Find all the stationary points of the curve: y = (2/3)x^3 – (1/2)x^2 – 3x + 7/6 and determine their classifications.
The first thing to do is write down y’ and y’’ neatly since these are the main two equations we will be working with… y’ = 2x2 – x – 3 & y’’ = 4x – 1… To find the stationary point we must find where y’ = 0. The best method for this quadratic, in this case, is factorising… 0 = (2x – 3)(x + 1)… Therefore x = -1 and x = 3/2 are the two solutions… To see what types of stationary point these are we plug these x values in to y’’… For x=-1 we get y’’=-5. This being negative means we have a maximum here… For x=3/2 we get y’’=5. This being positive means we have a minimum here… Last thing to do is plug the x values back in to the original equation to find the corresponding y values… For x=-1 we get y=3 and for x=3/2 we get y=-53/24… In summary we have a maximum at (-1,3) and a minimum at (3/2, -53/24)
