Show that the equation x3+10x=2x^3 + 10x = 2x3+10x=2 has a solution between x=0x = 0x=0 and x=1x = 1x=1.
Show that the equation x3+10x=2x^3 + 10x = 2x3+10x=2 can be rearranged to give: x=15−x310\displaystyle x = \frac{1}{5} - \frac{x^3}{10}x=51−10x3
Starting with x0=0x_0 = 0x0=0, use the iteration formula xn+1=15−xn310\displaystyle x_{n+1} = \frac{1}{5} - \frac{x_n^3}{10}xn+1=51−10xn3 twice to find an estimate for the solution to x3+10x=2x^3 + 10x = 2x3+10x=2.
