Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{957 x^{3}}{1000} - 4$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{957 x_{n}^{3}}{1000} + 4 + e^{- x_{n}}}{- \frac{2871 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{957 (1.0000000000)^{3}}{1000} + 4 + e^{- (1.0000000000)}}{- \frac{2871 (1.0000000000)^{2}}{1000} - e^{- (1.0000000000)}} = 2.0531047861$ $LaTeX: x_{2} = (2.0531047861) - \frac{- \frac{957 (2.0531047861)^{3}}{1000} + 4 + e^{- (2.0531047861)}}{- \frac{2871 (2.0531047861)^{2}}{1000} - e^{- (2.0531047861)}} = 1.7134679713$ $LaTeX: x_{3} = (1.7134679713) - \frac{- \frac{957 (1.7134679713)^{3}}{1000} + 4 + e^{- (1.7134679713)}}{- \frac{2871 (1.7134679713)^{2}}{1000} - e^{- (1.7134679713)}} = 1.6398119583$ $LaTeX: x_{4} = (1.6398119583) - \frac{- \frac{957 (1.6398119583)^{3}}{1000} + 4 + e^{- (1.6398119583)}}{- \frac{2871 (1.6398119583)^{2}}{1000} - e^{- (1.6398119583)}} = 1.6365513140$ $LaTeX: x_{5} = (1.6365513140) - \frac{- \frac{957 (1.6365513140)^{3}}{1000} + 4 + e^{- (1.6365513140)}}{- \frac{2871 (1.6365513140)^{2}}{1000} - e^{- (1.6365513140)}} = 1.6365451005$