Numerical Analysis

If a function f(x) is continuous between a and b, and f(a) and f(b) are of opposite signs, then there exists at least one root between a and b. For definiteness, let f(a) be negative and f(b) be positive. Then the root lies between a and b and let its approximate value be given by x 0 = (a+b)/2. If f(x 0 )=0, we conclude that x 0 is a root of the equation f(x) =0. Otherwise, the root lies either between x 0 and b, or between x 0 and a depending on whether f(x 0 ) is negative or positive. The method is shown graphically on the below Example: - Find a real root of the equation f(x) = x 3 – x – 1 = 0. Sol n : - Given, f(x) = x 3 – x – 1 And f(1) = -1 f(2) = 5 Since f(1) is negative and f(2) is positive, a root lies between 1 and 2, Step-1: Here, a = 1 and b = 2 So, xo =(1+2)/2=1.5 f(x o ) = 0.875 x o =1.5 Step-2: a = 1 , b = 1.5 x 1 = (1+1.5)/2 = 1.25 f (x 1 ) = -0.296875 x 1 = 1.25