Problem 26 - Essential Calculus Early Transcendentals

Question

Find a formula for the inverse of the function. $$ y=\frac{e^{x}}{1+2 e^{x}} $$

Step-by-Step Solution

Answer

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

Step 1: Set up the equation for the inverse

The first step in finding the inverse of the function is to replace the function notation with variables. We start with:\[ y = \frac{e^x}{1 + 2e^x} \]Now, swap \(y\) and \(x\):\[ x = \frac{e^y}{1 + 2e^y} \]

Step 2: Clear the fraction

To solve for \( y \) in terms of \( x \), we need to eliminate the fraction. Multiply both sides by \(1 + 2e^y\):\[ x(1 + 2e^y) = e^y \]This simplifies to:\[ x + 2xe^y = e^y \]

Step 3: Rearrange the equation

Group terms to isolate \( e^y \):\[ 2xe^y - e^y = -x \]Factor out \(e^y\):\[ e^y(2x - 1) = -x \]

Step 4: Solve for \( e^y \)

Divide both sides by \(2x - 1\) to isolate \( e^y \):\[ e^y = \frac{-x}{2x - 1} \]

Step 5: Solve for \( y \)

Take the natural logarithm of both sides to solve for \( y \):\[ y = \ln\left(\frac{-x}{2x - 1}\right) \]

Step 6: State the inverse function

The inverse function is obtained from our expression for \(y\). Thus, the inverse function is:\[ f^{-1}(x) = \ln\left(\frac{-x}{2x - 1}\right) \]

Key Concepts

Natural Logarithm

The natural logarithm, often written as \(\ln(x)\), is a logarithm to the base \(e\), where \(e\) is an irrational constant approximately equal to 2.71828. This special logarithm has unique properties that make it particularly useful in calculus and mathematical analysis.

• Inverse of Exponential: The natural logarithm is the inverse function of the exponential function \(e^x\). This relationship means that if you have \(e^y = x\), then \(y = \ln(x)\).


• Simplifying:\(\ln(e^x) = x\): This is a direct result of the logarithm and exponential having inversing actions, thus simplifying expressions like \(\ln(e^x)\) directly to \(x\).


• Domain:\(\ln(x)\) is undefined for negative values or zero: You can only take the natural logarithm of a positive number, which is crucial when working with functions involving logarithms.

Understanding these properties of the natural logarithm helps in solving equations where the variable is inside an exponential function. Taking the natural logarithm can often simplify and solve such equations, just like in the problem provided.

Exponential Functions

Exponential functions have the form \(f(x) = a^{x}\), where \(a\) is a constant and \(x\) is an exponent. They are characterized by rapid growth or decay, depending on whether \(a\) is greater or lesser than one.

• Base \(e\) Exponential Function: The exponential function with base \(e\) is particularly important in natural processes and equations, noted as \(e^x\). This function shows the rate of continuous growth or decay.


• Inverse Relationship: Exponential functions have a unique inverse, which is the natural logarithm. If you have \(e^y = x\), solving for \(y\) means finding \(\ln(x)\).


• Real-life Applications: These functions model numerous real-world scenarios such as population growth, radioactive decay, and interest calculations, making them fundamental in both science and finance.

The inverse function property of exponential functions was utilized in the problem to solve for the original input given the output, as seen in steps involving swapping the roles of \(x\) and \(y\). Understanding exponential functions and their inverses is essential for dealing efficiently with growth models and equations.

Solving Equations

Solving equations involves finding the values of variables that satisfy the equation. In the context of inverse functions, this often requires isolating the variable of interest. Here are some key approaches used in the exercise you reviewed:

• Isolating the Variable: After swapping \(x\) and \(y\), there was a need to rearrange the equation by moving terms around to isolate \(e^y\).


• Removing Fractions: Fractions were cleared by multiplying both sides with the denominator, which is a standard technique to simplify equations.


• Using Logarithms: To solve for \(y\), taking the natural logarithm of both sides was crucial, as it allowed the expression \(e^y\) to simplify directly to \(y\).

Solving equations, especially those involving inverse functions, often requires strategically maneuvering terms to simplify and isolate variables. By following logical algebraic steps, such as those demonstrated, a solution can be efficiently reached.