Problem 23 - Precalculus with Calculus Previews

Question

Express the given composition of functions as a rational function of \(x\), where \(x>0\). $$ \operatorname{coth}(\ln 2 x) $$

Step-by-Step Solution

Answer

The composition is expressed as \(\frac{4x^2 + 1}{4x^2 - 1}\).

Step 1: Recall the Definitions

First, recall the definition of the hyperbolic cotangent function, which is given by:\[\operatorname{coth}(u) = \frac{\cosh(u)}{\sinh(u)}\]where \(\cosh(u) = \frac{e^u + e^{-u}}{2}\) and \(\sinh(u) = \frac{e^u - e^{-u}}{2}\).

Step 2: Substitute the Argument

Next, substitute \(u = \ln(2x)\) into the definitions of \(\cosh(u)\) and \(\sinh(u)\):\[\cosh(\ln(2x)) = \frac{e^{\ln(2x)} + e^{-\ln(2x)}}{2} = \frac{2x + \frac{1}{2x}}{2}\] \[sinh(\ln(2x)) = \frac{e^{\ln(2x)} - e^{-\ln(2x)}}{2} = \frac{2x - \frac{1}{2x}}{2}\]where \(e^{\ln(2x)} = 2x\) and \(e^{-\ln(2x)} = \frac{1}{2x}\).

Step 3: Write the Expression for \(\operatorname{coth}(\ln(2x))\)

Now substitute these expressions for \(\cosh(\ln(2x))\) and \(\sinh(\ln(2x))\) into the formula for \(\operatorname{coth}(\ln(2x))\):\[\operatorname{coth}(\ln(2x)) = \frac{\cosh(\ln(2x))}{\sinh(\ln(2x))} = \frac{\frac{2x + \frac{1}{2x}}{2}}{\frac{2x - \frac{1}{2x}}{2}}\]Simplify this expression by multiplying both the numerator and the denominator by 2:\[\operatorname{coth}(\ln(2x)) = \frac{2x + \frac{1}{2x}}{2x - \frac{1}{2x}}\].

Step 4: Simplify the Resulting Rational Function

To simplify \(\frac{2x + \frac{1}{2x}}{2x - \frac{1}{2x}}\), combine the terms in the numerator and the denominator over a common denominator of \(2x\):\[\frac{2x + \frac{1}{2x}}{2x - \frac{1}{2x}} = \frac{\frac{4x^2 + 1}{2x}}{\frac{4x^2 - 1}{2x}} = \frac{4x^2 + 1}{4x^2 - 1}\]Cancel the common denominator \(2x\) since it appears in both the numerator and the denominator.

Key Concepts

Rational Functions

A rational function is essentially a fraction where both the numerator and the denominator are polynomials. This type of function is expressed as \( \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomial functions, and \( Q(x) eq 0 \). One of the main purposes behind studying rational functions is to analyze their behavior, which involves understanding how they change across different values of \( x \).
When dealing with the composition of functions, such as hyperbolic functions, you might be led to creating a rational function as we convert more complex forms into simpler, fractional representations.
In the given exercise, we've formed a rational expression from the composition function \( \operatorname{coth}(\ln(2x)) \) by substituting and simplifying, ultimately obtaining \( \frac{4x^2 + 1}{4x^2 - 1} \). This is a classic example of how learning to manipulate expressions into rational forms can aid in their analysis and comprehension.

Function Composition

Function composition involves applying one function to the results of another. In mathematical terms, if you have two functions \( f(x) \) and \( g(x) \), the composition is written as \( f(g(x)) \).
This is crucial because it allows us to build new functions from existing ones and is a powerful tool in calculus for creating more complex forms from simpler ones. In our exercise, we first utilized the function \( \ln(2x) \) within the hyperbolic cotangent function, forming \( \operatorname{coth}(\ln(2x)) \).
By substituting \( \ln(2x) \) into the hyperbolic identities for \( \cosh(u) \) and \( \sinh(u) \), we effectively composed these functions, allowing us to simplify the expression into a rational function.
This showcases how function composition is essential to manipulate and transform expressions into more comprehensible formats using known identities.

Precalculus Techniques

Precalculus involves a series of techniques and skills that prepare students for calculus, developing an understanding of functions, complex numbers, trigonometry, and more. This foundations course provides critical skills for simplifying and manipulating expressions, which are used throughout calculus.
In the provided solution, precalculus techniques—such as manipulating and simplifying expressions, utilizing identities, and understanding logarithmic properties—came into full play. Given the hyperbolic function \( \operatorname{coth}(\ln(2x)) \), we recalled definitions for hyperbolic functions, transformed logarithmic terms using exponential rules, and eventually simplified the expression into \( \frac{4x^2 + 1}{4x^2 - 1} \).
Understanding, applying, and bridging these precalculus formalisms are crucial for solving complicated equations and expressions in later stages of calculus.