Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{201 x^{3}}{1000} - 2$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{201 x_{n}^{3}}{1000} + 2 + e^{- x_{n}}}{- \frac{603 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{201 (3.0000000000)^{3}}{1000} + 2 + e^{- (3.0000000000)}}{- \frac{603 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 2.3833587303$ $LaTeX: x_{2} = (2.3833587303) - \frac{- \frac{201 (2.3833587303)^{3}}{1000} + 2 + e^{- (2.3833587303)}}{- \frac{603 (2.3833587303)^{2}}{1000} - e^{- (2.3833587303)}} = 2.2045442018$ $LaTeX: x_{3} = (2.2045442018) - \frac{- \frac{201 (2.2045442018)^{3}}{1000} + 2 + e^{- (2.2045442018)}}{- \frac{603 (2.2045442018)^{2}}{1000} - e^{- (2.2045442018)}} = 2.1903256773$ $LaTeX: x_{4} = (2.1903256773) - \frac{- \frac{201 (2.1903256773)^{3}}{1000} + 2 + e^{- (2.1903256773)}}{- \frac{603 (2.1903256773)^{2}}{1000} - e^{- (2.1903256773)}} = 2.1902401579$ $LaTeX: x_{5} = (2.1902401579) - \frac{- \frac{201 (2.1902401579)^{3}}{1000} + 2 + e^{- (2.1902401579)}}{- \frac{603 (2.1902401579)^{2}}{1000} - e^{- (2.1902401579)}} = 2.1902401548$