We want to find a root to the equation [math: x3+x+1=0]x^3+x+1=0. Starting from the interval [math: [−1,0]][-1,0], apply the bisection method for [math: 3] times to obtain an approximation to the root. (Correct the answer to 2 decimal places.)

Question

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We are given the function f(x) = x^3 + x + 1 and the interval [-1, 0]. Step 1: Evaluate the function at the endpoints to ensure a sign change. f(-1) = (-1)^3 + (-1) + 1 = -1 + -1 + 1 = -1 < 0 f(0) = 0^3 + 0 + 1 = 1 > 0 Since f(-1) and f(0) have opposite signs, there is at least one root in [-1, 0], so the bisection method can be applied. Step 2: First bisection (3 iterations total will be performed). - Midpoint m1 = (-1 + 0)/2 = -0.5 f(-0.5) = (-0.5)^3 + (-0.5) + 1 = -0.125 - 0.5 + 1 = 0.375 > 0 Since f(-1) < 0 and f(-0.5) > 0, the root lies in [-1, -0.5]. New interval: [-1, -0.5] Step 3: Second bisection. - Midpoint m2 = (-1 + (-0.5))/2 = -0.75 f(-0.75) = (-0.75)^3 + (-0.75) + 1 = -0.421875 - 0.75 + 1 = -0.171875 < 0 Since f(-0.5) > 0 and f(-0.75) < 0, the root lies in [-0.75, -0.5]. New interval: [-0.75, -0.5] Step 4: Third bisection. - Midpoint m3 = (-0.75 + (-0.5))/2 = -0.625 f(-0.625) = (-0.625)^3 + (-0.625) + 1 = -0.244140625 - 0.625 + 1 = 0.130859375 > 0 Since f(-0.75) < 0 and f(-0.625) > 0, the root lies in [-0.75, -0.625]. Final interval after three iterations: [-0.75, -0.625] Step 5: Approximate the root from the final interval. - The bisection method typically takes the midpoint of the final interval as the approximation. - Midpoint = (-0.75 + -0.625)/2 = -1.375/2 = -0.6875 Step 6: Round to two decimal places as requested. -0.6875 rounded to two decimals is -0.69. Why the provided answer -0.38 is not correct: - The sequence of midpoints and the sign checks above show the root after three iterations lies between -0.75 and -0.625, with the midpoint -0.6875. Rounding gives -0.69, not -0.38. - A value like -0.38 would imply a different interval or an incorrect number of iterations or miscalculation of f(x) at the midpoints. Summary: After performing three bisection steps starting from [-1, 0] for f(x) = x^3 + x + 1, the approximate root is -0.69 (to two decimals).

We want to find a root to the equation [math: x3+x+1=0]x^3+x+1=0. Starting from the interval [math: [−1,0]][-1,0], apply the bisection method for [math: 3] times to obtain an approximation to the root. (Correct the answer to 2 decimal places.)简答题

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