Order-preserving function

Question

I have an order on $\mathbb R ^ \mathbb R: f \le g $ iff $\forall x \in \mathbb R: f(x) \le g(x)$. Now I have a function $\mathbb R ^ \mathbb R \to \mathbb R ^ \mathbb R: F (f)(x)= f(x^2)+1$. I have to decide, if this is function is order-preserving or not. My train of thought: $\forall x \in \mathbb R: f(x) \le g(x) \Rightarrow f(x^2) \le g(x^2)$(just call $x^2 = y \in \mathbb R, $so$ f(y) \le g(y)$) $ \Rightarrow f(x^2)+1 \le g(x^2)+1$, so this is function is order-preserving, but I can't decide if my explanation sufficient or not.

Answer 1

What you say is correct but a bit confusing. Why not just say, supposing $f\leq g$, that for $x\in\Bbb R$ one has $$ F(f)(x)=f(x^2)+1\leq g(x^2)+1=F(g)(x) $$ so (since this holds for all $x\in\Bbb R$) one has $F(f)\leq F(g)$, and $F$ is order preserving.