Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{51 x^{3}}{500} - 4$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{51 x_{n}^{3}}{500} + 4 + e^{- x_{n}}}{- \frac{153 x_{n}^{2}}{500} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{51 (3.0000000000)^{3}}{500} + 4 + e^{- (3.0000000000)}}{- \frac{153 (3.0000000000)^{2}}{500} - e^{- (3.0000000000)}} = 3.4621560185$ $LaTeX: x_{2} = (3.4621560185) - \frac{- \frac{51 (3.4621560185)^{3}}{500} + 4 + e^{- (3.4621560185)}}{- \frac{153 (3.4621560185)^{2}}{500} - e^{- (3.4621560185)}} = 3.4076696463$ $LaTeX: x_{3} = (3.4076696463) - \frac{- \frac{51 (3.4076696463)^{3}}{500} + 4 + e^{- (3.4076696463)}}{- \frac{153 (3.4076696463)^{2}}{500} - e^{- (3.4076696463)}} = 3.4068105088$ $LaTeX: x_{4} = (3.4068105088) - \frac{- \frac{51 (3.4068105088)^{3}}{500} + 4 + e^{- (3.4068105088)}}{- \frac{153 (3.4068105088)^{2}}{500} - e^{- (3.4068105088)}} = 3.4068102975$ $LaTeX: x_{5} = (3.4068102975) - \frac{- \frac{51 (3.4068102975)^{3}}{500} + 4 + e^{- (3.4068102975)}}{- \frac{153 (3.4068102975)^{2}}{500} - e^{- (3.4068102975)}} = 3.4068102975$