If y = 2/3 x^3 + x^2; a) What is dy/dx? b) Where are the turning points? c) What are the nature of the turning points?

a) By simple parametric differentiation of each term, dy/dx = 2x^2 + 2x b) The condition for a turning point is the gradient (dy/dx) at that point is zero. 0 = x(2x + 2) so either x=0 or 2x+2=0. Therefore the turning points are at x=0 or x=-1. c) The nature of the turning point is how the gradient is changing at that point so we have to find d^2y/dx^2. If the value is positive it is a minimum, negative is a maximum and zero may be a stationary point. From the equation for dy/dx, this is 4x+2 (simple differentiation). We substitute in the turning points x=0,-1) and find at x=0, it is 2 so minimum and at x=-1 it is -2 which means it's a maximum.