(i) By Secant Method Let, f ( x ) = e-x - x f ( x0 ) = f ( 0 ) = 1 > 0 f ( x1 ) = f ( 1 ) = -0.6321 < 0 Therefore, root lies between [ 0, 1 ]. By Secant Method: x n+1 = x n - x n- x n-1 f ( x n ) - f ( x n-1 ) f ( x n ) , n ≥ 1. x 2 = x 1 - x 1- x 0 f ( x 1 ) - f ( x 0 ) f ( x 1 ) x 2 = 1 - 1- 0 -0.6321 - 1 ( -0.6321 ) x 2 = 0.6127 x0 > 0 x1 < 0 fx0 fx1 x n+1 = x n - x n- x n-1 f ( x n ) - f ( x n-1 ) f ( x n ) f ( x n+1 ) 0 1 1 -0.6321 0.6127 -0.0708 1 0.6127 -0.6321 -0.0708 0.5638 0.0052 0.6172 0.5638 -0.0708 0.0052 0.5671 0.0000 Therefore, x = 0.5671 is the required root.
