Let $f(x) = 2 − |2x − 1|$. Show that there is no value of $c$ such that $$f(3) − f(0) = f'(c)(3 − 0).$$ Why does this not contradict the Mean Value Theorem?
Note: I am not asking how to solve this problem, I'm asking a question about this problem.
EX1: For the answer of this question, $x$ is not differentiable at $1/2$ because the left slope $2$ is not equal to the right slope $-2$. So there is no value $c$ where $f'(c)$ has a slope of $-4/3$ on the interval $(0,3)$.
EX2: My question is that what if you restricted the domain to $[3,9]$? Then the function would be continuous as before but also DIFFERENTIABLE this time. However, there is still no value $c$ where $f'(c)$ has a slope of $-4/3$.
I know my question may sound confusing but in short, what does it really mean when it says "contradict the mean value theorem?" Is EX2 a contradiction to MVT and EX1 isn't because EX1 has an extra condition where it's not differentiable on the domain $(0,3)$?