Problem 108 - Precalculus

Question

The hyperbolic tangent is defined by tanh \(x=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\) Find its inverse function \(\tanh ^{-1} x\).

Step-by-Step Solution

Answer

The inverse function is \(\tanh^{-1}(x) = \frac{1}{2} \ln\left(\frac{x + 1}{1 - x}\right)\).

Step 1: Define the Equation

Start by stating the equation for the hyperbolic tangent: \[ y = \tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} \] We need to find the inverse, \(\tanh^{-1}(x)\), such that \(x = \tanh^{-1}(y)\).

Step 2: Swap Variables

To find the inverse function, we swap the variables \(x\) and \(y\), resulting in:\[ x = \frac{e^y - e^{-y}}{e^y + e^{-y}} \] This represents \(y\) in terms of \(x\).

Step 3: Solve for Exponential Terms

Multiply both sides by \(e^y + e^{-y}\) to eliminate the fraction:\[ x(e^y + e^{-y}) = e^y - e^{-y} \] Simplify this to:\[ xe^y + xe^{-y} = e^y - e^{-y} \]

Step 4: Rearrange to Consolidate Terms

Bring like terms to one side:\[ xe^y - e^y = - xe^{-y} - e^{-y} \] Factor out \(e^y\) on the left and \(e^{-y}\) on the right:\[ (x - 1)e^y = -(x + 1)e^{-y} \]

Step 5: Express in Terms of One Exponential

Multiply the whole equation by \(e^y\) to express everything in terms of \(e^{2y}\):\[ (x - 1)e^{2y} = -(x + 1) \] Now isolate \(e^{2y}\):\[ e^{2y} = \frac{-(x + 1)}{x - 1} \]

Step 6: Solve for \(y\)

Take the natural logarithm on both sides to solve for \(2y\):\[ 2y = \ln\left(\frac{x + 1}{1 - x}\right) \] Finally, solve for \(y\) by dividing both sides by 2:\[ y = \frac{1}{2} \ln\left(\frac{x + 1}{1 - x}\right) \] Thus, the inverse function is \(\tanh^{-1}(x) = \frac{1}{2} \ln\left(\frac{x + 1}{1 - x}\right)\).

Key Concepts

Hyperbolic Tangent

The hyperbolic tangent function, denoted as \( \tanh(x) \), represents a relationship in hyperbolic geometry similar to the tangent function in trigonometry. This function is defined by:

• \( \tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} \)

Here, \( e \) is the base of the natural logarithm, which is approximately 2.718.

The hyperbolic tangent function is used to map real numbers within the range -1 to 1, making it very useful in various fields like engineering and physics.
A few properties of the hyperbolic tangent function include:

• Range: Values between -1 and 1.
• Odd Function: Since \( \tanh(-x) = -\tanh(x) \).
• Smoothness: It is a continuous and smooth curve.

These properties make the hyperbolic tangent function ideal for mathematical modeling, especially in neural network algorithms where non-linear activation functions are vital.

Inverse Functions

Inverse functions reverse the roles of inputs and outputs. For the function \( y = \tanh(x) \), the inverse is finding \( x \) given \( y \). This means solving \( y = \frac{e^x - e^{-x}}{e^x + e^{-x}} \) for \( x \).

To find the inverse function of \( \tanh(x) \):

• First, swap \( x \) and \( y \) in the equation to express \( x \) in terms of \( y \).
• Rearrange to solve for \( y \) by eliminating fractions and combining like terms.

The result derives the inverse function:

• \( \tanh^{-1}(x) = \frac{1}{2} \ln\left(\frac{x + 1}{1 - x}\right) \)

This process shows how an inverse function gives us the original input from an output, which is crucial for reverting transformations in equations and calculating original data values.

Exponential Functions

Exponential functions involve the constant, \( e \), raised to the power of a variable and are expressed as \( e^x \). These functions describe growth or decay processes, which is a fundamental concept in areas such as biology, finance, and physics.

In the context of the hyperbolic tangent, exponential components \( e^x \) and \( e^{-x} \) appear in the numerator and denominator:

• \( \tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} \)

Solving the inverse of the hyperbolic tangent involves multiple steps using exponential equations:

• Eliminate the fraction involving \( e^x \) and \( e^{-x} \) by multiplying both sides.
• Rewriting terms using \( e^{2y} \) transforms the problem into a solvable logarithmic form.

The manipulation of exponential functions in this context highlights their versatility in solving complex mathematical expressions and obtaining the inverse of a hyperbolic function.