Let $f:[0,2]\to \mathbb{R}$ be a continuous function with no roots. Prove that the function is not surjective and $f(0)\cdot f(2)>0$ Let $f:[0,2]\to \mathbb{R}$ be a continuous function with no roots
i) Prove that the function is not surjective

ii) Show that $f(0)\cdot f(2)&gt;0$
Got no ideas, maybe $y=f(x)$? Doesn't seem helpful though.
Also, is it possible to change the interval from $[0, 2]$ to $[a, b]$? I remember asking about a problem here for a particular 'case' and it ended up being true for every 'case': How to prove the following integral equation? $\int_{0}^{c}x^2f(x)=0$ (hope it makes sense what I'm saying)
Thanks in advance!
Since it has no roots there is no $x \in [0,2]$ such that $f(x) = 0 \in \mathbb R$
(Updated) Assume $f(0)\cdot f(2) \le 0$, but this implies either at least one of $f(0)$ or $f(2)$ is $0$ (contradiction) or they have different signs which means there is a zero between $0$ and $2$ (contradiction). Therefore $f(0)\cdot f(2) &gt; 0$.
i) $f^{-1}(0) = \emptyset$.
ii) Is not true, consider $f \equiv 1$.