To understand inverse of a function let us assume that there are two functions \(f(x)\) and \(g(x)\) such that \(f\left( x \right){\rm{ }} = {x^2}\) and \(g\left( x \right){\rm{ }} = {x^{1/2}}\), now 

\[f\left( {g\left( x \right)} \right){\rm{ }} = f\left( {{x^{1/2}}} \right){\rm{ }} = {\rm{ }}{\left( {{x^{1/2}}} \right)^2} = x\]

\[g\left( {f\left( x \right)} \right){\rm{ }} = g\left( {{x^2}} \right)\; = {\rm{ }}{\left( {{x^2}} \right)^{1/2}} = x\]

We observe that \(f(x)\) and \(g(x)\) cancel the effect of each other. There are so many similar examples which follow this property, such as \(lo{g_e}_{}x\) and \({e^x},{\rm{ }}lo{g_{10}}x\) and \({10^x}\). 

If \(f(g(x)) = x\)  for every \(x\) in the domain of \(g\) and  \(g(f(x)) = x\) for every \(x\) in the domain of  \(f\), then \(f\) and \(g\) are known as inverse functions of each other. Inverse of \(f\) is represented by \({f^{-{\bf{1}}}}\). Note that  \({f^{-1}}\)  does not mean \(\frac{1}{f}\).

Example 13:  Find the inverse of the function \(f\left( x \right) = \left[ {\sqrt {{x^3} + 3}  + 3} \right]\)

Solution:  Suppose \(y = \left[ {\sqrt {{x^3} + 3}  + 3} \right]\) 

Hence inverse of \(f(x)\) is \({\left[ {{{\left( {x - 3} \right)}^2} - 3} \right]^{\frac{1}{3}}}\), [ replacing \(y\) with \(x\)]

Example 14: If \(f(x)\) and \(g(x)\) are invertible and 

\(f(2x + 3) = 4g(3x + 2)\), find the value of the function \(\left[ {\frac{{{f^{ - 1}}(x) - 3}}{{{g^{ - 1}}\left( {\frac{x}{4}} \right) - 2}}} \right]\)

Solution: suppose \(f(2x + 3) = 4g(3x + 2) =k\), then

\(2x + {\rm{ }}3{\rm{ }} = {f^{-1}}\left( k \right)\) and \(3x + 2 = {g^{ - 1}}\left( {\frac{k}{4}} \right)\)

Or \(x = \frac{{{f^{ - 1}}(k) - 3}}{2} = \frac{{{g^{ - 1}}\left( {\frac{k}{4}} \right) - 2}}{3}\)

Replacing \(k\) with \(x\) we get,

\(\frac{{{f^{ - 1}}(x) - 3}}{{{g^{ - 1}}\left( {\frac{x}{4}} \right) - 2}} = \frac{2}{3}\)