10/17/10

This week we had exams and then on Thursday we started Chapter 3. We learned how to find critical numbers, extreme values, max and mins, and horizontal tangents..simply by taking the derivative and setting it equal to zero! Also, to determine the max or min on an interval, you plug in the numbers of your interval and the numbers of your endpoint into the original equation. Overall, the concepts this week were pretty easy..I just need to practice with it some more. So here are some examples:

Ex. 1) Find any critical numbers of the function f(x) = 4x / x^2+1
*Since they're asking you to find critical numbers, the first thing you have to do is take the derivative of the function
*So, using the quotient rule, you get:
*(x^2+1)(4) - [(4x)(2x)] / (x^2+1)^2
Simplifying that you get:
-4x^2+4 / (x^2+1)^2
*To find the critical numbers, you take the numerator of the fraction and set it equal to zero.
So you get:
-4x^2 + 4 = 0
x^2 = 1
*So your critical numbers are x = 1, -1

Ex. 2) Locate the absolute extrema of the function on the closed interval for f(x) = (2x+5)/3 on the interval [0,5]
*The first thing you want to do for this problem is find your extreme values (which is the same as finding critical numbers) Sooooo, take the derivative of the function and set it equal to zero
*So for the derivative you get:
(3)(2) - [(2x+5)(0)] / 9
=6/9
f'(x) = 2/3 (*Since your derivative is a number and not a function, you have NO critical numbers)
*So you just need to plug in the numbers of your interval into the original equation to find the max and min
*So you get:
f(0) = 2(0)+5 / 3 = 5/3
f(5) = 2(5)+5 / 3 = 5
*So your answers are:
(0,5/3) min
(5,5) max

Ex. 3) Locate the absolute extrema of the function on the closed interval for f(x) = x^3-3/2x^2
on the interval [-1,2]
*First thing you have to do is take the derivative and set it equal to zero to find your critical numbers/extreme values
*So you get:
f'(x) = 3x^2 - 3x
*3x^2 - 3x = 0
3x(x-1) = 0
x = 0, 1 (extreme values)
*Now you plug in the numbers in the interval and the numbers of your endpoint into the original equation
*f(-1) = -1-3/2(-1)^2 = -5/2
f(2) = 8 - 3/2(2)^2 = 2
f(0) = 0
f(1) = 1-3/2 = -1/2
*So your answers are:
(2,2) max
(-1,-5/2) min

**So pretty much everything we learned this week was easy for me. The only thing I don't understand is how to find critical numbers and everything else when you are just given a graph to look at in the book...Alsoooo, I need to refresh my memory on Chapter 8 from Advanced math, hahah