Problem 11 - Thomas Calculus

Question

Let \(f(x)=x-3, \quad g(x)=\sqrt{x}, \quad h(x)=x^{3},\) and \(j(x)=2 x .\) Express each of the functions in Exercises 11 and 12 as a composite involving one or more of \(f, g, h,\) and \(j .\) $$\begin{array}{ll}{\text { a. } y=\sqrt{x}-3} & {\text { b. } y=2 \sqrt{x}} \\\ {\text { c. } y=x^{1 / 4}} & {\text { d. } y=4 x} \\ {\text { e. } y=\sqrt{(x-3)^{3}}} & {\text { f. } y=(2 x-6)^{3}}\end{array}$$

Step-by-Step Solution

Answer

a. \(f(g(x))\), b. \(j(g(x))\), c. \(g(h(x))\), d. \(j(j(x))\), e. \(g(h(f(x)))\), f. \(h(f(j(x)))\).

Step 1: Identify the Core Functions

Examine each given function and identify which of the base functions \(f(x)=x-3\), \(g(x)=\sqrt{x}\), \(h(x)=x^{3}\), and \(j(x)=2x\) can be used in combination to form it.

Step 2: Solve Part A

The function is \(y=\sqrt{x}-3\). This is equivalent to \(f(g(x))\) because it can be broken down as follows: apply \(g(x)=\sqrt{x}\) and then apply \(f(x)=x-3\).

Step 3: Solve Part B

The function is \(y=2\sqrt{x}\). This can be expressed as \(j(g(x))\) because it involves applying \(g(x) = \sqrt{x}\) first and then applying \(j(x) = 2x\).

Step 4: Solve Part C

The function is \(y=x^{1/4}\). This can be viewed as \(g(h(x))\) by rewriting it as \((x^{1/2})^{1/3}\). Apply \(h(x)=x^3\) and then take the square root using \(g(x)=\sqrt{x}\), or think of it as the inverse by switching the operations of \(g\) and \(h\).

Step 5: Solve Part D

The function is \(y=4x\). Though none of the given functions directly yield \(4x\), it can be formed by compositing \(j(x)=2x\) twice or realizing \(j(2x)\). Use the combination as \(j(j(x))\).

Step 6: Solve Part E

The function is \(y=\sqrt{(x-3)^3}\). Recognize this as \(g(h(f(x)))\). First, apply \(f(x)=x-3\), then cube the result with \(h(x)=x^3\), and finally apply the square root using \(g(x)=\sqrt{x}\).

Step 7: Solve Part F

The function is \(y=(2x-6)^3\). This is equivalent to \(h(f(j(x)))\). First, apply \(j(x)=2x\), then apply \(f(x)=x-3\), and finally cube the result using \(h(x)=x^{3}\).

Key Concepts

Function Composition

Function composition is like putting one function inside another to create a new, combined function. Think of it as a step-by-step process where each function modifies what's put into the next. For example, if you have two basic functions, say \( f(x) \) and \( g(x) \), composing them could form \( f(g(x)) \), meaning you first apply \( g \) to \( x \), and then apply \( f \) to the result.

• In our exercise, for part a, we used function composition to express \( y = \sqrt{x} - 3 \).
• To achieve the desired expression, we first employed \( g(x) = \sqrt{x} \) and then added \( f(x) = x - 3 \), forming \( f(g(x)) \).

This clear, step-by-step approach helps in systematically processing transformations until the desired formula is formed. Function composition is essential for solving complex functional relationships by breaking them down into simpler operations.

Basic Functions

Basic functions are the building blocks of more complicated expressions. Understanding these is vital because they serve as the foundation for function composition and other algebraic transformations. In our exercise, we worked with functions like \( f(x) = x - 3 \), \( g(x) = \sqrt{x} \), \( h(x) = x^3 \), and \( j(x) = 2x \). Let's break down their roles:

• \( f(x) = x - 3 \) shifts any value of \( x \) by subtracting 3, moving it leftward on a number line.
• \( g(x) = \sqrt{x} \) transforms \( x \) by finding its square root, reducing large numbers and increasing small fractions.
• \( h(x) = x^3 \) stretches \( x \) by cubing it, making small numbers smaller and large numbers larger very quickly.
• \( j(x) = 2x \) doubles any chosen \( x \), scaling it by increasing size at a constant rate.

By understanding these functions separately, you can see how they can be creatively recombined or composed to solve exercises involving composite functions. They involve simple operations that, when combined, lead to deceptively complex behaviors.

Algebraic Manipulation

Algebraic manipulation is the process of rewriting expressions into alternative, often simpler forms, which allows for easier analysis and solution of mathematical problems. It involves operations like expanding, factoring, and simplifying expressions. It is a fundamental skill in the proper execution of function composition. In the exercise, one must often rearrange terms to better match them with the combinations of basic function compositions.

• For instance, when handling \( y = x^{1/4} \), it helps to deconstruct it as \( x^{1/4} = (x^{1/2})^{1/2} \) or inversely associate it with the overlap of \( g \) and \( h \).
• Similarly, for \( y = (2x - 6)^3 \), we utilize \( j(x) \) and \( f(x) \) to rewrite the inner expression \( 2x - 6 \), before applying \( h(x) \).

This skill is pivotal when resolving composite functions because it involves breaking down transformations into manageable steps aligned with known base functions. Clear algebraic manipulation underpins successful solutions to complex function relationships.