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

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- Sep 13th 2007, 08:53 PMtttcomraderFixed Point Problem
Define $\displaystyle f:[1, \infty ] \rightarrow R $ by $\displaystyle f(x) = 1+ \sqrt{x}$ for all $\displaystyle x \geq 1$. Show that f has exactly one fixed point.

- Sep 14th 2007, 03:52 AMPlato
You do know that $\displaystyle f$ is strictly increasing on $\displaystyle [1,\infty )$?

The find a solution to $\displaystyle 1 + \sqrt t = t$ where $\displaystyle t>1$. - Sep 14th 2007, 09:35 AMtttcomrader
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 AMPlato
- Sep 14th 2007, 12:26 PMCaptainBlack
If $\displaystyle x \in [0, \infty)$ is a fixed point of $\displaystyle f$ then:

$\displaystyle

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

$

Now introduce $\displaystyle y=\sqrt{x}$, 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