y = (x^3)/3 - 4x^2 + 12x find the stationary points of the curve and determine their nature.

first we find the first derivative of the function. Here dy/dx = x^2-8x+12. We set this to zero and factorise to obtain the roots of the function. Such that dy/dx = (x-6)(x-2)= 0. This gives the stationary points as x=6 and x=2. By substituting our x values into our function we can obtain the coordinates of the points. These are (6,0) and (2,32/3) To determine the nature of these points we take the second derivative of the function. d^2y/dx^2= 2x-8. By substituting our values of x into the second derivative we obtain d^2y/dx^2= 4 for x=6 and d^2y/dx^2= -4 for x=2. If the second derivative is larger than zero the point is a minima, if smaller than zero the point is a maxima. Thus (6,0) is a mimima and (2,32/3) is a maxima.