Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{87 x^{3}}{250} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{87 x_{n}^{3}}{250} + 6 + e^{- x_{n}}}{- \frac{261 x_{n}^{2}}{250} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{87 (3.0000000000)^{3}}{250} + 6 + e^{- (3.0000000000)}}{- \frac{261 (3.0000000000)^{2}}{250} - e^{- (3.0000000000)}} = 2.6457454622$ $LaTeX: x_{2} = (2.6457454622) - \frac{- \frac{87 (2.6457454622)^{3}}{250} + 6 + e^{- (2.6457454622)}}{- \frac{261 (2.6457454622)^{2}}{250} - e^{- (2.6457454622)}} = 2.5950530842$ $LaTeX: x_{3} = (2.5950530842) - \frac{- \frac{87 (2.5950530842)^{3}}{250} + 6 + e^{- (2.5950530842)}}{- \frac{261 (2.5950530842)^{2}}{250} - e^{- (2.5950530842)}} = 2.5940735398$ $LaTeX: x_{4} = (2.5940735398) - \frac{- \frac{87 (2.5940735398)^{3}}{250} + 6 + e^{- (2.5940735398)}}{- \frac{261 (2.5940735398)^{2}}{250} - e^{- (2.5940735398)}} = 2.5940731788$ $LaTeX: x_{5} = (2.5940731788) - \frac{- \frac{87 (2.5940731788)^{3}}{250} + 6 + e^{- (2.5940731788)}}{- \frac{261 (2.5940731788)^{2}}{250} - e^{- (2.5940731788)}} = 2.5940731788$