The curve a x^3+b x^2+c x+d has a point of minima at x=1 , then

Step-by-step solution

Let f(x) = ax^3 + bx^2 + cx + d . The derivative is given by f'(x) = 3ax^2 + 2bx + c . Since the curve has a point of minima at x=1 , we must have f'(1) = 0 . Thus, 3a(1)^2 + 2b(1) + c = 0 , which implies 3a + 2b + c = 0 . For a local minimum at x=1 , the second derivative must satisfy f''(1) > 0 . The second derivative is f''(x) = 6ax + 2b . Substituting x=1 , we get f''(1) = 6a + 2b > 0 . Dividing the inequality by 2 , we obtain 3a + b > 0 . Answer: 3 a+b>0