Question

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

Short answer

The formula for the inverse of the function \(y = \frac{{{e^x}}}{{1 + 2{e^x}}}\) is \({f^{ - 1}}(x) = \ln \left( {\frac{x}{{1 - 2x}}} \right)\).

Step-by-step solution

Given data

The given function is \(y = \frac{{{e^x}}}{{1 + 2{e^x}}}\).

Concept of functions

The simplest definition is an equation will be a function if, for any \({\rm{x}}\) in the domain of the equation (the domain is all the \({\rm{x}}\)'s that can be plugged into the equation), the equation will yield exactly one value of \({\rm{y}}\) when we evaluate the equation at a specific \({\rm{X}}\).

Solve the equation

Solve this equation for \(x\) as shown below.

\(\begin{array}{c}y\left( {1 + 2{e^x}} \right) = {e^x}\\y + 2y\left( {{e^x}} \right) - {e^x} = 0\\{e^x}(2y - 1) = 0 - y\\{e^x} = \frac{{ - y}}{{2y - 1}}\end{array}\)

Take natural log on both sides and simplify further as follows:

\(\begin{array}{c}\ln {e^x} = \ln \left( {\frac{y}{{1 - 2y}}} \right)\\x = \ln \left( {\frac{y}{{1 - 2y}}} \right)\end{array}\)

Then, interchange \(x\) and \(y\), obtain the inverse function, \(y = \ln \left( {\frac{x}{{1 - 2x}}} \right)\).

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