagine you have a function \(p(x)\) and you want to express it as a combination of two simpler functions: \(f(x)\) and \(g(x)\). This process involves identifying a part of \(p(x)\) that can be considered as \(g(x)\) and another part that can be \(f(x)\), such that \(p(x) = f(g(x))\).

When we decompose functions, it helps break down complex expressions into manageable pieces. For example, in exercise (a) with \(p(x) = \frac{2}{(x^2 + x + 1)^3}\), we choose \(g(x) = x^2 + x + 1\) — a simpler polynomial expression. Then, the remaining part \(f(u) = \frac{2}{u^3}\), where \(u = g(x)\), completes the decomposition. This method is beneficial for analyzing and integrating complex algebraic functions.

Function Composition

Function composition is the reverse of decomposition. It’s about combining two functions into one. Specifically, you take an input, apply the first function \(g\) to it, then apply the second function \(f\) to the result. Mathematically, this is represented as \(f(g(x))\). Think of it as a two-step process:

• Step 1: Compute \(g(x)\)
• Step 2: Compute \(f(g(x))\)

For example, in exercise (b), \(g(x) = x^3 + 3x\) is applied first, followed by \(f(u) = \frac{1}{u}\), meaning we first compute \(x^3 + 3x\) and then take the reciprocal of the result, giving us \(p(x) = \frac{1}{x^3 + 3x}\). When composed, functions can take inputs through complex sequences, creating powerful mathematical outcomes.

Algebraic Functions

Algebraic functions like \(p(x) = \frac{2}{(x^2 + x + 1)^3}\) or \(p(x) = \frac{1}{x^3 + 3x}\), involve computations with polynomials and rational operations. These functions are central in algebra and precalculus where students learn to manipulate expressions involving sums, products, powers, and roots of variables.

Understanding algebraic functions is about mastering the basics of polynomial expressions. Factorization, division, and understanding function behaviors (like domains and asymptotes) become essential skills. As seen with the decomposition exercises, breaking down these functions helps reveal hidden structures, allowing for simpler computation and analysis. The ability to see both the forest and the trees is crucial in mathematics.

Precalculus Concepts

Precalculus acts as a bridge between basic algebra and more advanced calculus. Key precalculus groundwork involves mastering function operations — such as composition and decomposition — which prepare students for calculus concepts like limits and derivatives.

Students often delve into topics like polynomial and rational functions, their properties, and transformations. This includes understanding the behavior of complex functions, how they can be transformed (stretched, compressed, shifted), and stitched together via composition and decomposition. These exercises foster a deeper insight into how functions interact and pave the way for tackling calculus challenges head-on. Understanding these concepts solidifies mathematical foundations, ensuring readiness for future mathematical exploration.