Solved the following function

        If `\f\left(x\right) = \sqrt{x - 3}` then verify `\left(fof^{-1}\right)\left(x\right) = \left(f^- 1of\right)\left(x\right) = x` 

         Answer: 

                Here `f\left(x \right) = \sqrt {x - 3}` 

                Let assume `f\left(x\right) = y` 

                then `y = \sqrt {x - 3}` 

`y^2 = x - 3` 

`x = y^2 + 3` 

                We know that `\(f^ - 1\left(y\right) = x` 

`\(f^- 1\left(y\right) = y^2 + 3` 

                By replacing y by x then

 

`\(f^- 1\left(x\right) = x^2 + 3` 

                First we solve `\left(fof^- 1\right)\left(x\right)` 

`\left(fof^- 1\right)\left(x \right) = f\left(f^ - 1\left(x\right)\right)` 

`\left(fof^- 1\right)\left( x \right) = f\left(x^2 + 3\right)` 

`\left(fof^- 1\right)\left(x\right) = \sqrt{\left(x^2 + 3\right)- 3}`

 
`\left(fof^- 1\right)\left(x\right) = \sqrt x^2` 

                                           `\left(fof^ - 1\right)\left(x\right) = x                                  - - - - - - - \left(1\right)`

                Now we solve  `\left(f^- 1of\right)\left(x\right)` 

`\left(f^- 1of\right)\left(x\right) = f^- 1\left(f\left(x\right)\right)` 

`\left(f^- 1of\right)\left(x\right) = f^- 1\left(\sqrt {x - 3}\right)` 

`\left(f^- 1of\right)\left(x\right) = \left(\sqrt{x - 3}\right)^2 + 3` 

`\left(f^- 1of\right)\left( x \right) = x - 3 + 3` 

                                        `\left(f^- 1of\right)\left(x\right) = x              - - - - - - - - - -\left( 2 \right)` 

                By (1) and (2) we can prove that 

`\left(fof^- 1\right)\left(x\right) = \left(f^- 1of \right)\left(x\right) = x`