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(B) To find the fixed point, we set $f(c) = c$.$1 + \sqrt{c} = c$$\sqrt{c} = c - 1$Squaring both sides (where $c \geq 1$):$c = (c - 1)^2$$c = c^2 - 2c

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$f$ has a unique fixed point in $[1, \infty)$

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(B) To find the fixed point, we set $f(c) = c$.$1 + \sqrt{c} = c$$\sqrt{c} = c - 1$Squaring both sides (where $c \geq 1$):$c = (c - 1)^2$$c = c^2 - 2c + 1$$c^2 - 3c + 1 = 0$Using the quadratic formula $c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$c = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}$Since $c \geq 1$, we check the values:$\frac{3 + \sqrt{5}}{2} \approx \frac{3 + 2.236}{2} = 2.618 \geq 1$ (Valid)$\frac{3 - \sqrt{5}}{2} \approx \frac{3 - 2.236}{2} = 0.382 Thus, there is only one fixed point in the domain $[1, \infty)$.

Given that $f: S \rightarrow R$ is said to have a fixed point at $c \in S$ if $f(c)=c$. Let $f:[1, \infty) \rightarrow R$ be defined by $f(x)=1+\sqrt{x}$. Then:

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