Published: f (1) =1; Let x be 0, F(F(0))=F(0)=a, f (a) = f (0) = a; What does the second question X0 mean? And this f(x0? What do you mean?

I see. X0, right?

II: solution: because for any x∈R, there is,

Since there is only one real number x0, f(x0)= x0.

So for any x∈R, there is the square of f(x)-X +X = x0.

Let x= x0 and the square of f in the above formula be (x0)-X0 +X0 = x0.

And because f(x0)= x0, x0-x02=0, x0=0 or x0= 1.

If x0=0, the square of f(x)-X +X =0, that is

F(x)= X squared -X

But the equation = x0 has two different real roots, which contradicts the conditions of the topic, so x0≠0.

If x0= 1, then there is the square of F(x)-x +x = 1, which means it is easy to verify that the function satisfies the conditions.

To sum up, the function is f(x)= x-x squared+1 (x∈R).

And upstairs, are you still a teacher? If you can't, don't mislead the students, ok? I'll take the exam!