Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{411 x^{3}}{1000} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{411 x_{n}^{3}}{1000} + 1 + e^{- x_{n}}}{- \frac{1233 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{411 (1.0000000000)^{3}}{1000} + 1 + e^{- (1.0000000000)}}{- \frac{1233 (1.0000000000)^{2}}{1000} - e^{- (1.0000000000)}} = 1.5977211129$ $LaTeX: x_{2} = (1.5977211129) - \frac{- \frac{411 (1.5977211129)^{3}}{1000} + 1 + e^{- (1.5977211129)}}{- \frac{1233 (1.5977211129)^{2}}{1000} - e^{- (1.5977211129)}} = 1.4562474474$ $LaTeX: x_{3} = (1.4562474474) - \frac{- \frac{411 (1.4562474474)^{3}}{1000} + 1 + e^{- (1.4562474474)}}{- \frac{1233 (1.4562474474)^{2}}{1000} - e^{- (1.4562474474)}} = 1.4435569017$ $LaTeX: x_{4} = (1.4435569017) - \frac{- \frac{411 (1.4435569017)^{3}}{1000} + 1 + e^{- (1.4435569017)}}{- \frac{1233 (1.4435569017)^{2}}{1000} - e^{- (1.4435569017)}} = 1.4434608459$ $LaTeX: x_{5} = (1.4434608459) - \frac{- \frac{411 (1.4434608459)^{3}}{1000} + 1 + e^{- (1.4434608459)}}{- \frac{1233 (1.4434608459)^{2}}{1000} - e^{- (1.4434608459)}} = 1.4434608404$