Monotonically increasing function and supremes

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ a monotonically increasing function and $A \subset \mathbb{R}$ where $A \neq \emptyset$ and boundend.

i) If f is continuous function, prove that $f(\sup (A))= \sup (f(A))$

ii) Find a function $f: \mathbb{R} \rightarrow \mathbb{R}$ wich do not fulfilled i).

For i), I thought because $A$ is boundend it has supreme,
$$ x \le \sup A$$In other hand, because $f$ is monotonically increasing, if $x \le \sup A$ then $f(x) \le f(\sup A)$ for $x \in A$.

As $f$ is continuous and monotonically increasing in a boundend set, the set $f(A)$ has a supreme, i.e $f(y) \le \sup f(A)$ for $y \in A$, so $\sup f(A)$ and $f(\sup A)$ are upper bounds.

But I don't know what else I can do.