A quadratic polynomial f(x) satisfies f(0) = 3 , f'(2) = 0 , and f''(x) = 2 for all x . If the area of the region bounded by y = f(x) and y = x - 1, 5 - 3x is A , then the value of 6A is

Step-by-step solution

First, we determine the quadratic polynomial f(x) . Given f''(x) = 2 , integrating yields f'(x) = 2x + c₁ . Using f'(2) = 0 , we get 2(2) + c₁ = 0 c₁ = -4 . Thus, f'(x) = 2x - 4 . Integrating again gives f(x) = x^2 - 4x + c₂ . Using f(0) = 3 , we get c₂ = 3 . Therefore, f(x) = x^2 - 4x + 3 . Next, we find the area of the region bounded by y = x^2 - 4x + 3 and y = x - 1, 5 - 3x . Find the intersection of the two lines inside the min function: x - 1 = 5 - 3x 4x = 6 x = 3 2 . For x 3 2 , x - 1 5 - 3x , so the upper bo