The root of the function, f(x) = x³ - 2x² + 3x - 1 in the interval [0, 1] using bisection method after two iterations, is ______. (round off up to 2 decimals)

Steps to Solve

1. Define the function and interval We have the function ( f(x) = x^3 - 2x^2 + 3x - 1 ) and the interval ([0, 1]).

2. Calculate initial function values Evaluate the function at the endpoints of the interval:

   • ( f(0) = 0^3 - 2(0^2) + 3(0) - 1 = -1 )
   • ( f(1) = 1^3 - 2(1^2) + 3(1) - 1 = 1 )

3. Check domain Since ( f(0) < 0 ) and ( f(1) > 0 ), we confirm there is a root in the interval ([0, 1]).

4. First iteration

   • Calculate midpoint: ( c_1 = \frac{0 + 1}{2} = 0.5 )
   • Evaluate ( f(c_1) ): $$ f(0.5) = (0.5)^3 - 2(0.5)^2 + 3(0.5) - 1 = 0.125 - 0.5 + 1.5 - 1 = 0.125 $$
   • Since ( f(0) < 0) and ( f(0.5) > 0 ), the new interval is ([0, 0.5]).

5. Second iteration

   • Calculate new midpoint: ( c_2 = \frac{0 + 0.5}{2} = 0.25 )
   • Evaluate ( f(c_2) ): $$ f(0.25) = (0.25)^3 - 2(0.25)^2 + 3(0.25) - 1 = 0.015625 - 0.125 + 0.75 - 1 = -0.359375 $$
   • Since ( f(0.25) < 0 ) and ( f(0.5) > 0 ), the new interval is ([0.25, 0.5]).