Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{87 x^{3}}{500} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{87 x_{n}^{3}}{500} + 5 + e^{- x_{n}}}{- \frac{261 x_{n}^{2}}{500} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{87 (3.0000000000)^{3}}{500} + 5 + e^{- (3.0000000000)}}{- \frac{261 (3.0000000000)^{2}}{500} - e^{- (3.0000000000)}} = 3.0740949548$ $LaTeX: x_{2} = (3.0740949548) - \frac{- \frac{87 (3.0740949548)^{3}}{500} + 5 + e^{- (3.0740949548)}}{- \frac{261 (3.0740949548)^{2}}{500} - e^{- (3.0740949548)}} = 3.0723808383$ $LaTeX: x_{3} = (3.0723808383) - \frac{- \frac{87 (3.0723808383)^{3}}{500} + 5 + e^{- (3.0723808383)}}{- \frac{261 (3.0723808383)^{2}}{500} - e^{- (3.0723808383)}} = 3.0723799042$ $LaTeX: x_{4} = (3.0723799042) - \frac{- \frac{87 (3.0723799042)^{3}}{500} + 5 + e^{- (3.0723799042)}}{- \frac{261 (3.0723799042)^{2}}{500} - e^{- (3.0723799042)}} = 3.0723799042$ $LaTeX: x_{5} = (3.0723799042) - \frac{- \frac{87 (3.0723799042)^{3}}{500} + 5 + e^{- (3.0723799042)}}{- \frac{261 (3.0723799042)^{2}}{500} - e^{- (3.0723799042)}} = 3.0723799042$