Problem 2 - Entrance Test

To find the local maxima and minima of the function f(x), we need to find the critical points, which are the points where the derivative of the function is equal to zero. The derivative of f(x) is f'(x) = 3x^2 - 12x + 11. Setting this equal to zero, we get 3x^2 - 12x + 11 = 0, which can be solved using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a, where a = 3, b = -12, and c = 11. Plugging these values into the formula, we get x = (12 ± √((-12)^2 - 4*3*11)) / 2*3, which simplifies to x = (12 ± √(144 - 132)) / 6, or x = (12 ± √12) / 6, which is x = (12 ± 2√3) / 6, or x = 2 ± (√3)/3. To determine which of these points corresponds to a local maximum and which corresponds to a local minimum, we can use the second derivative test. The second derivative of f(x) is f''(x) = 6x - 12. Plugging in the critical points, we get f''(2 + (√3)/3) = 6(2 + (√3)/3) - 12 = 12 + 2√3 - 12 = 2√3, which is positive, so this point corresponds to a local minimum, and f''(2 - (√3)/3) = 6(2 - (√3)/3) - 12 = 12 - 2√3 - 12 = -2√3, which is negative, so this point corresponds to a local maximum. Therefore, the value of a is 2 - (√3)/3 and the value of b is 2 + (√3)/3, so the value of a + b is (2 - (√3)/3) + (2 + (√3)/3) = 4.