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.
 A: By the Intermediate Value Theorem there exists a point $c \in (1, 2)$ such that
$$f(c) = f(0),$$
since by hypothesis $f(1) < f(0) < f(2)$.
Applying Rolle's theorem on the interval $[0, c]$ yields the desired result.

By Weierstrass' Extreme Value Theorem, the continuous function $f$ attains either a maximum or a minimum $m$ in $[0, 2]$. By hypothesis $m \in (0, 2)$. Since $f$ is differentiable, by Fermat's theorem it must satisfy
$$f'(m) = 0.$$
A: 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 $f'(x_0)=0$.
A: After discussion with Daniel Fischer I decided to leave this as an answer.
First thing to notice is that $f$ is not differentiable only at endpoints, hence it's differentiable everywhere in the interior of $I=(a,b)$. 
Secondly, since $f(1)<f(0)$ and $f(1) <f(2)$, either of two intervals exist: 
$$
I_1 = [\alpha, \beta] \ \text{s.t.} 0<\alpha<\beta \leq 1\\
I_2 = [\gamma, \delta] \ \text{s.t.} 1 \leq \gamma <\delta < 1
$$
where $f$ has a critical point, and at this critical point it is differentiable (it can also be $f(1)$, hence the weak inequality above). Since it is differentiable, $f'$ is continuous and changes sign $\to \ f'$ has a solution in at least 1 point in $I$.
