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.