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Consider a differentiable fuction $f: \mathbb{R} \rightarrow \mathbb{R}$ for which $f'(t) \neq 1$ for every $t \in \mathbb{R}$. Is it true that $f$ has exactly one fixed point? It is clear to me why it can have no more than one, but how does one show that $f$ can't have none?

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What you are trying to show is false: Consider $f (x) = x - e^x$. Then $f'(x) = 1 - e^x < 1$ for all $x $, but if $f $ had a fixed point, then $x = f (x) = x - e^x $, and hence $e^x =0$, a contradiction.

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The upper bound follows from Rolle's theorem: a fixed point of $f$ is a root of $g(t)=f(t)-t$, and the assumption and Rolle's theorem gives that there can only be one of those.

I think the lower bound is not true: take $g(x)$ to be any strictly monotone continuously differentiable function with no roots, like $e^x$ or $\arctan(x)-10$. Then $f(x)=x+g(x)$ has no fixed point but $f'$ is never $1$.

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