Find the stationary points on y = x^3 + 3x^2 + 4 and identify whether these are maximum or minimum points.

Differentiate wrt x. This leaves dy/dx = 3x^2 + 6x and equate this to 0 as we are looking for stationary points.So, 3x^2 + 6x = 0. Factorise to get x(3x +6) = 0. So x = 0 and x = -2 are the two solutions.Now differentiate wrt x once again, to give d^2y/dx^2 = 6x + 6. Now substitute in the values of x we have: For x = 0, d^2y/dx^2 = 6(0) + 6 = 6. Since 6 > 0, this can be identified as a minimum point.For x = -2, d^2y/dx^2 = 6(-2) + 6 = -6. Since -6 < 0, this can be identified as a maximum point.Now to complete the question, substitute our values of x for the stationary points into the original equation to find the y coordinates.So for x = 0, y = (0)^3 + 3(0)^2 + 4 = 4. This tells us that the minimum point is (0,4)And for x = -2, y = (-2)^3 + 3(-2)^2 + 4 = 8. This tells us that the maximum point is (-2,8)