Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{3 x^{3}}{100} - 7$ using $LaTeX: \displaystyle x_0=7$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{3 x_{n}^{3}}{100} + 7 + e^{- x_{n}}}{- \frac{9 x_{n}^{2}}{100} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 7$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (7.0000000000) - \frac{- \frac{3 (7.0000000000)^{3}}{100} + 7 + e^{- (7.0000000000)}}{- \frac{9 (7.0000000000)^{2}}{100} - e^{- (7.0000000000)}} = 6.2543292167$ $LaTeX: x_{2} = (6.2543292167) - \frac{- \frac{3 (6.2543292167)^{3}}{100} + 7 + e^{- (6.2543292167)}}{- \frac{9 (6.2543292167)^{2}}{100} - e^{- (6.2543292167)}} = 6.1585066964$ $LaTeX: x_{3} = (6.1585066964) - \frac{- \frac{3 (6.1585066964)^{3}}{100} + 7 + e^{- (6.1585066964)}}{- \frac{9 (6.1585066964)^{2}}{100} - e^{- (6.1585066964)}} = 6.1570038935$ $LaTeX: x_{4} = (6.1570038935) - \frac{- \frac{3 (6.1570038935)^{3}}{100} + 7 + e^{- (6.1570038935)}}{- \frac{9 (6.1570038935)^{2}}{100} - e^{- (6.1570038935)}} = 6.1570035276$ $LaTeX: x_{5} = (6.1570035276) - \frac{- \frac{3 (6.1570035276)^{3}}{100} + 7 + e^{- (6.1570035276)}}{- \frac{9 (6.1570035276)^{2}}{100} - e^{- (6.1570035276)}} = 6.1570035276$