b2 = 1.14205. In the third iteration, we use inverse quadratic interpolation between (a2, f(a2)) = (−4, −25) and (b1, f(b1)) = (1.23256, 0.22891) and (b2, f(b2)) = (1.14205, 0.083582). This yields 1.09032, which lies between (3a2 + b2) / 4 and b2. But here Brent's additional condition kicks in: the inequality |1.09032 − b2| ≤ |b1 − b0| / 2 is not satisfied, so this value is rejected. Instead, the midpoint m = −1.42897 of the interval [a2, b2] is computed. We have f(m) = 9.26891, so we set a3 = a2 and b3 = −1.42897. In the fourth iteration, we use inverse quadratic interpolation between (a3, f(a3)) = (−4, −25) and (b2, f(b2)) = (1.14205, 0.083582) and (b3, f(b3)) = (−1.42897, 9.26891). This yields 1.15448, which is not in the interval between (3a3 + b3) / 4 and b3). Hence, it is replaced by the midpoint m = −2.71449. We have f(m) = 3.93934, so we set a4 = a3 and b4 = −2.71449. In the fifth iteration, inverse quadratic interpolation yields −3.45500, which lies in the required interval. However, the previous iteration was a bisection step, so the inequality |−3.45500 − b4| ≤ |b4 − b3| / 2 need to be satisfied. This inequality is false, so we use the midpoint m = −3.35724. We have f(m) = −6.78239, so m becomes the new contrapoint (a5 = −3.35724) and the iterate remains the same (b5 = b4). In the sixth iteration, we cannot use inverse quadratic interpolation because b5 = b4. Hence, we use linear interpolation between (a5, f(a5)) = (−3.35724, −6.78239) and (b5, f(b5)) = (−2.71449, 3.93934). The result is s = −2.95064, which satisfies all the conditions. But since the iterate did not change in the previous step, we reject this result and fall back to bisection. We update s = -3.03587, and f(s) = -0.58418. In the seventh iteration, we can again use inverse quadratic interpolation. The result is s = −3.00219, which satisfies all the conditions. Now, f(s) = −0.03515, so we set a7 = b6 and b7 = −3.00219 (a7 and b7 are exchanged so that the condition |f(b7)| ≤ |f(a7)| is satisfied). (Correct : linear interpolation ) In the eighth iteration, we cannot use inverse quadratic interpolation because a7 = b6. Linear interpolation yields s = −2.99994, which is accepted. (Correct : ) In the following iterations, the root x =