Question

Let `f` be a real function defined by `f(x)=sqrt(x-1.)` Find `(fofof)(x)dot` Also, show that `fof!=f^2dot`

Answer 1

`f(x) = sqrt(x-1)`
`(fofof)(x) = fof(f(x)) = fof(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`
`fof(x) = f(sqrt(x-1)) = sqrt(sqrt(x-1)-1)`
`:. f^2 != fof`.