Check whether the function defined by `f(x+lambda)=1+sqrt(2f(x)-f^2(x))` `AAx in R` is periodic or not. If yes, then find its period `(lambda>0)dot`

Explanation

For the function to be true,
`2f(x)-f^(2)(x) ge 0`
or `f(x)[f(x)-2] le 0 " or " 0 le f(x) le 2 " (1) " `
and from the given function,
`f(x+lambda) ge 1 " or " f(x) ge 1 " (2)" `
From (1) and (2), we have `1 le f(x) le 2.`
Again, we have
`{f(x+lambda)-1}^(2)=2f(x)-f^(2)(x)`
or `{f(x+lambda)-1}^(2)=1+{2 f(x)-f^(2)(x)-1}`
or ` {f(x+lambda)-1}^(2)=1-{f(x)-1}^(2)`
Replacing x by `x+lambda`, we get
`{f(x+2lambda)-1}^(2)=1-{f(x+lambda)-1}^(2)`
Subtracting (3) from (4), we get
`{f(x+2lambda)-1}^(2)={f(x)-1}^(2)`
or ` f(x+2lambda)-1=f(x)-1 " " ( :' 1 le f(x) le 2)`
` :. f(x+2 lambda)=f(x)`
So, `f` is periodic with period ` 2 lambda.`