Let's solve the problem step by step: ### Step 1: **Existence of the Root on [0, 1]** Given function: \( f(x) = x^3 + x - 1 \) Check the values at the endpoints: - \( f(0) = 0^3 + 0 - 1 = -1 \) - \( f(1) = 1^3 + 1 - 1 = 1 \) Since \( f(0) = -1 \) and \( f(1) = 1 \), the function changes sign over \([0, 1]\). **By the Intermediate Value Theorem**, there **must be at least one root in \([0, 1]\)**. **Explanation:** The function is continuous, and it changes sign from negative to positive on the interval. So, a root exists between 0 and 1. --- ### Step 2: **Four Steps of the Bisection Method** #### **Initial interval:** \([a, b] = [0, 1]\) **Step 1:** Midpoint: \( m_1 = \frac{0 + 1}{2} = 0.5 \) \( f(0.5) = (0.5)^3 + 0.5 - 1 = 0.125 + 0.5 - 1 = -0.375 \) - \( f(0) = -1 \) and \( f(0.5) = -0.375 \) (both negative) - So, root is in \([0.5, 1]\) **Step 2:** New interval: \([0.5, 1]\) Midpoint: \( m_2 = \frac{0.5 + 1}{2} = 0.75 \) \( f(0.75) = (0.75)^3 + 0.75 - 1 = 0.422 + 0.75 - 1 = 0.172 \) - \( f(0.5) = -0.375 \), \( f(0.75) = 0.172 \) (sign change) - So, root is in \([0.5, 0.75]\) **Step 3:** New interval: \([0.5, 0.75]\) Midpoint: \( m_3 = \frac{0.5 + 0.75}{2} = 0.625 \) \( f(0.625) = (0.625)^3 + 0.625 - 1 = 0.244 + 0.625 - 1 = -0.131 \) - \( f(0.625) = -0.131 \), \( f(0.75) = 0.172 \) (sign change) - So, root is in \([0.625, 0.75]\) **Step 4:** New interval: \([0.625, 0.75]\) Midpoint: \( m_4 = \frac{0.625 + 0.75}{2} = 0.6875 \) \( f(0.6875) = (0.6875)^3 + 0.6875 - 1 = 0.325 + 0.6875 - 1 = 0.0125 \) - \( f(0.625) = -0.131 \), \( f(0.6875) = 0.0125 \) (sign change) - So, root is in \([0.625, 0.6875]\) **Explanation:** Each step, we halve the interval and choose the subinterval where the function changes sign, progressively narrowing down where the root is. --- ### **Final Answer:** - **(a)** Yes, a root exists on \([0, 1]\) because the function changes sign. - **(b)** After 4 steps, the root lies in the interval \([0.625, 0.6875]\).

Let's proceed step-by-step as requested: --- ### **(a) Determine the existence of the root on [0, 1]** **Step 1:** Evaluate the function at the endpoints: \[ f(0) = 0^3 + 0 - 1 = -1 \\ f(1) = 1^3 + 1 - 1 = 1 \] Since \(f(0) = -1\) (negative) and \(f(1) = 1\) (positive), and the function \(f(x) = x^3 + x - 1\) is continuous on \([0, 1]\), **by the Intermediate Value Theorem**, **there exists at least one root in \([0, 1]\)**. **Explanation:** The function crosses zero somewhere between 0 and 1 because it takes values of opposite signs at the endpoints, confirming the existence of at least one root in the interval. --- ### **(b) Perform 4 steps of the bisection method** **Initial interval:** \([a, b] = [0, 1]\) --- **Step 1:** Calculate the midpoint: \[ m_1 = \frac{a + b}{2} = \frac{0 + 1}{2} = 0.5 \] Evaluate \(f(0.5)\): \[ f(0.5) = (0.5)^3 + 0.5 - 1 = 0.125 + 0.5 - 1 = -0.375 \] Since \(f(0) = -1\) and \(f(0.5) = -0.375\) are both negative, the sign change occurs in \([0.5, 1]\). **Interval after step 1:** \([0.5, 1]\) --- **Step 2:** Midpoint: \[ m_2 = \frac{0.5 + 1}{2} = 0.75 \] Evaluate \(f(0.75)\): \[ f(0.75) = (0.75)^3 + 0.75 - 1 = 0.422 + 0.75 - 1 = 0.172 \] Since \(f(0.5) = -0.375\) and \(f(0.75) = 0.172\), the sign change occurs in \([0.5, 0.75]\). **Interval after step 2:** \([0.5, 0.75]\) --- **Step 3:** Midpoint: \[ m_3 = \frac{0.5 + 0.75}{2} = 0.625 \] Evaluate \(f(0.625)\): \[ f(0.625) = (0.625)^3 + 0.625 - 1 = 0.244 + 0.625 - 1 = -0.131 \] Since \(f(0.625) = -0.131\) (negative) and \(f(0.75) = 0.172\) (positive), the root lies in \([0.625, 0.75]\). **Interval after step 3:** \([0.625, 0.75]\) --- **Step 4:** Midpoint: \[ m_4 = \frac{0.625 + 0.75}{2} = 0.6875 \] Evaluate \(f(0.6875)\): \[ f(0.6875) = (0.6875)^3 + 0.6875 - 1 \approx 0.325 + 0.6875 - 1 = 0.0125 \] Since \(f(0.625) = -0.131\) and \(f(0.6875) = 0.0125\), the sign change occurs in \([0.625, 0.6875]\). --- ### **Final answer:** - **Existence:** Confirmed; a root exists in \([0, 1]\). - **Approximate interval after 4 steps:** \(\boxed{[0.625, 0.6875]}\) This interval contains the root to the desired accuracy after 4 iterations of the bisection method.