Blog #8

This week wasn't too bad at all, it passed by kinda really fast for me. Anyway, we started Chapter 3, finally, and we started off learning the guidelines for finding extrema on a closed interval.

To find the extrema of a continuous function f on a closed interval [a,b], use the following steps.

1. Find the critical numbers of f in (a,b).

2. Evaluate f at each critical number in (a,b).

3. Evaluate f at each endpoint of [a,b].

4. The least of these values is the minimum. The greatest is the maximum.

There are different terms in 3.1 that we should learn.

Extreme values - mins and maxes

absolute max - is the highest point on an interval

absolute min - in the lowest point on an interval

relative max (local) - any max not on an interval

relative min (local) - any min not on an interval

critical numbers - any max. or min. typically referred to as x=c

To find a critical number, extreme value, max, min, or horizontal tangent, take derivative and set it equal to 0.

Example:

Find the critical values of the function

f(x) = 4x^3 - 36x

12x^2 - 36

12x^2 - 36 = 0

12x^2 = 36

x^2 = 3

x = +/- sqrt 3 <------FINAL ANSWER

It's that simple.

Now, I'm not too sure about how to do extrema with trig functions and stuff. It's a little complicated when you have to use formulas, identities, trig functions, and other stuff all at the same time.