Let `f` be a real function given by `f(x)=sqrt(x-2)` . Find `fofdot` Also, show that `fof!=f^2`

Explanation

`f(x)=sqrt{x-1}`
`
(f o f o f)(x)=f o f(f(x))=f o f(sqrt{x-1})=f(sqrt{sqrt{x-1}-1})`
`
=(sqrt{sqrt{sqrt{x-1}-1}-1})`

Now, `f^{2}(x)=(sqrt{x-1})^{2}=x-1`
`
f o f(x)=f(sqrt{x-1})=sqrt{sqrt{x-1}-1}`
`
therefore f^{2} != f o f`