Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{31 x^{3}}{40} - 7$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{31 x_{n}^{3}}{40} + 7 + e^{- x_{n}}}{- \frac{93 x_{n}^{2}}{40} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{31 (3.0000000000)^{3}}{40} + 7 + e^{- (3.0000000000)}}{- \frac{93 (3.0000000000)^{2}}{40} - e^{- (3.0000000000)}} = 2.3384813449$ $LaTeX: x_{2} = (2.3384813449) - \frac{- \frac{31 (2.3384813449)^{3}}{40} + 7 + e^{- (2.3384813449)}}{- \frac{93 (2.3384813449)^{2}}{40} - e^{- (2.3384813449)}} = 2.1188056063$ $LaTeX: x_{3} = (2.1188056063) - \frac{- \frac{31 (2.1188056063)^{3}}{40} + 7 + e^{- (2.1188056063)}}{- \frac{93 (2.1188056063)^{2}}{40} - e^{- (2.1188056063)}} = 2.0949703040$ $LaTeX: x_{4} = (2.0949703040) - \frac{- \frac{31 (2.0949703040)^{3}}{40} + 7 + e^{- (2.0949703040)}}{- \frac{93 (2.0949703040)^{2}}{40} - e^{- (2.0949703040)}} = 2.0947036517$ $LaTeX: x_{5} = (2.0947036517) - \frac{- \frac{31 (2.0947036517)^{3}}{40} + 7 + e^{- (2.0947036517)}}{- \frac{93 (2.0947036517)^{2}}{40} - e^{- (2.0947036517)}} = 2.0947036186$