Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{107 x^{3}}{1000} - 3$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{107 x_{n}^{3}}{1000} + 3 + e^{- x_{n}}}{- \frac{321 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{107 (3.0000000000)^{3}}{1000} + 3 + e^{- (3.0000000000)}}{- \frac{321 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 3.0547120511$ $LaTeX: x_{2} = (3.0547120511) - \frac{- \frac{107 (3.0547120511)^{3}}{1000} + 3 + e^{- (3.0547120511)}}{- \frac{321 (3.0547120511)^{2}}{1000} - e^{- (3.0547120511)}} = 3.0537828725$ $LaTeX: x_{3} = (3.0537828725) - \frac{- \frac{107 (3.0537828725)^{3}}{1000} + 3 + e^{- (3.0537828725)}}{- \frac{321 (3.0537828725)^{2}}{1000} - e^{- (3.0537828725)}} = 3.0537826008$ $LaTeX: x_{4} = (3.0537826008) - \frac{- \frac{107 (3.0537826008)^{3}}{1000} + 3 + e^{- (3.0537826008)}}{- \frac{321 (3.0537826008)^{2}}{1000} - e^{- (3.0537826008)}} = 3.0537826008$ $LaTeX: x_{5} = (3.0537826008) - \frac{- \frac{107 (3.0537826008)^{3}}{1000} + 3 + e^{- (3.0537826008)}}{- \frac{321 (3.0537826008)^{2}}{1000} - e^{- (3.0537826008)}} = 3.0537826008$