SOLUTION 2: We are given the equation $x^3=x^2+2$, or $x^3-x^2-2=0$, so let's define function $f(x) = x^3-x^2-2$, whose graph is given below.

The derivative of $f$ is $f'(x) = 3x^2-2x$. Now use Newton's Method: $$x_{n+1} = x_{n} - { f(x_{n}) \over f'(x_{n}) } \ \ \ \ \longrightarrow$$ $$x_{n+1} = x_{n} - { x_{n}^3-x_{n}^2-2 \over 3x_{n}^2-2x_{n} } \ \ \ \ \longrightarrow$$ (Let's simplify the right-hand side of this equation. First get a common denominator.) $$x_{n+1} = x_{n} \ { 3x_{n}^2-2x_{n} \over 3x_{n}^2-2x_{n} } - { x_{n}^3-x_{n}^2-2 \over 3x_{n}^2-2x_{n} } \ \ \ \ \longrightarrow$$ $$x_{n+1} = { 3x_{n}^3-2x_{n}^2 \over 3x_{n}^2-2x_{n} } - { x_{n}^3-x_{n}^2-2 \over 3x_{n}^2-2x_{n} } \ \ \ \ \longrightarrow$$ $$x_{n+1} = { 3x_{n}^3-2x_{n}^2 - ( x_{n}^3-x_{n}^2-2 ) \over 3x_{n}^2-2x_{n} } \ \ \ \ \longrightarrow$$ $$x_{n+1} = { 2x_{n}^3-x_{n}^2+2 \over 3x_{n}^2-2x_{n} }$$ a.) Let $x_{0}=1$. Then using Newton's Method formula we get that $$x_{1} = { 2x_{0}^3-x_{0}^2+2 \over 3x_{0}^2-2x_{0} } = { 2(1)^3-(1)^2+2 \over 3(1)^2-2(1) } = 3$$ and $$x_{2} = { 2x_{1}^3-x_{1}^2+2 \over 3x_{1}^2-2x_{1} } = { 2(3)^3-(3)^2+2 \over 3(3)^2-2(3) } = { 47 \over 21 } \approx 2.238095238$$ Using Newton's Method formula for 7 iterations in a spreadsheet results in :

Thus the solution $r$ to the original equation to five decimal places is $r \approx 1.69562$.

b.) Let $x_{0}=2$. Then using Newton's Method formula we get that $$x_{1} = { 2x_{0}^3-x_{0}^2+2 \over 3x_{0}^2-2x_{0} } = { 2(2)^3-(2)^2+2 \over 3(2)^2-2(2) } = { 14 \over 8 } = 1.75$$ and $$x_{2} = { 2x_{1}^3-x_{1}^2+2 \over 3x_{1}^2-2x_{1} } = { 2(1.75)^3-(1.75)^2+2 \over 3(1.75)^2-2(1.75) } = { 618 \over 364 } \approx 1.697802198$$ Using Newton's Method formula for 5 iterations in a spreadsheet results in :

Thus the solution $r$ to the original equation to five decimal places is $r \approx 1.69562$.

c.) Let $x_{0}=0.5$. Then using Newton's Method formula we get that $$x_{1} = { 2x_{0}^3-x_{0}^2+2 \over 3x_{0}^2-2x_{0} } = { 2(0.5)^3-(0.5)^2+2 \over 3(0.5)^2-2(0.5) } = -8$$ and $$x_{2} = { 2x_{1}^3-x_{1}^2+2 \over 3x_{1}^2-2x_{1} } = { 2(-8)^3-(-8)^2+2 \over 3(-8)^2-2(-8) } = { -1086 \over 208 } \approx -5.221153846$$ Using Newton's Method formula for 13 iterations in a spreadsheet results in :

Thus the solution $r$ to the original equation to five decimal places is $r \approx 1.69562$.

d.) Let $x_{0}=1000$. Then using Newton's Method formula we get that (Use technology.) $$x_{1} = { 2x_{0}^3-x_{0}^2+2 \over 3x_{0}^2-2x_{0} } = { 2(1000)^3-(1000)^2+2 \over 3(1000)^2-2(1000) } \approx 666.7778526$$ and $$x_{2} = { 2x_{1}^3-x_{1}^2+2 \over 3x_{1}^2-2x_{1} } = { 2(666.7778526 )^3-(666.7778526 )^2+2 \over 3(666.7778526 )^2-2(666.7778526 ) } = \approx 444.6297922$$ Using Newton's Method formula for 21 iterations in a spreadsheet results in :

Thus the solution $r$ to the original equation to five decimal places is $r \approx 1.69562$.

e.) Let $x_{0}=-500$. Then using Newton's Method formula we get that (Use technology.) $$x_{1} = { 2x_{0}^3-x_{0}^2+2 \over 3x_{0}^2-2x_{0} } = { 2(-500)^3-(-500)^2+2 \over 3(-500)^2-2(-500) } \approx -333.2223675$$ and $$x_{2} = { 2x_{1}^3-x_{1}^2+2 \over 3x_{1}^2-2x_{1} } = { 2(-333.2223675)^3-(-333.2223675)^2+2 \over 3(-333.2223675)^2-2(-333.2223675) } = \approx -222.0373498$$ Using Newton's Method formula for 25 iterations in a spreadsheet results in :

Thus the solution $r$ to the original equation to five decimal places is $r \approx 1.69562$.