Finding $m$ such that $f(x) = x^2+2mx+m^2+1$ satisfies $f(\Im_f)=\Im_f$

Question

Problem:

Let $f : \mathbb{R} \to \mathbb{R}$ such that $f(x) = x^2+2mx+m^2+1$. Find $m \in \mathbb{R}$ such that $f(\Im_f) = \Im_f$, where $\Im_f$ indicates the image of $f$.

My question is, what's the actual condition for $f(\Im_f) = \Im_f$ ?

I'm confused.

Answer 1

Note that $f(x)=(x+m)^2+1\geq 1$, so $f(\mathbb{R})=[1,\infty)$. If we want $f([1,\infty))=[1,\infty)$, $(x+m)^2$ must equal zero for some $x\in [1,\infty)$, i.e. $f(x)$ has a minimum on $[1,\infty)$. This can only happen if $m\leq -1$.