Problem
Prove or disprove the following statement: Let $f(x)$ be differentiable over $[a,b]$, and $f'(x)$ be continuous over $(a,b)$. $f'(a)=f'(b)=0$. Then there exist $x_1,x_2$ satisfying $a<x_2<x_1<b$ such that $f'(x_1)=f'(x_2),$ namely, $f'(x)$ can not be injective over $(a,b)$.
Proof
Consider proving by contradiction. Assume the conclusion does not hold. Then $f'(x)$ is injective over $(a,b)$. Taking the fact $f(x)$ is continuous over $(a,b)$ into account, we may obtain $f'(x)$ is strictly monotonic over $(a,b)$. Without loss of generality, we assume $f'(x)$ is strictly increasing over $(a,b)$. Obviously, there exsits $c \in (a,b)$ such that $f'(c) \neq 0$. Without loss of generality, we assume $f'(c)>0$ and consider the interval $(c,b)$. (If $f'(c)<0$, we may consider the interval $(a,c)$. The reasoning is similar.) Notice that $f'(c)>0$ and $f'(b)=0$. According to Darboux's theorem, there exists $\xi \in (c,b)$ such that $f'(\xi)=\dfrac{1}{2}f'(c)$. But, since $f'(x)$ is strictly increasing over $(c,b)$, we may have $\dfrac{1}{2}f'(c)=f'(\xi)>f'(c)>0$, which contradicts. Please correct me if I'm wrong.
This follows directly from the intermediate value property of derivatives. Continuity of derivative is not required.
If $f'$ is constant on $[a, b] $ then we are done. So let there be a $c\in(a, b) $ with $f'(c) \neq 0$. Let $k$ be any number which lies between $0$ and $f'(c) $ (for example we can take $k=f'(c) /2$). By intermediate value property there exists an $x_1\in(a,c)$ such that $f'(x_1)=k$ and an $x_2\in(c,b)$ such that $f'(x_2)=k$.