Problem 49 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Let \(f(x)=\frac{1}{x}+x\) and \(g(x)=\frac{2 x}{x^{2}+1} .\) Evaluate the following expressions. (a) \((f \circ f)(x)\) (b) \((f \circ f \circ f)(x)\)

Step-by-Step Solution

Answer

(a) \((f \circ f)(x)\) simplifies to \(\frac{x}{1 + x^{2}} + \frac{1}{x} + x\), and (b) \((f \circ f \circ f)(x)\) simplifies to \(\frac{x}{x^4 + 2x^2 + 1} + \frac{x}{1 + x^{2}} + \frac{1}{x} + x.\)

Step 1: (a) Calculate \((f \circ f)(x)\)

First, let's find \((f \circ f)(x)\). This means that wherever we see an \(x\) in the function \(f(x)\), we need to instead put the whole function \(f(x)\). So,\[(f \circ f)(x) = f(f(x)) = f\left(\frac{1}{x}+x\right) = \frac{1}{(\frac{1}{x}+x)} + \frac{1}{x} + x.\]

Step 2: Simplifying \((f \circ f)(x)\)

We now simplify this expression:\[\frac{1}{(\frac{1}{x}+x)} + \frac{1}{x} + x = \frac{x}{1 + x^{2}} + \frac{1}{x} + x.\]

Step 3: (b) Calculate \((f \circ f \circ f)(x)\)

The term \((f \circ f \circ f)(x)\) means that we compose the function \(f(x)\) with itself twice. Now that we have \((f \circ f)(x)\), we can express this as \(f((f \circ f)(x))\), this means that wherever we see an \(x\) in the function \(f(x)\), we need to put the whole function \((f \circ f)(x)\):\[(f \circ f \circ f)(x) = f((f \circ f)(x)) = f\left(\frac{x}{1 + x^{2}} + \frac{1}{x} + x\right) = \frac{1}{\left(\frac{x}{1 + x^{2}} + \frac{1}{x} + x\right)} + \frac{x}{1 + x^{2}} + \frac{1}{x} + x.\]

Step 4: Simplifying \((f \circ f \circ f)(x)\)

We now simplify this expression:\[\frac{1}{\left(\frac{x}{1 + x^{2}} + \frac{1}{x} + x\right)} + \frac{x}{1 + x^{2}} + \frac{1}{x} + x = \frac{x}{x^4 + 2x^2 + 1} + \frac{x}{1 + x^{2}} + \frac{1}{x} + x.\]

Key Concepts

Function Composition

Function composition is like stacking functions on top of one another. It involves inserting one function inside another. For example, when we talk about \(f \circ g\)(x), it means "apply function \(g\) to \(x\) first, and then apply function \(f\) to the result." In the context of the problem, we had functions \(f(x)\) and \(g(x)\). The exercise asked us to find \( (f \circ f)(x)\) and \( (f \circ f \circ f)(x)\). "(f ∘ f)(x)" means applying the function \(f\) to itself. More simply, it means taking the result of \(f(x)\) and plugging it back into \(f(x)\). So, you're essentially using the output of a function as the input for itself. This is a creative aspect of math that teaches us how functions can be reused in different ways.
To break it down further:

• The process begins with applying the given function to an initial value or expression.
• Then, the output of that first calculation becomes the new input for the next layer of the function application.
• This can be layered multiple times to create more complex computations.

Understanding this concept helps one see how various mathematical operations interact with each other, which is crucial in advanced levels of math.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying expressions to make calculations more manageable. When working with compositions like \( (f \circ f)(x)\), simplification is necessary to achieve a clearer form.
In the given solution, after using function composition, the result included several fractions. Algebraic manipulation allows us to simplify these into a neater form by following a few steps:

• Combining like terms: Look for terms in the expression that can be combined or simplified.
• Fraction simplification: When dealing with complex fractions, rewrite them as simpler expressions.
• Canceling terms: If possible, cancel terms that appear in both the numerator and denominator to make things simpler.

Through these steps, rationalizing the initial composition leads to a more readable and usable form, which is crucial for further applications or additional operations.

Rational Functions

Rational functions are expressed as fractions where both the numerator and the denominator are polynomials. In this exercise, both given functions, \(f(x)\) and \(g(x)\), qualify as rational functions. To solve problems involving these functions, one must understand how to manage and manipulate their expressions.
Here are some key points about working with rational functions:

• Identifying the function: Make sure to recognize that a function is rational if it can be written as a fraction of two polynomials.
• Domain concerns: Rational functions can have restrictions on their domains, primarily where the denominator is zero. Always consider where the expression becomes undefined.
• Simplification: Like with algebraic manipulation, simplifying rational functions can sometimes allow you to see solutions more clearly through reduced and manageable expressions.

These concepts are foundational in math, providing a thorough understanding of how functions behave and interact through compositions and manipulations.