# Fixed Point Problem

• Sep 13th 2007, 08:53 PM
Fixed 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 AM
Plato
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 AM
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:

Okay, the solution I found was t=1.

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

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.
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.