Let $f:[0,2] \to \mathbb R$ be differentiable with $f(0)=0,f(1)=2$, and $f(2)=1$. I must prove that there exists $c \in (0,2)$ such that $f'(c)=-\frac12$.
I have nearly completed the proof, but there is one claim which I am not certain I can make. For now, here is the proof:
We apply the Mean Value Theorem to obtain that there exists $c_1 \in (0,1)$ such that $f(1)-f(0)=f'(c_1)(1-0) \iff f'(c_1)=2$. We apply M.V.T. again to find that there exists $c_2 \in (1,2)$ such that $f(2)-f(1)=f'(c_2)(2-1) \iff f'(c_2)=-1$. This implies that $k=-\frac12$ is between $f'(0)$ and $f'(2)$, and hence, by Darboux's theorem, there exists $c \in (0,2)$ such that $f'(c)=-\frac12$.
My problem is with claiming that $k=-\frac12$ is between $f'(0)$ and $f'(2)$. I am not sure if this necessarily follows from the previous result. It would be fine if the question allowed $f$ to be continuously differentiable, but as it is, we can potentially have $f'$ to not be continuous for all $x \in [0,2]$. Is there any way I could fix/improve my proof, or is it fine as is? Thank you!
Best Answer
Indeed, you don't have enough information to determine $f'(0)$ and $f'(2)$ (which would have to be one-sided derivatives, at best). So you're wrong to claim $-1/2$ is between $f'(0)$ and $f'(2)$.
But this is no problem; you can do even better. You know $f'(c_2)<f'(c_1)$ and $k$ is between these two values; you also know $0<c_1<1<c_2<2$. So there is a $c$ in $(c_1,c_2)$ such that $f'(c)=k$. But if $c\in(c_1,c_2)$, then $c\in(0,2)$, because $(c_1,c_2)\subset(0,2)$.