so I have the following function $f: x \in \mathbb{R} \rightarrow x^{3}-3x$. Let $A$ be the biggest Interval, so that $0 \in A$ and $f$ is monotone and convex in $A$.
Show that $3 \sup A-7 \inf A=3$.
So i am pretty unclear what it exactly means for me that the function in this interval should be convex and monotone. Does it mean $f'(x)\geq 0$. Because i solved this and got $x\geq 1$ and $x\leq-1$ Or does it mean that $f''(x)>0$ therefore $x>0$. However in either of this cases I don't get the value $3$ which I am supposed to get.
You have to take both $f'(x) \leq 0$ and $f''(x) \geq 0$ into account. [There is no interval containing $0$ where $f'(x) \geq 0$ so we have to considering only decreasing $f$]. So we get $0 \leq x \leq 1$. This means $A=[0,1]$. So $3 \sup A-7 \inf A=3-0=3$.