Check whether the function defined by $f(x+\lambda)=1+\sqrt{2f(x)-f(x)^2}$ for all $x$ belonging to the set of real numbers is periodic or not. If yes, then find its period ($\lambda >0$).
I can't understand how to approach this question can anyone please help.
Answer: Period is $2\lambda$.
$$f(x+\lambda)=1+\sqrt{2f(x)-(f(x))^2}$$
$$\Rightarrow(f(x+\lambda)-1)^2=2f(x)-(f(x))^2$$
$$\Rightarrow(f(x+\lambda))^2-2f(x+\lambda)=-(f(x)-1)^2$$
$$\Rightarrow f(x)=1+\sqrt{2f(x+\lambda)-(f(x+\lambda))^2}$$
Substituting $x$ by $x+\lambda$ in this equation, we get
$$\Rightarrow f(x+\lambda)=1+\sqrt{2f(x+2\lambda)-(f(x+2\lambda))^2}$$
On comparing this with the given equation, we get
$$\Rightarrow f(x)=f(x+2\lambda)$$
$\Rightarrow f(x)$ is periodic with period $2\lambda$.