Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{61 x^{3}}{500} - 8$ using $LaTeX: \displaystyle x_0=5$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{61 x_{n}^{3}}{500} + 8 + e^{- x_{n}}}{- \frac{183 x_{n}^{2}}{500} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 5$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (5.0000000000) - \frac{- \frac{61 (5.0000000000)^{3}}{500} + 8 + e^{- (5.0000000000)}}{- \frac{183 (5.0000000000)^{2}}{500} - e^{- (5.0000000000)}} = 4.2089691662$ $LaTeX: x_{2} = (4.2089691662) - \frac{- \frac{61 (4.2089691662)^{3}}{500} + 8 + e^{- (4.2089691662)}}{- \frac{183 (4.2089691662)^{2}}{500} - e^{- (4.2089691662)}} = 4.0424890745$ $LaTeX: x_{3} = (4.0424890745) - \frac{- \frac{61 (4.0424890745)^{3}}{500} + 8 + e^{- (4.0424890745)}}{- \frac{183 (4.0424890745)^{2}}{500} - e^{- (4.0424890745)}} = 4.0355016964$ $LaTeX: x_{4} = (4.0355016964) - \frac{- \frac{61 (4.0355016964)^{3}}{500} + 8 + e^{- (4.0355016964)}}{- \frac{183 (4.0355016964)^{2}}{500} - e^{- (4.0355016964)}} = 4.0354896916$ $LaTeX: x_{5} = (4.0354896916) - \frac{- \frac{61 (4.0354896916)^{3}}{500} + 8 + e^{- (4.0354896916)}}{- \frac{183 (4.0354896916)^{2}}{500} - e^{- (4.0354896916)}} = 4.0354896915$