Question

Find a formula for the inverse of the function.

30. \(y = \frac{{1 - {e^{ - x}}}}{{1 + {e^{ - x}}}}\)

Short answer

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

Step-by-step solution

Condition to find the inverse function of a one-to-one function

Step 1: Write the equation as \(y = f\left( x \right)\).

Step 2: If possible, solve the equation in terms of \(y\).

Step 3: Express the inverse \({f^{ - 1}}\) as a function of \(x\), for that interchange\(x\) and \(y\). The equation becomes \(y = {f^{ - 1}}\left( x \right)\).

Determine the formula for the inverse of the function 

Solve the given equation for \(x\) as shown below:

\(\begin{aligned}y &= \frac{{1 - {e^{ - x}}}}{{1 + {e^{ - x}}}}\\y\left( {1 + {e^{ - x}}} \right) &= 1 - {e^{ - x}}\\y + y{e^{ - x}} &= 1 - {e^{ - x}}\\{e^{ - x}} + y{e^{ - x}} &= 1 - y\\{e^{ - x}}\left( {1 + y} \right) &= 1 - y\\{e^{ - x}} &= \frac{{1 - y}}{{1 + y}}\end{aligned}\)

Take log on both sides of the above equation as shown below:

\(\begin{aligned} - x &= \ln \frac{{1 - y}}{{1 + y}}\\x &= - \ln \frac{{1 - y}}{{1 + y}}\\x &= \ln {\left( {\frac{{1 - y}}{{1 + y}}} \right)^{ - 1}}\\x &= \ln \left( {\frac{{1 + y}}{{1 - y}}} \right)\end{aligned}\)

Interchange \(x\) and \(y\) in the above equation as shown below:

\(y = \ln \left( {\frac{{1 + x}}{{1 - x}}} \right)\)

Thus, the inverse function is \(y = \ln \left( {\frac{{1 + x}}{{1 - x}}} \right)\).