Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{147 x^{3}}{250} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{147 x_{n}^{3}}{250} + 1 + e^{- x_{n}}}{- \frac{441 x_{n}^{2}}{250} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{147 (1.0000000000)^{3}}{250} + 1 + e^{- (1.0000000000)}}{- \frac{441 (1.0000000000)^{2}}{250} - e^{- (1.0000000000)}} = 1.3658177972$ $LaTeX: x_{2} = (1.3658177972) - \frac{- \frac{147 (1.3658177972)^{3}}{250} + 1 + e^{- (1.3658177972)}}{- \frac{441 (1.3658177972)^{2}}{250} - e^{- (1.3658177972)}} = 1.2972926502$ $LaTeX: x_{3} = (1.2972926502) - \frac{- \frac{147 (1.2972926502)^{3}}{250} + 1 + e^{- (1.2972926502)}}{- \frac{441 (1.2972926502)^{2}}{250} - e^{- (1.2972926502)}} = 1.2940505046$ $LaTeX: x_{4} = (1.2940505046) - \frac{- \frac{147 (1.2940505046)^{3}}{250} + 1 + e^{- (1.2940505046)}}{- \frac{441 (1.2940505046)^{2}}{250} - e^{- (1.2940505046)}} = 1.2940435045$ $LaTeX: x_{5} = (1.2940435045) - \frac{- \frac{147 (1.2940435045)^{3}}{250} + 1 + e^{- (1.2940435045)}}{- \frac{441 (1.2940435045)^{2}}{250} - e^{- (1.2940435045)}} = 1.2940435044$