Let $f(x)$ be continuous on $[0,2]$, and differentiable on $(0,2)$ such that $0
Let $f(x)$ be continuous on $[0,2]$, and differentiable on $(0,2)$ such that $0<f(1)<f(0)<f(2)$. Prove that $f'$ has a solution on $(0,2)$.
Here's a little crappy sketch:

My attempt:
From $f(1)<f(0)<f(2)$ and continuity, there's a point $c\in (1,2)$ such that $f(c)=f(0)$, $f$ is continuous on $[0,c](\subseteq[0,2])$, differentiable on $(0,c)(\subseteq(0,2))$ so from Rolle's we know that there's some $k\in (0,c)$ such that $f'(k)=0$.
Is this alright? Is there another way to do this? Maybe with Lagrange's MVT?
Note: no integration or Taylor's.

Since $f$ is continuous and $[0,2]$ is compact, $f$ attains its global minimum at some point $x_0\in[0,2]$. As $f(1)<f(0)$ and $f(1)<f(2)$, we see that in fact $x_0\in(0,2)$. As we have a minimum in an open intervall, we conclude $f'(x_0)=0$.
