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.
RonL