Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{123 x^{3}}{250} - 4$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{123 x_{n}^{3}}{250} + 4 + e^{- x_{n}}}{- \frac{369 x_{n}^{2}}{250} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{123 (3.0000000000)^{3}}{250} + 4 + e^{- (3.0000000000)}}{- \frac{369 (3.0000000000)^{2}}{250} - e^{- (3.0000000000)}} = 2.3074575974$ $LaTeX: x_{2} = (2.3074575974) - \frac{- \frac{123 (2.3074575974)^{3}}{250} + 4 + e^{- (2.3074575974)}}{- \frac{369 (2.3074575974)^{2}}{250} - e^{- (2.3074575974)}} = 2.0630491693$ $LaTeX: x_{3} = (2.0630491693) - \frac{- \frac{123 (2.0630491693)^{3}}{250} + 4 + e^{- (2.0630491693)}}{- \frac{369 (2.0630491693)^{2}}{250} - e^{- (2.0630491693)}} = 2.0329307158$ $LaTeX: x_{4} = (2.0329307158) - \frac{- \frac{123 (2.0329307158)^{3}}{250} + 4 + e^{- (2.0329307158)}}{- \frac{369 (2.0329307158)^{2}}{250} - e^{- (2.0329307158)}} = 2.0324989083$ $LaTeX: x_{5} = (2.0324989083) - \frac{- \frac{123 (2.0324989083)^{3}}{250} + 4 + e^{- (2.0324989083)}}{- \frac{369 (2.0324989083)^{2}}{250} - e^{- (2.0324989083)}} = 2.0324988204$