Suppose $f:[0,1] \to [0,1]$ is continuous on $[0,1]$ and differentiable on $(0,1)$. If $f'(x)\neq 1$ for all $x\in(0,1)$. Prove there is at most one $c\in[0,1]$ such that $f(c)=c$.

  • $\begingroup$ Put latex between dollars. $\endgroup$
    – ajotatxe
    Nov 10, 2014 at 18:51
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    $\begingroup$ What happens if there are two such numbers? $\endgroup$
    – Simon S
    Nov 10, 2014 at 18:51
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    $\begingroup$ @ajotatxe : To call it LaTeX is highly misleading. It's MathJax. In LaTeX one does not work only with mathematical notation. $\endgroup$ Nov 10, 2014 at 18:53

1 Answer 1


If there is another, say $d$, then there exists some $\xi$ such that $$f'(\xi)=\frac{f(d)-f(c)}{d-c}=\frac{d-c}{d-c}$$

  • $\begingroup$ This notation is just what I was taught in eighth and ninth grades. Only later did I learn that that "prime" is also used for derivatives. But I suspect many student in calculus courses would think that by $c'$ you mean the derivative of $c$. ${}\qquad{}$ $\endgroup$ Nov 10, 2014 at 18:56

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