Let's solve the equation: **Equation:** \( 3x^3 + x^2 - 2x + 1 = 0 \) **Interval:** \( -2 \leq r \leq -1 \) Let's use the Intermediate Value Theorem and Newton's Method (or bisection) to approximate the root to two decimal places. Let's check values at the endpoints: - For \( x = -2 \): \( 3(-2)^3 + (-2)^2 - 2(-2) + 1 = 3(-8) + 4 + 4 + 1 = -24 + 4 + 4 + 1 = -15 \) - For \( x = -1 \): \( 3(-1)^3 + (-1)^2 - 2(-1) + 1 = 3(-1) + 1 + 2 + 1 = -3 + 1 + 2 + 1 = 1 \) The function changes sign in the interval, so a root exists between \(-2\) and \(-1\). Now, let's try the **bisection method**: ### Step 1: Midpoint \( x = \frac{-2 + (-1)}{2} = -1.5 \) \( 3(-1.5)^3 + (-1.5)^2 - 2(-1.5) + 1 = 3(-3.375) + 2.25 + 3 + 1 = -10.125 + 2.25 + 3 + 1 = -3.875 \) So at \( x = -1.5 \), the value is \(-3.875\). Now, the sign change is between \( x = -1.5 \) (\(-3.875\)) and \( x = -1 \) (\(1\)). ### Step 2: Next Midpoint \( x = \frac{-1.5 + (-1)}{2} = -1.25 \) \( 3(-1.25)^3 + (-1.25)^2 - 2(-1.25) + 1 \) \( = 3(-1.953125) + 1.5625 + 2.5 + 1 \) \( = -5.859375 + 1.5625 + 2.5 + 1 \) \( = -0.796875 \) So at \( x = -1.25 \), the value is \(-0.796875\). Now, the sign change is between \( x = -1.25 \) (\(-0.80\)) and \( x = -1 \) (\(1\)). ### Step 3: Next Midpoint \( x = \frac{-1.25 + (-1)}{2} = -1.125 \) \( 3(-1.125)^3 + (-1.125)^2 - 2(-1.125) + 1 \) \( = 3(-1.423828125) + 1.265625 + 2.25 + 1 \) \( = -4.271484375 + 1.265625 + 2.25 + 1 \) \( = 0.244140625 \) Now, the sign change is between \( x = -1.25 \) (\(-0.80\)) and \( x = -1.125 \) (\(0.24\)). ### Step 4: Next Midpoint \( x = \frac{-1.25 + (-1.125)}{2} = -1.1875 \) \( 3(-1.1875)^3 + (-1.1875)^2 - 2(-1.1875) + 1 \) \( = 3(-1.674808502) + 1.41015625 + 2.375 + 1 \) \( = -5.024425507 + 1.41015625 + 2.375 + 1 \) \( = -0.239269257 \) So, at \( x = -1.1875 \), the value is \(-0.24\). Now, the sign change is between \( x = -1.1875 \) (\(-0.24\)) and \( x = -1.125 \) (\(0.24\)). ### Step 5: Next Midpoint \( x = \frac{-1.1875 + (-1.125)}{2} = -1.15625 \) \( 3(-1.15625)^3 + (-1.15625)^2 - 2(-1.15625) + 1 \) \( = 3(-1.545610428) + 1.337158203 + 2.3125 + 1 \) \( = -4.636831284 + 1.337158203 + 2.3125 + 1 \) \( = 0.012827919 \) Now, the sign change is between \( x = -1.1875 \) (\(-0.24\)) and \( x = -1.15625 \) (\(0.01\)). ### Step 6: Next Midpoint \( x = \frac{-1.1875 + (-1.15625)}{2} = -1.171875 \) \( 3(-1.171875)^3 + (-1.171875)^2 - 2(-1.171875) + 1 \) \( = 3(-1.609502852) + 1.373286247 + 2.34375 + 1 \) \( = -4.828508555 + 1.373286247 + 2.34375 + 1 \) \( = -0.111472308 \) Now, the sign change is between \( x = -1.171875 \) (\(-0.11\)) and \( x = -1.15625 \) (\(0.01\)). ### Step 7: Next Midpoint \( x = \frac{-1.171875 + (-1.15625)}{2} = -1.1640625 \) \( 3(-1.1640625)^3 + (-1.1640625)^2 - 2(-1.1640625) + 1 \) \( = 3(-1.577437935) + 1.355190754 + 2.328125 + 1 \) \( = -4.732313805 + 1.355190754 + 2.328125 + 1 \) \( = -0.049998051 \) So, the sign change is between \(-1.164\) (\(-0.05\)) and \(-1.156\) (\(0.01\)). ### Step 8: Next Midpoint \( x = \frac{-1.1640625 + (-1.15625)}{2} = -1.16015625 \) \( 3(-1.16015625)^3 + (-1.16015625)^2 - 2(-1.16015625) + 1 \) \( = 3(-1.561523237) + 1.346171677 + 2.3203125 + 1 \) \( = -4.684569711 + 1.346171677 + 2.3203125 + 1 \) \( = -0.018085534 \) We're very close to zero and our answer should be rounded to two decimal places. ### Let's check values at \(-1.16\) and \(-1.15\): - \( x = -1.16 \Rightarrow 3(-1.16)^3 + (-1.16)^2 - 2(-1.16) + 1 \) \( = 3(-1.561) + 1.346 + 2.32 + 1 = -4.683 + 1.346 + 2.32 + 1 = -0.017 \) - \( x = -1.15 \Rightarrow 3(-1.15)^3 + (-1.15)^2 - 2(-1.15) + 1 \) \( = 3(-1.522) + 1.3225 + 2.3 + 1 = -4.566 + 1.3225 + 2.3 + 1 = 0.0565 \) The sign changes between \(-1.16\) and \(-1.15\). So, to two decimal places, the solution is: \[ \boxed{-1.16} \] **Final answer:** \[ r \approx -1.16 \]