Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{239 x^{3}}{250} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{239 x_{n}^{3}}{250} + 1 + e^{- x_{n}}}{- \frac{717 x_{n}^{2}}{250} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{239 (1.0000000000)^{3}}{250} + 1 + e^{- (1.0000000000)}}{- \frac{717 (1.0000000000)^{2}}{250} - e^{- (1.0000000000)}} = 1.1272851627$ $LaTeX: x_{2} = (1.1272851627) - \frac{- \frac{239 (1.1272851627)^{3}}{250} + 1 + e^{- (1.1272851627)}}{- \frac{717 (1.1272851627)^{2}}{250} - e^{- (1.1272851627)}} = 1.1157997146$ $LaTeX: x_{3} = (1.1157997146) - \frac{- \frac{239 (1.1157997146)^{3}}{250} + 1 + e^{- (1.1157997146)}}{- \frac{717 (1.1157997146)^{2}}{250} - e^{- (1.1157997146)}} = 1.1156961846$ $LaTeX: x_{4} = (1.1156961846) - \frac{- \frac{239 (1.1156961846)^{3}}{250} + 1 + e^{- (1.1156961846)}}{- \frac{717 (1.1156961846)^{2}}{250} - e^{- (1.1156961846)}} = 1.1156961763$ $LaTeX: x_{5} = (1.1156961763) - \frac{- \frac{239 (1.1156961763)^{3}}{250} + 1 + e^{- (1.1156961763)}}{- \frac{717 (1.1156961763)^{2}}{250} - e^{- (1.1156961763)}} = 1.1156961763$