Proving $f'(c_1)+f'(c_2)=2$ for $f$ such that $f(a) = a$ and $f(b) = b$ Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$. If $f(a) = a$ and $f(b) = b$, show that there exist distinct $c_1$, $c_2$ $\in$ $(a,b)$ such that $f'(c_1)+f'(c_2)=2.$
My try: 
By applying Mean Value Theorem on interval $(a,b)$ one can see that for some $c_1\in(a,b), \,f'(c_1)=1$ but how do I prove that another point $c_2$ lies in the given interval for which $f'(c_2)=1$
PS: One possible way I considered was to prove that for some $x\in(a,b),f(x)=x$ but I could not do so (it's false most probably anyway)
 A: Let $d=\frac{a+b}{2}$. By the mean value theorem,
$$f'(c_1)=\frac{f(d)-f(a)}{d-a}=\frac{2\big(f(d)-f(a)\big)}{b-a}$$ and $$f'(c_2)=\frac{f(b)-f(d)}{b-d}=\frac{2\big(f(b)-f(d)\big)}{b-a}$$ for certains $c_1\in]a,d[$ and $c_2\in]d,b[$. Then
$$f'(c_1)+f'(c_2)=\frac{2\big(f(d)-f(a)\big)+2\big(f(b)-f(d)\big)}{b-a}=\frac{2\big(f(b)-f(d)+f(d)-f(a)\big)}{b-a}=\frac{2\big(f(b)-f(a)\big)}{b-a}=\frac{2\big(b-a\big)}{b-a}=2$$
A: Consider the function $g(x)=f(x)-x$. Then $g(a)=g(b)=0$, and we need to show that $g'(c_1)+g'(c_2)=0$ for distinct $c_1,c_2$. If $g=0$, we are done, so suppose this is not the case. Since $g$ has the desired property if and only if $-g$ has the same property, we can without loss of generality assume that there is an $x_0\in(a,b)$ such that $g(x_0)>0$. It follows that $g$ has a global maximum $m\in (a,b)$ and thus $g'(m)=0$. Observe now by the mean value theorem that there is a point $x_1\in(a,m)$ such that $g'(x_1)>0$ and a point $x_2\in(m,b)$ such that $g'(x_2)<0$. If $|g'(x_1)|= |g'(x_2)|$, we are done. Assume without loss of generality that $|g'(x_1)|< |g'(x_2)|$. By Darboux's Theorem, we conclude that there is an $x_3\in(m,x_2)$ such that $g'(x_1)+g'(x_3)=0$. Now take $c_1=x_1, c_2=x_3$ to finish the argument in this final case.
