y=20x-x^2-2x^3. Curve has a stationary point at the point M where x=-2. Find the x coordinate of the other stationary point of the curve and the value of the second derivative of both of these point, hence determining their nature.

Differentiate to get dy/dx=20-2x-6x^2Then stationary points occur when dy/dx = 0 so 0 = 20-2x-6x^2 Factorise to get x= -2, x=5/3Differentiate dy/dx to get second derivative = -2-12x at x=5/3 is -22 so max pointat x=-2 second derivative is 24>0 so min point.