Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{47 x^{3}}{50} - 4$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{47 x_{n}^{3}}{50} + 4 + e^{- x_{n}}}{- \frac{141 x_{n}^{2}}{50} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{47 (1.0000000000)^{3}}{50} + 4 + e^{- (1.0000000000)}}{- \frac{141 (1.0000000000)^{2}}{50} - e^{- (1.0000000000)}} = 2.0752851557$ $LaTeX: x_{2} = (2.0752851557) - \frac{- \frac{47 (2.0752851557)^{3}}{50} + 4 + e^{- (2.0752851557)}}{- \frac{141 (2.0752851557)^{2}}{50} - e^{- (2.0752851557)}} = 1.7268081679$ $LaTeX: x_{3} = (1.7268081679) - \frac{- \frac{47 (1.7268081679)^{3}}{50} + 4 + e^{- (1.7268081679)}}{- \frac{141 (1.7268081679)^{2}}{50} - e^{- (1.7268081679)}} = 1.6496758167$ $LaTeX: x_{4} = (1.6496758167) - \frac{- \frac{47 (1.6496758167)^{3}}{50} + 4 + e^{- (1.6496758167)}}{- \frac{141 (1.6496758167)^{2}}{50} - e^{- (1.6496758167)}} = 1.6461168327$ $LaTeX: x_{5} = (1.6461168327) - \frac{- \frac{47 (1.6461168327)^{3}}{50} + 4 + e^{- (1.6461168327)}}{- \frac{141 (1.6461168327)^{2}}{50} - e^{- (1.6461168327)}} = 1.6461094721$