Iteration for solving $x=g(x)$. $g(x) = \frac{x^2}{3}$ $P=3$ $p_0 = 3.5$ 1) Graph $g(x)$, the line $y=x$, and the fixed point $P$ ( **done** ) 2) Using the given starting value $p_0$, compute $p_1$ and $p_2$ ( **the answer might be** $p_1 = 4.083333,p_2 = 5.537869$) Determine geometrically if fixed point iteration converges ( **answer: diverges** ) We might decide this using the analogical graphing structure: !enter image description here Please, help!

Writing $g(x)=x\cdot \frac{x}{3}$, we see that if $x>3$ then $g(x)>x$ and if $0\leq x<3$, then $g(x)