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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)

Dmoreno
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Daniel250
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  • Just an advice to make things a bit less complicated: start by proving it for $a=0$ and $b=1$. This special case can be generalized. – drhab Aug 18 '14 at 11:53
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    Let $g(x)=f(x)-x$. Then $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$, just like $f$ is. However $g(a)=g(b)=0$ and $g^\prime = f^\prime-1$, so the equality you have to prove is equivalent to $g^\prime(c_1) + g^\prime(c_2) = 0$. If $g$ is constant, the equation is satisfied by any two points, otherwise there is a subinterval, where $g$ is increasing (and $g^\prime$ is positive) and another subinterval where $g$ is decreasing (and $g^\prime$ is negative). HTH. – CiaPan Aug 18 '14 at 12:19
  • Generally, in questions like this, using $g(x)=f(x)-x$ gives some leads. So here, your question reduces to proving that there exist $c_1,c_2 \in (a,b)$ such that $g'(c_1)+g'(c_2)=0$, where $g(a)=g(b)=0$. This immediately strikes me that its a special case of the more general version of Rolle's, which robjhon proves here: http://math.stackexchange.com/questions/237884/rolles-theorem . Keep in mind that theorem next time :) – Sawarnik Aug 18 '14 at 14:28

2 Answers2

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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 $$f'(c_1)+f'(c_2)=\frac{2\big(f(d)-f(a)\big)+2\big(f(b)-f(d)\big)}{b-a}=\frac{2\big(f(b)-f(d)+f(d)-f(a)\big)}{b-a}=\frac{2\big(f(b)-f(a)\big)}{b-a}=\frac{2\big(b-a\big)}{b-a}=2$$

idm
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0

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.

Jonas Dahlbæk
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