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If $f(x)$ is continuous in the closed interval $[a,b]$ and differentiable in the open interval $a<x<b$, and if $f(a)=a$, $f(b)=b$, prove there exist points $x_1$ and $x_2$ with $a<x_1<x_2<b$ for which the following equation is true: $1/f'(x_1)+1/f'(x_2)=2$.

Hakim
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mark
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  • What are your thoughts on this problem? – Jason Knapp Oct 30 '14 at 15:22
  • I am stuck...it is supposedly an exercise on the mean value theorem in the Advanced Calculus book by A.E.Taylor. – mark Oct 30 '14 at 15:24
  • If the curve y=f(x) crosses the line y=x between x=a and x=b, then the MVT shows that there are two points x_1 and x_2 where f'=1 and the problem is trivial. The interesting case is where there is only ONE point where f'=1...and I am stuck. – mark Oct 30 '14 at 15:28
  • Or, more simply, if f'=1 in two such points, for whatever reason, the problem is trivial. – mark Oct 30 '14 at 15:30
  • Is the image of $f$ the interval $[a,b]$ also, or the problem doesn't mention this? – Vinícius Novelli Oct 30 '14 at 15:39
  • I've given the complete statement of the problem. I've looked at parabolic examples like y=x(2-x) for a=0 and b=1, but to no avail. – mark Oct 30 '14 at 15:43

1 Answers1

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Since $f(x)$ is continuous in the closed interval $[a,b]$ $\exists c\in (a,b)$ s.t $f(c)=\frac{f(a)+f(b)}{2}=\frac{a+b}{2}$ by intermediate value theorem.

Then by mean value theorem,

$\exists x_1\in (a,c)$s.t. $f'(x_1)=\frac{\frac{a+b}{2}-f(a)}{c-a}=\frac{\frac{a+b}{2}-a}{c-a}=\frac{b-a}{2(c-a)}$

$\exists x_2\in (c,b)$s.t. $f'(x_2)=\frac{f(b)-\frac{a+b}{2}}{b-c}=\frac{b-\frac{a+b}{2}}{b-c}=\frac{b-a}{2(b-c)}$

Hence $1/f'(x_1)+1/f'(x_2)=2$.

John
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