Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{327 x^{3}}{500} - 5$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{327 x_{n}^{3}}{500} + 5 + e^{- x_{n}}}{- \frac{981 x_{n}^{2}}{500} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{327 (1.0000000000)^{3}}{500} + 5 + e^{- (1.0000000000)}}{- \frac{981 (1.0000000000)^{2}}{500} - e^{- (1.0000000000)}} = 3.0232289096$ $LaTeX: x_{2} = (3.0232289096) - \frac{- \frac{327 (3.0232289096)^{3}}{500} + 5 + e^{- (3.0232289096)}}{- \frac{981 (3.0232289096)^{2}}{500} - e^{- (3.0232289096)}} = 2.2989863562$ $LaTeX: x_{3} = (2.2989863562) - \frac{- \frac{327 (2.2989863562)^{3}}{500} + 5 + e^{- (2.2989863562)}}{- \frac{981 (2.2989863562)^{2}}{500} - e^{- (2.2989863562)}} = 2.0271345194$ $LaTeX: x_{4} = (2.0271345194) - \frac{- \frac{327 (2.0271345194)^{3}}{500} + 5 + e^{- (2.0271345194)}}{- \frac{981 (2.0271345194)^{2}}{500} - e^{- (2.0271345194)}} = 1.9885529534$ $LaTeX: x_{5} = (1.9885529534) - \frac{- \frac{327 (1.9885529534)^{3}}{500} + 5 + e^{- (1.9885529534)}}{- \frac{981 (1.9885529534)^{2}}{500} - e^{- (1.9885529534)}} = 1.9878204423$