Problem 100 - Precalculus

Question

A function called the hyperbolic sine is defined by \(f(x)=\frac{e^{x}-e^{-x}}{2} .\) Find its inverse.

Step-by-Step Solution

Answer

The inverse of the hyperbolic sine function is \( \text{arsinh}(y) = \ln(y + \sqrt{y^2 + 1}) \).

Step 1: Understand the Problem

We need to find the inverse of the function defined by the hyperbolic sine, which is \( f(x) = \frac{e^x - e^{-x}}{2} \). This means solving for \( x \) in terms of \( y = f(x) \).

Step 2: Set the Function Equal to y

Start by replacing \( f(x) \) with \( y \) so that you have the equation:\[y = \frac{e^x - e^{-x}}{2}\]

Step 3: Eliminate the Fraction

Multiply both sides of the equation by 2 to get rid of the fraction:\[2y = e^x - e^{-x}\]

Step 4: Rewrite the Equation in Terms of e^x

Let \( z = e^x \). Then \( e^{-x} = \frac{1}{z} \), so the equation becomes:\[2y = z - \frac{1}{z}\]

Step 5: Multiply Through by z

Multiply both sides by \( z \) to eliminate the fraction:\[2yz = z^2 - 1\]

Step 6: Rearrange to Form a Quadratic Equation

Rearrange the terms to form a quadratic equation:\[z^2 - 2yz - 1 = 0\]

Step 7: Solve the Quadratic Equation

Use the quadratic formula to solve for \( z \):\[z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \( a = 1 \), \( b = -2y \), and \( c = -1 \). Substitute these values into the formula:\[z = \frac{2y \pm \sqrt{(2y)^2 + 4}}{2}\]Simplifying, we have:\[z = y \pm \sqrt{y^2 + 1}\]

Step 8: Determine the Correct Branch

Since \( z = e^x \) is always positive, choose the positive branch:\[z = y + \sqrt{y^2 + 1}\]

Step 9: Find x from z

Knowing \( z = e^x \), we solve for \( x \) by taking the natural logarithm:\[x = \ln(z) = \ln(y + \sqrt{y^2 + 1})\]

Step 10: Write the Inverse Function

Finally, express the inverse function, which is sometimes called the "area hyperbolic sine," as:\[\text{arsinh}(y) = \ln(y + \sqrt{y^2 + 1})\]

Key Concepts

Understanding Hyperbolic Sine

The hyperbolic sine function is a mathematical function that resembles the familiar sine function but is adapted for hyperbolic geometry. It is often written as \( \sinh(x) \) and defined by:

• \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)

In this equation, \( e^x \) represents the exponential function, which is a fundamental component of calculus and algebra. By considering both \( e^x \) and \( e^{-x} \), the hyperbolic sine provides an analogy to sinusoids in trigonometry but within the context of hyperbolic functions.
Hyperbolic functions offer solutions to various differential equations and are used in calculus to model physical systems, like hanging cables or the shape of a ship hull. They are essential in areas that require the concept of "growth" that is more complex than what typical trigonometric functions can describe. The hyperbolic sine function, especially, is often used because it provides a smooth transition and can take on values from negative to positive infinity.

Inverse Function: Arsinh

The inverse hyperbolic sine function, also known as "arsinh" or "area hyperbolic sine," is the reverse of the hyperbolic sine function. Essentially, when you have \( \sinh(x) \) and want to find \( x \) itself, you use the inverse function. The notation for the inverse hyperbolic sine is usually \( \text{arsinh}(y) \).

The formula for calculating \( \text{arsinh}(y) \) is:

• \( \text{arsinh}(y) = \ln(y + \sqrt{y^2 + 1}) \)

This formula was derived through a series of mathematical steps, including recognizing the quadratic form in the hyperbolic sine expression and using logarithms.
Thus, anything that originally followed the hyperbolic sine transformation can be "inversed" or "unpacked" using this function. Arsinh is vital in mathematical transformations and solutions where reversing the hyperbolic sine operation is necessary, particularly in calculus and its applications.

The Role of Quadratic Equations

Quadratic equations are fundamental mathematical expressions that involve terms up to the second degree. A standard form of a quadratic equation is:

• \( ax^2 + bx + c = 0 \)

where \( a, b, \) and \( c \) are constants, and \( x \) is the variable to solve for. In finding the inverse of the hyperbolic sine, a quadratic equation naturally emerges from manipulating the function.

In the exercise given:

• The equation \( z^2 - 2yz - 1 = 0 \) represents this quadratic form.

Solving such a quadratic involves using the quadratic formula:

• \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

By substituting values for \( a, b, \) and \( c \), you can find the potential values for \( z \), which is related to \( e^x \) in the solution process.
Understanding quadratic equations allows us to tackle a wide array of problems in mathematics, from simple factorizations to more complex relationships, such as those involving exponential and hyperbolic functions.