I am completely stuck with finding the fixed point of $g(x)$. The question i am stuck is below.
$$x_n=\frac{x_{n-1}}{2}+\frac{1}{x_{n-1}},\qquad n=1,2,\dots$$
This question has two parts. First to find $g(x)$ and then the fixed point of $g(x)$. It would be great if some can help me in finding the fixed point of $g(x)$. Thanks in advance.
According to the recursive equation, we have $g(x) = \dfrac{x}{2} + \dfrac{1}{x}$, and to find the fixed point of it, we set $g(x) = x \implies \dfrac{x}{2} +\dfrac{1}{x} = x \implies \dfrac{1}{x} = \dfrac{x}{2} \implies x^2 = 2 \implies x = \pm \sqrt{2}$. Thus we have $2$ fixed points.