Proving $f(x) = x^2 + 1$ is surjective Let $f(x) = x^2 + 1$, where $x$ is a real number. Prove that $f$ maps $ \mathbb R$ onto $[1, \infty)$.


We must show that if $y \in Y$, then there exists an $x$ such that $f(x) = y$.

I am tempted to use the property $f(x) = y$ to replace $f(x)$ in  $f(x) = x^2 + 1$ with $y$ and solve $y = x^2 + 1$ for $x$. Would that be permissible what with the property $f(x) = y$ being in the conclusion  of the statement, rather than in the hypothesis and given conditions?
Let $y \in [1,\infty]$.  Then, choose $x=\sqrt{y-1}$, so that $f(x)=(\sqrt{y-1})^2+1=y-1+1=y$, and f is surjective.
Draw a rough sketch of f(x). It is an upward parabola with no real root. So, f(x) is an one-to-one function, but not onto, because co-domain of f(x)=R, whereas range of f(x)=Z(+). So, it is a surjective function.