Note: In order to broaden our understanding of Maple and critical points, we will be doing the same problem from many different approaches. You may find some to be quicker and "easier" than others. You may also find some to be mo re informative and precise. The way this notebook is set up is to facilitate comparison along these lines. Find what works best for you, and enjoy.
Finding the Maximum, Minimum, and Inflection Points of a Function Using Maple's Capabilities. source information
The Function
Enter the function you wish to examine, giving it a name. In this case I called the function "y."
> y := 2*x^33*x^212*x+5;
Next, you might want to see the plot of the function to get a feel for what the function looks like.
> plot( y(x), x= 3..3 );
Maple's Easy Step:
Finding the y value for the maximum and minimum.
>
readlib(extrema):
extrema( y, {}, x );
This means that the maximum and minimum occur at y12 and y=15.
Now, we have d rawn the function so we know that when y=12 we have a maximum and that when y=15 we have a minimum. Our next step then is to find the x component for the ordered pair.
Finding the x component.
To find the x component for the ordered pair of maximum and minimum, we substitute y=12 and y=15 into the function y.
> solve( y=15, x );
> solve( y=12, x );
Looking at the graph we know that the maximum of y is (1, 12), and the minimum is (2, 15).
We are now going to pretend we had not graphed y and that we have no way of knowing whether or not y=12 or y=15 is a maximum or a minimum. Now to find the maximum and minimum we are going to have to take the second derivative of y and use the second derivative test.
Some explanation is needed before we continue. When finding the critical points of a function by usng the first and second derivatives, one solves for the x values that make the function equal to zero. However, Maple found the y component of the extrema instead of the x component. This has made our work more complicated because when we substituted the y value into the function y, we are given three values for x. In order to proceed we had to look at the graph and find which value actually made sence. Since we are now going to pretend that we have no graph to use, we must compute the answer using the first and second derivatives. In essence, this means beginning the problem again, but with a different approach. Using first and second derivatives.
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