Problem 24 - Precalculus with Calculus Previews

Question

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

Step-by-Step Solution

Answer

The rational function is \( \frac{2x}{x^2 + 1} \).

Step 1: Understanding the Function Composition

The function given is a composition of two functions: \(\operatorname{sech}(u)\) where \(u = \ln x\). The hyperbolic secant function is defined as \(\operatorname{sech}(u) = \frac{1}{\cosh(u)}\). Here, we need to find \(\operatorname{sech}(\ln x)\).

Step 2: Express \( \cosh(\ln x) \) in Terms of \( x \)

The hyperbolic cosine function is defined as \(\cosh(u) = \frac{e^u + e^{-u}}{2}\). Substituting \(u = \ln x\), we have:\[\cosh(\ln x) = \frac{e^{\ln x} + e^{-\ln x}}{2} = \frac{x + \frac{1}{x}}{2}\].

Step 3: Calculate \( \operatorname{sech}(\ln x) \)

The hyperbolic secant of \(\ln x\) is:\[\operatorname{sech}(\ln x) = \frac{1}{\cosh(\ln x)}\].Using the expression from Step 2 for \(\cosh(\ln x)\), we have:\[\operatorname{sech}(\ln x) = \frac{1}{\frac{x + \frac{1}{x}}{2}} = \frac{2}{x + \frac{1}{x}}\].

Step 4: Simplify the Expression

To simplify \(\frac{2}{x + \frac{1}{x}}\), multiply the numerator and denominator by \(x\):\[\frac{2}{x + \frac{1}{x}} = \frac{2x}{x^2 + 1}\].This is the simplified expression of the rational function.

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for a hyperbola rather than a circle. An example is the hyperbolic secant, denoted as \( \operatorname{sech}(u) \), which is defined as \( \operatorname{sech}(u) = \frac{1}{\cosh(u)} \). Here, \( \cosh(u) \) is the hyperbolic cosine, given by the formula \( \cosh(u) = \frac{e^u + e^{-u}}{2} \). These functions find applications in areas such as calculus and complex analysis. They can be quite useful in solving problems that involve exponential growth and decay, similar to the roles trigonometric functions play in periodic phenomena.In our exercise, we are tasked with finding \( \operatorname{sech}(\ln x) \). Understanding the relationship between hyperbolic functions and exponential functions is key to simplifying and expressing such compositions accurately.

Rational Functions

Rational functions are expressions that represent the ratio of two polynomials. They take the form \( \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials. These functions are widely studied in algebra and calculus because they appear frequently in real-world problems.In the given exercise, we derive a rational function from the composition \( \operatorname{sech}(\ln x) \). After simplifying, we arrive at the expression \( \frac{2x}{x^2 + 1} \). This is a clear example of a rational function because it is expressed as a ratio of polynomials in \( x \).Understanding how to express complex compositions as rational functions helps students not only solve such problems but also recognize similar patterns across different mathematical contexts.

Function Simplification

Simplifying functions entails reducing them to their simplest or most precise form, often making them easier to evaluate or manipulate. This process often involves algebraic manipulation such as factoring, combining like terms, or using identities.In our exercise, simplification involves expressing \( \operatorname{sech}(\ln x) \) as a rational function of \( x \). The steps include converting the hyperbolic cosine \( \cosh(\ln x) \) into an expression in terms of \( x \). Then, by rationalizing the expression, we translate it to \( \frac{2x}{x^2 + 1} \), which is its simplified form.Simplification is a crucial skill in mathematics because it provides a way to deal with complex expressions more easily, improving both comprehension and computational efficiency.