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Let $f\colon\mathbb{R}\to\mathbb{R}$ be given by $f(x)= e^{-x}$.

Show that $f$ has a fixed point and determine what it is.

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    $\begingroup$ Any progress at all? $\endgroup$ – rschwieb Dec 15 '14 at 19:26
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$f(x) = x \iff e^{-x} = x \iff e^{-x} - x = 0$.

Consider $g(x) = e^{-x} - x$. We have: $g(0) = 1 > 0$, and $g(1) = e^{-1} -1 <0$. Since $g$ is continuous on $(0,1)$, MVT says there exists $c \in (0,1)$ such that $g(c) = 0 \Rightarrow e^{-c} - c = 0 \Rightarrow f(c) - c = 0 \Rightarrow f(c) = c \Rightarrow $ $c$ is the fixed point.

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  • $\begingroup$ So by using the Mean Value Theorem could I also use this to solve for a numerical value of c? $\endgroup$ – Christine Dec 15 '14 at 21:17
  • $\begingroup$ To solve for $c$, you would have to use Euler or Newton method. $\endgroup$ – DeepSea Dec 15 '14 at 21:58
  • $\begingroup$ I have used the Newton Method and used the equation f(x)=x-[(e^(-x)-x)/(-e^(-x)-1)]. After subbing in subsequent values I tested values 1 through 10 to determine the closest solution would be that c=0.0005 $\endgroup$ – Christine Dec 16 '14 at 15:42

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