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Multiple Choice
Let for in the interval . At which value(s) of does attain a local minimum?
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Verified step by step guidance
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Step 1: To find the local minimum of v(t), first compute the derivative v'(t) to determine the critical points. The derivative of v(t) = t^3 - 6t^2 + 9t + 2 is v'(t) = 3t^2 - 12t + 9.
Step 2: Set v'(t) = 0 to find the critical points. Solve the equation 3t^2 - 12t + 9 = 0. Factorize the quadratic equation to get 3(t - 1)(t - 3) = 0, which gives critical points t = 1 and t = 3.
Step 3: Use the second derivative test or the first derivative test to determine the nature of each critical point. Compute the second derivative v''(t) = 6t - 12. Evaluate v''(t) at t = 1 and t = 3.
Step 4: For t = 1, v''(1) = 6(1) - 12 = -6, which is negative. This indicates that v(t) has a local maximum at t = 1. For t = 3, v''(3) = 6(3) - 12 = 6, which is positive. This indicates that v(t) has a local minimum at t = 3.
Step 5: Verify the endpoints of the interval [0, 5] by evaluating v(t) at t = 0, t = 5, and the critical points t = 1 and t = 3. Compare the values to confirm that the local minimum occurs at t = 3.