Let $f$ be a continuously differentiable function on $[a,b]$ with $f(a)=f(b)$ and $f'(a)=f'(b)$. Then, do there exist $x_1,x_2\in(a,b)$ such that $f'(x_1)=f'(x_2)$ for $x_1\neq x_2$?

I think the answer is yes by intermediate value theorem, but am not getting the sufficient rigour to prove it. Any hints. Thanks beforehand.


By the mean value theorem, there exists $c \in (a,b)$ s.t. : $$0=\frac {f(b)-f(a)}{b-a}=f'(c).$$

Knowing that $f'$ is continuous, by IVT, there exists $x_1 \in (a,c)$ such that $f'(x_1)=\dfrac{f'(a)}2$. By the same reasonning, there exists $x_2 \in (c,b)$ such that $f'(x_2)=\dfrac{f'(b)}2$.

Since $f'(a)=f'(b)$, $f'(x_1)=f'(x_2)$ and $x_1 \neq x_2$ since they belong to disjoints intervals.

Note that we could have taken any value between $f'(a)$ and $0$. Choosing $\dfrac{f'(a)}2$ and $\dfrac{f'(b)}2$ was just for convenience.

  • $\begingroup$ great! just using both intermediate and mean value theorems produce so many wonderful results in calculus and analysis, isnt it? $\endgroup$ – vidyarthi Sep 20 '17 at 7:11

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