Roots Finding: Bracketing Methods

Problem 5.7. 

(a) Determine the roots of f(x) = -12 - 21x + 18x² - 2.75x³ graphically. In addition determine the first root of the function with (b) bisection and (c) false position. For (b) and (c), use x_l = -1 and x_u = 0, and a stopping criterion of 1%.

Solution:

(a) Plot the graph using MATLAB. The commands are shown as below:

>> x = linspace(-1,8);

>> fx = -12-21*x+18*x.^2-2.75*x.^3;

>> plot(x,fx),grid,xlabel('x'),ylabel('f(x)')

The graph generated is shown as above. Since it is a cubic equation, there are 3 roots. By maximizing the graph, the roots are estimated as -0.4135, 2.2 and 4.74. 

(b) The root estimate using bisection method is x_r = (x_l + x_u)/2. Begin the iteration with guesses of x_l = -1 and x_u = 0.

First iteration:

x_l = -1                              f(-1) = 29.75

x_u = 0                               f(0) = -12

x_r = (-1+0)/2 = -0.50     f(-0.50) = 3.3438 

f(-1)f(-0.50) = 99.4766 > 0. Therefore, the root lies in the upper interval, and x_r becomes the lower limit for the next iteration, x_l = -0.50. 

Second iteration:

x_l = -0.50                             f(-0.50) = 3.3438

x_u = 0                                   f(0) = -12

x_r = (-0.50+0)/2 = -0.25    f(-0.25) = -5.5820

f(-0.50)f(-0.25) = -18.66496 < 0. Therefore, the root lies in the lower interval, and x_r becomes the upper limit for the next iteration, x_u = -0.25. 

The approximate relative error is 100%. 

The remainder of the iterations are displayed in the following table. 

Hence, after 8 iterations, the approximate error finally falls below 1% and the computation can be terminated. The root estimate is -0.41797. 

(c) The root estimate using false position method is x_r = x_u - f(x_u)(x_l - x_u)/(f(x_l) - f(x_u)). Begin the iteration with guesses of x_l = -1 and x_u = 0.

First iteration:

x_l = -1                        f(-1) = 29.75

x_u = 0                        f(0) = -12

x_r = -0.2874             f(-0.2874) = -4.4116

f(-1)f(-0.2874) = -131.2731 < 0. Therefore, the root lies in the lower interval, and x_r becomes the upper limit for the next iteration, x_u = -0.2874. 

Second iteration:

x_l = -1                          f(-1) = 29.75

x_u = -0.2874               f(-0.2874) = -4.4116

x_r = -0.3795                f(-0.3795) = -1.2878

f(-1)f(-0.3795) = -38.3130 < 0. Therefore, the root lies in the lower interval, and x_r becomes the upper limit for the next iteration, x_u = -0.3795. 

The approximate relative error is 24.3%. 

The remainder of the iterations are displayed in the following table. 

Hence, after 5 iterations, the approximate error finally falls below 1% and the computation can be terminated. The root estimate is -0.41402.