Method Of False Position

Since, f(0) f(0.5) < 0, the root lies in the interval (0, 0.5).

Since, f(0) f(0.36364) < 0, the root lies in the interval (0, 0.36364).

Since, f(0) f(0.3487) < 0, the root lies in the interval (0, 0.34870).

Since, f(0) f(0.34741) < 0, the root lies in the interval (0, 0.34741).

Now, | x6 – x5 | = | 0.347306 – 0.34741 | ≈ 0.0001 < 0.0005.

The root has been computed correct to three decimal places. The required root can be taken as x ≈ x₆ = 0.347306. We may also give the result as 0.347, even though x₆ is more accurate. Note that the left end point x = 0 is fixed for all iterations.

Now, we compute the root in (1, 2). We have

Since, f(1.25) f(2) < 0, the root lies in the interval (1.25, 2).

Since, f(1.407407) f(2) < 0, the root lies in the interval (1.407407, 2).

Since f(1.482367) f(2) < 0, the root lies in the interval (1.482367, 2).

Since, f(1.513156) f(2) < 0, the root lies in the interval (1.513156, 2).