Problem 12 - 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 \(j .\) $$ \begin{array}{ll}{\text { a. } y=2 x-3} & {\text { b. } y=x^{3 / 2}} \\\ {\text { c. } y=x^{9}} & {\text { d. } y=x-6} \\ {\text { e. } y=2 \sqrt{x-3}} & {\text { f. } y=\sqrt{x^{3}-3}}\end{array} $$

Step-by-Step Solution

Answer

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

Step 1: Find Composition for y = 2x - 3

The function given is \( y = 2x - 3 \). To express it as a composite of the functions \( f, g, h, \) and \( j \), we can use \( j(x) = 2x \) and \( f(x) = x - 3 \). Substituting \( x \) with \( j(x) = 2x \) in \( f(x) \), we get: \( f(j(x)) = j(x) - 3 = 2x - 3 \). Thus, the composition is \( f(j(x)) \).

Step 2: Find Composition for y = x^{3/2}

The function given is \( y = x^{3/2} \), which can be expressed using \( g(x) = \sqrt{x} \) and \( h(x) = x^3 \). First, you apply \( g(x) \) to find the square root, then apply \( h(x) \), which equals raising to the power of 3, on the result of the square root: \( h(g(x)) = (\sqrt{x})^3 = x^{3/2} \). Thus, the composition is \( h(g(x)) \).

Step 3: Find Composition for y = x^9

The function is \( y = x^9 \). We can achieve this by taking \( h(x) = x^3 \) and compositing it three times. Apply \( h \) on \( x \) yielding \( x^3 \), then apply \( h \) again on \( x^3 \) yielding \( (x^3)^3 = x^9 \). Thus, the composition is \( h(h(x)) \).

Step 4: Find Composition for y = x - 6

The function is \( y = x - 6 \). We know \( f(x) = x - 3 \), so applying \( f \) twice on \( x \) gives \( f(f(x)) = f(x - 3) = (x - 3) - 3 = x - 6 \). Thus, the composition is \( f(f(x)) \).

Step 5: Find Composition for y = 2\sqrt{x-3}

The function is \( y = 2\sqrt{x-3} \). Start by taking \( f(x) = x - 3 \) and applying it to \( x \), then take \( g(x) = \sqrt{x} \) to find the square root of the result, and finally apply \( j(x) = 2x \): \( j(g(f(x))) = 2\cdot\sqrt{x-3} \). Thus, the composition is \( j(g(f(x))) \).

Step 6: Find Composition for y = \\sqrt{x^3 - 3}

For the function \( y = \sqrt{x^3 - 3} \), start with \( h(x) = x^3 \), then apply \( f(x) = x - 3 \), so the step becomes \( f(h(x)) = x^3 - 3 \), and finally apply \( g(x) = \sqrt{x} \): \( g(f(h(x))) = \sqrt{x^3 - 3} \). Thus, the composition is \( g(f(h(x))) \).

Key Concepts

Function Composition

In algebra, composite functions are a way to combine two or more functions into a single expression. Function composition means applying one function to the results of another function. This creates a chain or sequence of operations. For example, if you have functions like \( f(x) = x-3 \) and \( j(x) = 2x \), you can compose them to form a new function. The notation \( f(j(x)) \) means first apply \( j(x) \) and then use the result as the input for \( f(x) \).

• Understanding Order: The order in which you apply functions is crucial. \( f(g(x)) \) is different from \( g(f(x)) \).
• Calculating: Replace the variable in the outer function with the result of the inner function.
• Resulting Function: The result is often a completely new function that expresses the relationship between the two original functions.

Composite functions allow mathematicians and students like yourself to build complex calculations from simpler parts more intuitively.

Function Operations

Function operations involve performing basic mathematical processes on algebraic functions. These operations include addition, subtraction, multiplication, division, and, notably, composition as previously mentioned. When dealing with function operations:

• Addition and Subtraction: If you have \( f(x) \) and \( g(x) \), then \( (f + g)(x) = f(x) + g(x) \) and \( (f - g)(x) = f(x) - g(x) \).
• Multiplication and Division: These operations work similarly, \( (f \cdot g)(x) = f(x) \cdot g(x) \) and \( (f / g)(x) = f(x) / g(x) \) given \( g(x) eq 0 \).
• Composition: This is perhaps one of the more advanced operations where the output of one function becomes the input of another.

Function operations allow you to manipulate and transform functions much like regular algebraic expressions. It's essential to recognize and correctly apply these operations to solve function-related problems effectively.

Algebraic Functions

Algebraic functions are mathematical expressions defined using algebraic operations such as addition, subtraction, multiplication, division, and root extractions. These functions form the foundation of many mathematical concepts and are defined over real numbers or other systems.

• Types of Algebraic Functions: Linear functions, polynomials, and radical functions are all examples of algebraic functions.
• Properties: Each type exhibits unique properties. For example, a quadratic function forms a parabola, while a linear function has constant rates of change represented as straight lines.
• Simplification: Often, algebraic functions need simplification for solving or composing with other functions, just like in the exercise given.

Understanding how to work with algebraic functions is crucial for function composition and operations, helping bridge simple equations to more complex mathematical relationships.