Calculus: finding relative extrema

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 mention in an earlier post, one can find relative extrema where the derivative of a function is zero.

Therefore, to find the relative extremes of the continuing example, one would want to find where 6x^2 + 20x -11 is 0:

Solve.

6x^2 + 20x -11 = 0.

Using 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.

Therefore, 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.

Perhaps 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.

Source:

Larson, R.E., Hostetler, R.P. (1989). Calculus, part one, third edition. D.C. Heath and Company.