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Question: The inverse function of the function \(f(x) = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}\) is...

The inverse function of the function f(x)=ex−e−xex+e−xf(x) = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}} is
A.12log⁡1+x1−x\frac{1}{2}\log \frac{{1 + x}}{{1 - x}}
B.12log⁡2+x2−x\frac{1}{2}\log \frac{{2 + x}}{{2 - x}}
C.12log⁡1−x1+x\frac{1}{2}\log \frac{{1 - x}}{{1 + x}}
D.None of these

Explanation

Solution

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)=ex−e−xex+e−xf(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=ex−e−xex+e−xy = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}.
Step 2 : Now we know that e−x{e^{ - x}} is nothing other than 1ex\frac{1}{{{e^x}}}
Applying this we get,
y=ex−1exex+1exy = \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=e2x−1exe2x+1ex=e2x−1e2x+1y = \frac{{\frac{{{e^{2x}} - 1}}{{{e^x}}}}}{{\frac{{{e^{2x}} + 1}}{{{e^x}}}}} = \frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}
Cross multiplying we get

⇒y(e2x+1)=e2x−1 ⇒ye2x+y=e2x−1  \Rightarrow y\left( {{e^{2x}} + 1} \right) = {e^{2x}} - 1 \\\ \Rightarrow y{e^{2x}} + y = {e^{2x}} - 1 \\\

Bringing e2x{e^{2x}} terms to one side and others to the other side we get

⇒ye2x−e2x=−1−y ⇒e2x(y−1)=−(1+y) ⇒e2x=−(1+y)−(1−y)=1+y1−y  \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

 ⇒log⁡e2x=log⁡1+y1−y  \\\ \Rightarrow \log {e^{2x}} = \log \frac{{1 + y}}{{1 - y}} \\\

By the property log⁡ex=x\log {e^x} = x
⇒2x=log⁡1+y1−y\Rightarrow 2x = \log \frac{{1 + y}}{{1 - y}}
Which can be written as
⇒x=12log⁡1+y1−y\Rightarrow x = \frac{1}{2}\log \frac{{1 + y}}{{1 - y}}
Now in the place of y write x
Therefore
⇒f−1=12log⁡1+x1−x\Rightarrow {f^{ - 1}} = \frac{1}{2}\log \frac{{1 + x}}{{1 - x}}
The inverse of the function is 12log⁡1+x1−x\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 12log⁡1+x1−x\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)≠1f(x){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.