Find the root of the equation using Bisection Method, Engineering Mathematics

Given equation is y = f(x) = x3 – 9x + 1 and it is continuous in the interval 2 ≤ x ≤ 4. Now, first determine the value of a function at the points 2 and 4 i.e. f(2) = 23 – 9(2) + 1 = 8 – 18 + 1 = – 9 f(4) = 43 – 9(4) + 1 = 64 – 36 + 1 = +29 f(2) f(4) < 0 Since f(2) is negative and f(4) is a positive value, therefore the root f(x) lies in between 2 and 4. Now go to the First Approximation. First Approximation value of x x1 = (2+4)/2 = 3 Again, check the value of f(x) at x = 3, f(3) = 33 – 9(3) + 1 = 27 – 27 + 1 = 1 f(2)f(3) < 0 Since f(3) is a positive value and f(2) is a negative value, therefore the root of f(x) lies in between 2 and 3. Now go to Second Approximation. Second Approximation value of x x2 = (2+3)/2 = 2.5 Again, check the value of f(x) at x = 2.5, f(2.5) = 2.53 – 9(2.5) + 1 = – 5.87 f(2.5)f(3) < 0 Since f (3) is a positive value and f (2.5) is a negative value, therefore the root of f(x) lies in between 2.5 and 3. Now go to Third Approximation. Third Approximation value of x x3 = (2.5+3)/2 = 2.75 Again, check the value of f(x) at x = 2.75, f(2.75) = 2.753 – 9(2.75) + 1 = – 2.96 f(2.75)f(3) < 0 Since f(3) is a positive value and f(2.75) is a negative value, therefore the root of f(x) lies in between 2.75 and 3 and go to Fourth Approximation. Fourth Approximation value of x x4 = (2.75+3)/2 = 2.88 Again, check the value of f(x) at x = 2.88, f(2.88) = 2.883 – 9(2.88) + 1 = – 1.03 f(2.88)f(3) < 0 Since f(3) is a positive value and f(2.88) is a negative value, therefore the root of f(x) lies in between 2.88 and 3. Now go to Fifth Approximation. Fifth Approximation value of x x5 = (2.88+3)/2 = 2.94 Again, check the value of f(x) at x = 2.94, f(2.94) = 2.943 – 9(2.94) + 1 = – 0.05 f(2.94)f(3) < 0 Proceeding similarly we obtain x6 = 2.9375. Therefore, x = 2.9375 is our approximate root of f(x).