Use Newton's method and the modified Newton's method described in Eq. (2.13) to find a solution accurate to within 10−5 to the problem
e6x + 1.441e2x − 2.079e4x − 0.3330 = 0, for − 1 ≤ x ≤ 0
This is the same problem as 1(d) with the coefficients replaced by their four-digit approximations. Compare the solutions to the results in 1(d) and 2(d).