Question

The inverse function of the function $f(x) = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}$ is
A.$\frac{1}{2}\log \frac{{1 + x}}{{1 - x}}$
B.$\frac{1}{2}\log \frac{{2 + x}}{{2 - x}}$
C.$\frac{1}{2}\log \frac{{1 - x}}{{1 + x}}$
D.None of these

Answer 1

Hint: Lets take f(x) = y and rearrange the terms by cross multiplying and taking the common terms out till we get the value of x and after getting the value of x , change the variable y into x and hence the inverse is obtained

Complete step-by-step answer:
Step 1: We are given that $f(x) = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}$ .
To find the inverse of the function the first thing we need to do is take f(x) = y
Therefore we get, $y = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}$.
Step 2 : Now we know that ${e^{ - x}}$ is nothing other than $\frac{1}{{{e^x}}}$
Applying this we get,
\[y = \frac{{{e^x} - \frac{1}{{{e^x}}}}}{{{e^x} + \frac{1}{{{e^x}}}}}\]
Now by taking lcm in both the numerator and denominator, we get
\[y = \frac{{\frac{{{e^{2x}} - 1}}{{{e^x}}}}}{{\frac{{{e^{2x}} + 1}}{{{e^x}}}}} = \frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}\]
Cross multiplying we get
\[
   \Rightarrow y\left( {{e^{2x}} + 1} \right) = {e^{2x}} - 1 \\
   \Rightarrow y{e^{2x}} + y = {e^{2x}} - 1 \\
\]
Bringing \[{e^{2x}}\] terms to one side and others to the other side we get
\[
   \Rightarrow y{e^{2x}} - {e^{2x}} = - 1 - y \\
   \Rightarrow {e^{2x}}(y - 1) = - (1 + y) \\
   \Rightarrow {e^{2x}} = \frac{{ - (1 + y)}}{{ - (1 - y)}} = \frac{{1 + y}}{{1 - y}} \\
\]
Step 3 :
Now let's take log on both sides
\[
    \\
   \Rightarrow \log {e^{2x}} = \log \frac{{1 + y}}{{1 - y}} \\
\]
By the property \[\log {e^x} = x\]
\[ \Rightarrow 2x = \log \frac{{1 + y}}{{1 - y}}\]
Which can be written as
\[ \Rightarrow x = \frac{1}{2}\log \frac{{1 + y}}{{1 - y}}\]
Now in the place of y write x
Therefore
\[ \Rightarrow {f^{ - 1}} = \frac{1}{2}\log \frac{{1 + x}}{{1 - x}}\]
The inverse of the function is \[\frac{1}{2}\log \frac{{1 + x}}{{1 - x}}\]

The correct option is A

Note: Alternatively f(x) = tan h (x) . So its inverse will be \[\frac{1}{2}\log \frac{{1 + x}}{{1 - x}}\]
Now, be careful with the notation for inverses. The “-1” is NOT an exponent despite the fact that it sure does look like one. When dealing with inverse functions we have got to remember that.
${f^{ - 1}}(x) \ne \frac{1}{{f(x)}}$.
This is one of the more common mistakes that students make when first studying inverse functions.
The inverse of a function may not always be a function. The original function must be a one-to-one function to guarantee that its inverse will also be a function. A function is a one-to-one function if and only if each second element corresponds to one and only one first element.