Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{57 x^{3}}{1000} - 7$ using $LaTeX: \displaystyle x_0=5$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{57 x_{n}^{3}}{1000} + 7 + e^{- x_{n}}}{- \frac{171 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 5$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (5.0000000000) - \frac{- \frac{57 (5.0000000000)^{3}}{1000} + 7 + e^{- (5.0000000000)}}{- \frac{171 (5.0000000000)^{2}}{1000} - e^{- (5.0000000000)}} = 4.9723798947$ $LaTeX: x_{2} = (4.9723798947) - \frac{- \frac{57 (4.9723798947)^{3}}{1000} + 7 + e^{- (4.9723798947)}}{- \frac{171 (4.9723798947)^{2}}{1000} - e^{- (4.9723798947)}} = 4.9722267694$ $LaTeX: x_{3} = (4.9722267694) - \frac{- \frac{57 (4.9722267694)^{3}}{1000} + 7 + e^{- (4.9722267694)}}{- \frac{171 (4.9722267694)^{2}}{1000} - e^{- (4.9722267694)}} = 4.9722267647$ $LaTeX: x_{4} = (4.9722267647) - \frac{- \frac{57 (4.9722267647)^{3}}{1000} + 7 + e^{- (4.9722267647)}}{- \frac{171 (4.9722267647)^{2}}{1000} - e^{- (4.9722267647)}} = 4.9722267647$ $LaTeX: x_{5} = (4.9722267647) - \frac{- \frac{57 (4.9722267647)^{3}}{1000} + 7 + e^{- (4.9722267647)}}{- \frac{171 (4.9722267647)^{2}}{1000} - e^{- (4.9722267647)}} = 4.9722267647$