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.

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 $I$.
