Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{31 x^{3}}{250} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{31 x_{n}^{3}}{250} + 5 + e^{- x_{n}}}{- \frac{93 x_{n}^{2}}{250} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{31 (3.0000000000)^{3}}{250} + 5 + e^{- (3.0000000000)}}{- \frac{93 (3.0000000000)^{2}}{250} - e^{- (3.0000000000)}} = 3.5008515937$ $LaTeX: x_{2} = (3.5008515937) - \frac{- \frac{31 (3.5008515937)^{3}}{250} + 5 + e^{- (3.5008515937)}}{- \frac{93 (3.5008515937)^{2}}{250} - e^{- (3.5008515937)}} = 3.4376166082$ $LaTeX: x_{3} = (3.4376166082) - \frac{- \frac{31 (3.4376166082)^{3}}{250} + 5 + e^{- (3.4376166082)}}{- \frac{93 (3.4376166082)^{2}}{250} - e^{- (3.4376166082)}} = 3.4364615963$ $LaTeX: x_{4} = (3.4364615963) - \frac{- \frac{31 (3.4364615963)^{3}}{250} + 5 + e^{- (3.4364615963)}}{- \frac{93 (3.4364615963)^{2}}{250} - e^{- (3.4364615963)}} = 3.4364612157$ $LaTeX: x_{5} = (3.4364612157) - \frac{- \frac{31 (3.4364612157)^{3}}{250} + 5 + e^{- (3.4364612157)}}{- \frac{93 (3.4364612157)^{2}}{250} - e^{- (3.4364612157)}} = 3.4364612157$