Question

The inverse of the function $ f\left( x \right) = \dfrac{{{9^x} - {9^{ - x}}}}{{{9^x} + {9^{ - x}}}} $ is:
A. $ {f^{ - 1}}\left( x \right) = {\log _9}\left( {\dfrac{{1 + x}}{{1 - x}}} \right) $
B. $ {f^{ - 1}}\left( x \right) = \dfrac{1}{2}{\log _9}\left( {\dfrac{{1 + x}}{{1 - x}}} \right) $
C. $ {f^{ - 1}}\left( x \right) = \dfrac{1}{4}{\log _9}\left( {\dfrac{{1 + x}}{{1 - x}}} \right) $
D. $ {f^{ - 1}}\left( x \right) = \dfrac{1}{2}{\log _9}\left( {\dfrac{{2x}}{{2x - 1}}} \right) $

Answer 1

Hint: Inverse function is a function that reverses another function. If the value of $ f\left( x \right) $ is y, then applying the inverse function to y gives the value of x. So here consider the given function as y first and then solve for x. The value of x is considered as $ {f^{ - 1}}\left( y \right) $ . Then replace y with x in $ {f^{ - 1}}\left( y \right) $ , which gives our required inverse function $ {f^{ - 1}}\left( x \right) $

Complete step by step solution:
We are given a function $ f\left( x \right) = \dfrac{{{9^x} - {9^{ - x}}}}{{{9^x} + {9^{ - x}}}} $ and we have to find its inverse.
Let $ f\left( x \right) $ is equal to y.
 $ f\left( x \right) = y $ , this means $ y = \dfrac{{{9^x} - {9^{ - x}}}}{{{9^x} + {9^{ - x}}}} $
Now we are solving for the value of x
 $ \Rightarrow y = \dfrac{{\left( {{9^x} - \dfrac{1}{{{9^x}}}} \right)}}{{\left( {{9^x} + \dfrac{1}{{{9^x}}}} \right)}} $ (Since we know that $ {a^{ - m}} $ can also be written as $ \dfrac{1}{{{a^m}}} $ )
 $ \Rightarrow y = \dfrac{{\left( {\dfrac{{\left( {{9^x} \times {9^x}} \right) - 1}}{{{9^x}}}} \right)}}{{\left( {\dfrac{{\left( {{9^x} \times {9^x}} \right) + 1}}{{{9^x}}}} \right)}} $
 $ \Rightarrow y = \dfrac{{\left( {\dfrac{{{9^{2x}} - 1}}{{{9^x}}}} \right)}}{{\left( {\dfrac{{{9^{2x}} + 1}}{{{9^x}}}} \right)}} $ (since we know that $ {a^m} \times {a^n} $ can also be written as $ {a^{m + n}} $ )
 $ \Rightarrow y = \dfrac{{{9^{2x}} - 1}}{{{9^{2x}} + 1}} $
On cross multiplication, we get
 $ \Rightarrow y\left( {{9^{2x}} + 1} \right) = {9^{2x}} - 1 $
 $ \Rightarrow {9^{2x}}y + y = {9^{2x}} - 1 $
Putting exponential terms one side and remaining terms other side, we get
 $ \Rightarrow y + 1 = {9^{2x}} - {9^{2x}}y $
Taking out $ {9^{2x}} $ common from the left hand side
 $ \Rightarrow 1 + y = {9^{2x}}\left( {1 - y} \right) $
We need the value of x, so put the terms containing x at the left hand side and remaining right hand side.
 $ \Rightarrow {9^{2x}} = \dfrac{{1 + y}}{{1 - y}} $
Sending 9 present at LHS to the RHS results in a logarithm with base 9.
 $ \Rightarrow 2x = {\log _9}\left( {\dfrac{{1 + y}}{{1 - y}}} \right) $
 $ \therefore x = \dfrac{1}{2}{\log _9}\left( {\dfrac{{1 + y}}{{1 - y}}} \right) $
Therefore, the value of x is $ \dfrac{1}{2}{\log _9}\left( {\dfrac{{1 + y}}{{1 - y}}} \right) $ , this means $ {f^{ - 1}}\left( y \right) = \dfrac{1}{2}{\log _9}\left( {\dfrac{{1 + y}}{{1 - y}}} \right) $
Now replace y with x in $ {f^{ - 1}}\left( y \right) $ to get the inverse function $ {f^{ - 1}}\left( x \right) $
Therefore, $ {f^{ - 1}}\left( x \right) = \dfrac{1}{2}{\log _9}\left( {\dfrac{{1 + x}}{{1 - x}}} \right) $
So, the correct answer is “Option B”.

Note: We can also verify the resulting inverse function. Equate the inverse function $ {f^{ - 1}}\left( x \right) $ with y and find the new value of x. The result (the value of x in terms of y) should resemble the function $ f\left( y \right) $ or else the result is wrong. Inverse function is not a reciprocal of the original function. It is just another function that undoes whatever done by the original function (like add after subtracting). So please be careful.