Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{24 x^{3}}{25} - 9$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{24 x_{n}^{3}}{25} + 9 + e^{- x_{n}}}{- \frac{72 x_{n}^{2}}{25} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{24 (3.0000000000)^{3}}{25} + 9 + e^{- (3.0000000000)}}{- \frac{72 (3.0000000000)^{2}}{25} - e^{- (3.0000000000)}} = 2.3503907873$ $LaTeX: x_{2} = (2.3503907873) - \frac{- \frac{24 (2.3503907873)^{3}}{25} + 9 + e^{- (2.3503907873)}}{- \frac{72 (2.3503907873)^{2}}{25} - e^{- (2.3503907873)}} = 2.1398593487$ $LaTeX: x_{3} = (2.1398593487) - \frac{- \frac{24 (2.1398593487)^{3}}{25} + 9 + e^{- (2.1398593487)}}{- \frac{72 (2.1398593487)^{2}}{25} - e^{- (2.1398593487)}} = 2.1181532342$ $LaTeX: x_{4} = (2.1181532342) - \frac{- \frac{24 (2.1181532342)^{3}}{25} + 9 + e^{- (2.1181532342)}}{- \frac{72 (2.1181532342)^{2}}{25} - e^{- (2.1181532342)}} = 2.1179334838$ $LaTeX: x_{5} = (2.1179334838) - \frac{- \frac{24 (2.1179334838)^{3}}{25} + 9 + e^{- (2.1179334838)}}{- \frac{72 (2.1179334838)^{2}}{25} - e^{- (2.1179334838)}} = 2.1179334614$