I had to start by saying… did you know that Tom’s makes a CALCULUS SHOE???? Check it out here http://www.toms.com/calculus-101-men-s-classics. They make it in men’s, women’s, and children’s sizes! I would so totally get these and be a nerd if they weren’t so expensive. 

 

So most of this week we learned about how to find absolute extrema using a graph. Friday we learned how to find Absolute Extrema using an equation. It all seems a bit TOO easy compared to what we’ve already learned, but I think this is because our foundation has been so good, the new stuff makes perfect sense using our firmly planted knowledge leading up to it.  Oh, and here’s how our current area of study is relevant to real life:

So, I’ve decided to explain in my own words how you can find the absolute extrema for a function without having to graph it.

First you’re going to need a function. Let’s use, for example’s sake, the function f(x)=4x^4-32x^2+12. Typically you are also given end points such as [-5,5] or (0,32) or (-\infty,\infty). All of these have different meanings. [brackets] like these mean that you can use those two numbers as endpoints. (parenthesis) means you can use all the points leading UP to the numbers in parenthesis but not the actual number. When you get infinity to infinity you have to judge by the problem whether or not you’ve got an absolute max or min by looking at its characteristics, which could mean doing a sign chart.

Moving on.. for our problem, we’re going to solve it with [-5,5] as our end points for the ease of discussion.

Now that we’ve got our endpoints and equation we can follow these steps to get our absolute extrema:

1. Find the derivative of the function. For f(x)=4x^4-32x^2+12 the derivative is f'(x)=16x^3-64x

2. Set the derivative to Zero. 16x^3-64x=0 in order to find critical points.

3. Find Critical Points. 16x(x^2-4)=0 can be factored to 16x(x-2)(x+2)=0 so our critical points are x=0, x=-2, x=2.

4. Solve for f(x) using critical points and endpoints. It’s easier to use a table so that you can immediately see the comparative numbers for each critical and end point, therefore you can instantly identify the absolute max and min. Your table may look something like this:

X

F(x)

-5

1712

-2

-52

0

12

2

-52

5

1712

An easy way to get these numbers without having to put them all in for the same equation over and over, is to press the y= button on your graphing calculator, plug in the original function, and then push the 2nd button and the graph button (which is a table button when you push 2nd first). This will show you a nice table with your x-coordinates and their corresponding y coordinates. Saves a lot of time (especially for testing purposes). This was an idea presented by our professor.

5. Write out your absolute extrema. So for our function and the given end points, we know that our Absolute Maximum is 1712 at x=-5 and x=5, and our Absolute Minimum is  -52 at x=-2 and x=2.

It is important to remember that there isn’t always an absolute max or absolute min. Sometimes you may have one or the other or neither.

Hope you are all having a nice weekend. I’m off to enjoy the sunshine.