\nThe following is by my understanding.<\/p>\n
\nIn my last post<\/a> I mention finding the derivative of a polynomial: the derivative turned out 6x^2 + 20x -11.<\/p>\n \nAs I mention in an earlier post<\/a>, one can find relative extrema where the derivative of a function is zero.<\/p>\n \nTherefore, to find the relative extremes of the continuing example, one would want to find where 6x^2 + 20x -11 is 0:\n<\/p>\n Solve.<\/p>\n 6x^2 + 20x -11 = 0.<\/p>\n \nUsing the quadratic formula, one can determine the solutions (aka roots) of the above equation to be 0.481 or else -3.814, rounded to three decimal places.<\/p>\n \nTherefore, it will be at the x values 0.481 and -3.814 that the relative extrema will be found. To find the actual relative extrema, one would, in turn, evaluate the original function, y=2x^3 + 10x^2 -11x -14, at x=0.481 and at x=-3.814. Doing so, we get (0.481,-16.755) and (-3.814, 62.458). So, the relative maximum of the function is at (-3.841, 62.459), then its relative minimum follows at (0.481, -16.755). Such values suggest the expected shape of said cubic function.<\/p>\n \nPerhaps more coverage of solving quadratic equations and using the quadratic formula might be warranted, given this post. I plan to mention more in that direction, going forward.<\/p>\n \nSource:<\/p>\n \nLarson, R.E., Hostetler, R.P. (1989). Calculus,<\/em> part one, third edition. D.C. Heath and Company.<\/p>\nJack of Oracle Tutoring by Jack and Diane,<\/a> Campbell River, BC.\n\n\n\n <\/p>\n","protected":false},"excerpt":{"rendered":" Tutoring calculus, finding relative extrema is important. The tutor continues an example. The following is by my understanding. In my last post I mention finding the derivative of a polynomial: the derivative turned out 6x^2 + 20x -11. As I …<\/p>\n