Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{971 x^{3}}{1000} - 7$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{971 x_{n}^{3}}{1000} + 7 + e^{- x_{n}}}{- \frac{2913 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{971 (1.0000000000)^{3}}{1000} + 7 + e^{- (1.0000000000)}}{- \frac{2913 (1.0000000000)^{2}}{1000} - e^{- (1.0000000000)}} = 2.9497453521$ $LaTeX: x_{2} = (2.9497453521) - \frac{- \frac{971 (2.9497453521)^{3}}{1000} + 7 + e^{- (2.9497453521)}}{- \frac{2913 (2.9497453521)^{2}}{1000} - e^{- (2.9497453521)}} = 2.2461932798$ $LaTeX: x_{3} = (2.2461932798) - \frac{- \frac{971 (2.2461932798)^{3}}{1000} + 7 + e^{- (2.2461932798)}}{- \frac{2913 (2.2461932798)^{2}}{1000} - e^{- (2.2461932798)}} = 1.9828377987$ $LaTeX: x_{4} = (1.9828377987) - \frac{- \frac{971 (1.9828377987)^{3}}{1000} + 7 + e^{- (1.9828377987)}}{- \frac{2913 (1.9828377987)^{2}}{1000} - e^{- (1.9828377987)}} = 1.9455609921$ $LaTeX: x_{5} = (1.9455609921) - \frac{- \frac{971 (1.9455609921)^{3}}{1000} + 7 + e^{- (1.9455609921)}}{- \frac{2913 (1.9455609921)^{2}}{1000} - e^{- (1.9455609921)}} = 1.9448555749$