3. a. Bisection method In(x4) = 0.7 where: XL = 0.5, Xu = 2 . Find the value of the function at XL and xu @ XL = 0.5 In((0.54)) = -2.77 0.7 The function is continuous at the interval [0.5,2] Getting the midpoint(t) of the interval: t = 0.5+2 2 = 1.25 @ t = 1.25 In((1.254)) = 0.89 > 0.7 Since f(t) is positive we replace xu by t for the next iteration Iterations XL XU f ( XL) f( Xu) f(t) 0.5 2 1.25 -2.77 2.77 0.89 2 0.5 1.25 0.8 -2.77 0.89 -0.51 3 0.88 1.25 1.06 -0.53 1.25 0.23 4 0.88 1.06 0.97 -0.53 0.23 0.12 5 0.97 1.06 1.02 -0.12 0.23 0.08 6 0.97 1.02 1.00 -0.12 0.08 0