Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{18 x^{3}}{25} - 4$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{18 x_{n}^{3}}{25} + 4 + e^{- x_{n}}}{- \frac{54 x_{n}^{2}}{25} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{18 (1.0000000000)^{3}}{25} + 4 + e^{- (1.0000000000)}}{- \frac{54 (1.0000000000)^{2}}{25} - e^{- (1.0000000000)}} = 2.4430591039$ $LaTeX: x_{2} = (2.4430591039) - \frac{- \frac{18 (2.4430591039)^{3}}{25} + 4 + e^{- (2.4430591039)}}{- \frac{54 (2.4430591039)^{2}}{25} - e^{- (2.4430591039)}} = 1.9490449205$ $LaTeX: x_{3} = (1.9490449205) - \frac{- \frac{18 (1.9490449205)^{3}}{25} + 4 + e^{- (1.9490449205)}}{- \frac{54 (1.9490449205)^{2}}{25} - e^{- (1.9490449205)}} = 1.8066763736$ $LaTeX: x_{4} = (1.8066763736) - \frac{- \frac{18 (1.8066763736)^{3}}{25} + 4 + e^{- (1.8066763736)}}{- \frac{54 (1.8066763736)^{2}}{25} - e^{- (1.8066763736)}} = 1.7953467971$ $LaTeX: x_{5} = (1.7953467971) - \frac{- \frac{18 (1.7953467971)^{3}}{25} + 4 + e^{- (1.7953467971)}}{- \frac{54 (1.7953467971)^{2}}{25} - e^{- (1.7953467971)}} = 1.7952781574$