Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{141 x^{3}}{1000} - 5$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{141 x_{n}^{3}}{1000} + 5 + e^{- x_{n}}}{- \frac{423 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{141 (3.0000000000)^{3}}{1000} + 5 + e^{- (3.0000000000)}}{- \frac{423 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 3.3222337781$ $LaTeX: x_{2} = (3.3222337781) - \frac{- \frac{141 (3.3222337781)^{3}}{1000} + 5 + e^{- (3.3222337781)}}{- \frac{423 (3.3222337781)^{2}}{1000} - e^{- (3.3222337781)}} = 3.2937192830$ $LaTeX: x_{3} = (3.2937192830) - \frac{- \frac{141 (3.2937192830)^{3}}{1000} + 5 + e^{- (3.2937192830)}}{- \frac{423 (3.2937192830)^{2}}{1000} - e^{- (3.2937192830)}} = 3.2934761940$ $LaTeX: x_{4} = (3.2934761940) - \frac{- \frac{141 (3.2934761940)^{3}}{1000} + 5 + e^{- (3.2934761940)}}{- \frac{423 (3.2934761940)^{2}}{1000} - e^{- (3.2934761940)}} = 3.2934761764$ $LaTeX: x_{5} = (3.2934761764) - \frac{- \frac{141 (3.2934761764)^{3}}{1000} + 5 + e^{- (3.2934761764)}}{- \frac{423 (3.2934761764)^{2}}{1000} - e^{- (3.2934761764)}} = 3.2934761764$