# Fixed Point Problem

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• Sep 13th 2007, 08:53 PM
tttcomrader
Fixed Point Problem
Define $f:[1, \infty ] \rightarrow R$ by $f(x) = 1+ \sqrt{x}$ for all $x \geq 1$. Show that f has exactly one fixed point.
• Sep 14th 2007, 03:52 AM
Plato
You do know that $f$ is strictly increasing on $[1,\infty )$?
The find a solution to $1 + \sqrt t = t$ where $t>1$.
• Sep 14th 2007, 09:35 AM
tttcomrader
Okay, the solution I found was t=1, but you said t > 1. I can't find a solutoin that matches this condition.
• Sep 14th 2007, 10:16 AM
Plato
Quote:

Originally Posted by tttcomrader
Okay, the solution I found was t=1.

How in the world is t=1 a solution?
$1 + \sqrt 1 = 2 \ne 1.$
• Sep 14th 2007, 12:26 PM
CaptainBlack
Quote:

Originally Posted by tttcomrader
Define $f:[1, \infty ] \rightarrow R$ by $f(x) = 1+ \sqrt{x}$ for all $x \geq 1$. Show that f has exactly one fixed point.

If $x \in [0, \infty)$ is a fixed point of $f$ then:

$
x=f(x)=1+\sqrt{x}
$

Now introduce $y=\sqrt{x}$, solve the equation you get for y, and then sort the roots according to if they correspond to valid solutions of the original fixed point problem.

RonL