The question is suppose that $f(x)$ is continious in $[0,3]$ and have derivative in $(0,3)$. $f(0)+f(1)+f(2)=3$, $f(3)=1$, try to prove that there must exist a number $t \in (0,3)$,s.t. $f'(t)=0$.
My proof like this:
I supppose that if $f(3)$ is a maximum, then it means that $f(0),f(1),f(2) \leq 1$, so $f(0)=f(1)=f(2)=1$, and it is easy to see that there exists a number $t \in (0,1) or (1,2)$, s.t. $f'(t)=0$. Then, if $f(3)$ is minimum, the prove is similar.
Now, suppose that $f(3)$ is neither the maximum nor the minimum. Because $f(x)$ is continious in $[0,3]$, so there must have a minimum $m$ and a maximum $M$. If $f(0)$ is a maximum, then if it is also a minimum, the function will be constant, it is absolutely $f'(x)=0 \forall x \in(0,3)$, if $f(0)$ is not minimum, then there must exist a number $t \in (0,3)$, s.t. $f(t)$ is a minimum, and its derivative is $0$. If $f(0)$ is a minimum, the prove is similar.
I think that my prove is a bit sophisticated, hope that someone can tell me if my proof is right and if there exists some easier proof.
Since $f(0)+f(1)+f(2)=3$, there exists $a,b\in\{0,1,2\}$ such that $f(a)\leq 1$ and $f(b) \geq 1$. (Can you see why?)
If $a=b$, then you are done: $f(a)=1$, and apply Rolle's theorem to $f$ on $[a,3]$.
If $a\neq b$, then by the IVT there exists $c\in[a,b]$ such that $f(c)=1$. Apply Rolle's theorem to $f$ on $[c,3]$.