Is this a Darboux function?

Question

Let $f(x)=x$ if $0\leq x\leq 1$ and $f(x)=x-\frac{1}{2}$ if $1<x\leq 2$. This is a discontinuous function on $[0;2]$ but it satisfies the intermediate value theorem so it's a Darboux function.

Questions

Am I right? Is $f$ a Darboux function?

If yes, then why do authors give complicated examples of Darboux functions like $g(x)=\sin\left(\frac{1}{x}\right)$ if $x>0$ and $g(0)=0$ or $h(x)=\left(x^2\sin\left(\frac{1}{x}\right)\right)'$ which are hard to draw near $x=0$?. Even worse when they mention Conway base 13 function.

Answer 1

It doesn't satisfy the intermediate value theorem :

$$f(\frac{9}{10}) = \frac{9}{10}$$

$$f(\frac{11}{10}) = \frac{6}{10}$$

But there is no $x \in [\frac{9}{10}, \frac{11}{10}]$ such that $f(x) = \frac{8}{10}$