A cubic curve y = f(x) = ax^3 + bx^2 + cx + d touches the line y = -9x + 2 at the point (0, 2) . The slope of the curve is minimum at x = 1 , and the curve has a local minimum at x = 3 . The absolute maximum value of f(x) on the interval [-2, 4] is

Step-by-step solution

The curve touches the line y = -9x + 2 at (0, 2) . This means f(0) = 2 and f'(0) = -9 . From f(x) = ax^3 + bx^2 + cx + d , we get d = 2 . Differentiating, f'(x) = 3ax^2 + 2bx + c . Since f'(0) = -9 , we have c = -9 . The slope of the curve is given by f'(x) . For the slope to be minimum at x = 1 , its derivative must be zero at x = 1 . f''(x) = 6ax + 2b f''(1) = 6a + 2b = 0 b = -3a The curve has a local minimum at x = 3 , so f'(3) = 0 . f'(3) = 3a(3)^2 + 2b(3) - 9 = 27a + 6b - 9 = 0 Substituting b = -3a into the eq