find the coordinates of the turning points of the curve y = 2x^4-4x^3+3, and determine the nature of these points
To begin, we must first use the fact that turning points of a graph occur at points where the gradient is equal to zero, in other words, points where dy/dx =0. Differentiating the equation with and setting to zero gives dy/dx = 8x3-12x2=0, then, solving for x we get x = 0 and x = 3/2. Putting these x values back into the original equation will give us the coordinates of the turning points which are (0,3) and(3/2,-3/8).The second part of the question asks us to determine the nature of the turning points, for which we will have to use the second derivative. Differentiating dy/dx again gives d2y/dx2=24x2-24x. at x= 3/2, d2y/dx2= 18 which makes it a minimum point since d2y/dx2>0. x=0, d2y/dx2=0 which means it could either be a minimum, maximum, or point of inflection, we will have to run further tests to determine the nature of this point.
