Let $f : R \rightarrow R$ be differentiable, and $f(0) = 0$, $f(1) = f(2) = −1$. Prove that there are three points $a, b, c$ in $(0,2)$ such that $f'(a)=-1/2$, $f'(b)=-3/4$ and $f'(c)=-1/11$.
I don't have any idea to solve the problem. Please solve this problem.
Using the Mean Value Theorem, we have that there exists $m \in (0,1)$ and $n \in (1,2)$ such that $$f'(m)=\frac{f(1)-f(0)}{1-0}=-1$$$$f'(n)=\frac{f(2)-f(1)}{2-1}=0$$
Since we have $$-1<-\frac{3}{4}<-\frac{1}{2}<-\frac{1}{11}<0$$
Using Darboux's Theorem, we have that there exists $\{a,b,c\} \subset (m,n)$ such that $$f'(a)=-\frac{1}{2} ,\; f'(b)=-\frac{3}{4}, \; f'(c)=-\frac{1}{11}$$ Thus, our proof is done.