3. Use the Bisection Method to estimate the positive root of \( f(x)=x^{6}-x-1 \) to 5 decimal places, with accuracy of \( f(x)<0.01 \) and find the error.

To use the Bisection Method to estimate the positive root of the function \( f(x) = x^6 - x - 1 \), we first need to identify an interval \([a, b]\) where the function changes sign, indicating the presence of a root. ### Step 1: Identify the interval We will evaluate \( f(x) \) at a few points to find an appropriate interval. - Calculate \( f(1) \): \[ f(1) = 1^6 - 1 - 1 = 1 - 1 - 1 = -1 \] - Calculate \( f(2) \): \[ f(2) = 2^6 - 2 - 1 = 64 - 2 - 1 = 61 \] Since \( f(1) < 0 \) and \( f(2) > 0 \), we can choose the interval \([1, 2]\). ### Step 2: Apply the Bisection Method We will repeatedly bisect the interval and evaluate \( f(x) \) until we achieve the desired accuracy. 1. **First iteration**: - Midpoint: \( c_1 = \frac{1 + 2}{2} = 1.5 \) - Evaluate \( f(1.5) \): \[ f(1.5) = (1.5)^6 - 1.5 - 1 = 11.390625 - 1.5 - 1 = 8.890625 \] - Since \( f(1.5) > 0 \), we update the interval to \([1, 1.5]\). 2. **Second iteration**: - Midpoint: \( c_2 = \frac{1 + 1.5}{2} = 1.25 \) - Evaluate \( f(1.25) \): \[ f(1.25) = (1.25)^6 - 1.25 - 1 = 3.0517578125 - 1.25 - 1 = 0.8017578125 \] - Since \( f(1.25) > 0 \), we update the interval to \([1, 1.25]\). 3. **Third iteration**: - Midpoint: \( c_3 = \frac{1 + 1.25}{2} = 1.125 \) - Evaluate \( f(1.125) \): \[ f(1.125) = (1.125)^6 - 1.125 - 1 = 1.9736328125 - 1.125 - 1 = -0.1513671875 \] - Since \( f(1.125) < 0 \), we update the interval to \([1.125, 1.25]\). 4. **Fourth iteration**: - Midpoint: \( c_4 = \frac{1.125 + 1.25}{2} = 1.1875 \) - Evaluate \( f(1.1875) \): \[ f(1.1875) = (1.1875)^6 - 1.1875 - 1 \approx 2.267578125 - 1.1875 - 1 = 0.080078125 \] - Since \( f(1.1875) > 0 \), we update the interval to \([1.125, 1.1875]\). 5. **Fifth iteration**: - Midpoint: \( c_5 = \frac{1.125 + 1.1875}{2} = 1.15625 \) - Evaluate \( f(1.15625) \): \[ f(1.15625) = (1.15625)^6 - 1.15625 - 1 \approx 1.74755859375 - 1.15625 - 1 = -0.40869140625 \] - Since \( f(1.15625) < 0 \), we update the interval to \([1.15625, 1.1875]\). 6. **Sixth iteration**: - Midpoint: \( c_6 = \frac{1.15625 + 1.1875}{2} = 1.171875 \) - Evaluate \( f(1.171875) \): \[ f(1.171875) = (1.171875)^6 - 1.171875