Find the stationary points of the curve y (x)= 1/3x^3 - 5/2x^2 + 4x and classify them.

To find the stationary points of the curve y(x), you must first differentiate the equation for y(x) in terms of x. This gives d(y(x))/dx = x^2 -5x +4. Now set this differential equal to zero and solve for x (as at a stationary point the gradient of a curve is equal to zero), to find the x coordinates of the stationary points. This quadratic can be factorised to give (x-4)(x-1) = 0, so x is equal to 4 and 1 for the two stationary points respectively. To find the y coordinate of the first stationary point when x=1, simply put this value into the equation for y(x). This gives a stationary point of (1,11/6). For the stationary point when x=4, do the same to obtain the y coordinate. This gives a stationary point (4,-8/3).To classify the stationary points, the second order differential for y(x) must be found, which is d^2(y(x))/dx^2 = 2x -5. Now for the two stationary points, substitute into the second order differential the two x coordinates. For x=1, d^2(y(x))/dx^2 = -3, which is less than zero so (1,11/6) is a local maximum. When x=4, d^2(y(x))/dx^2 = 3, which is greater than zero meaning (4,-8/3) is a local minimum.