So, i think i got the wrong idea on this blog thing..so i'm going to do both in one.
My bad..i guess I should ask questions if i didn't understand, or check edline..but anyways, here goes nothing.
For POST #14
Today i'll explain absolute max and mins.
First, you must do the first derivative test which includes taking the derivative and setting it equal to zero, then solving for x.
Second, you must plug in those values, also known as critical values, into original function to get y-values.
Third, plug in endpoints to original function to get more y-values.
Lastly, your highest y-value is an absolute max and lowest y-value is an absolute min.
THINGS YOU SHOULD REMEMBER:
**Absolute max and mins are written as a POINT or simply as a y-value
**They always and only include the y-value
**NEVER involve the x!!! It ONLY involves the y-values!
So, example problem:
Find the absolute max or min of f(x) = 3x^4 - 4x^3 on [-1,2]
12x^3-12x62 = 0
12X^2(x-1)
x = 0,1
(-1, 0) u (0,1) u (1,2)
f`(-.5) = -ve
f`(.5) = -ve
f`(1.5) = +ve
Then plug in endpoints and you recieve the points:
(1, -1)
(-1, 7)
(2, 16)
The absolute min is (1, -1) or y = -1
The absolute max is (2, 16) or y = 16
The thing I do not understand is average speed, velocity, and definition of a derivative. Could anyone help me because it is something that i can never quite grasp..
POST #15
One thing i am comfortable with is Rolle's theorem. In order to classify to be functionally correct with rolles, it must be continuous and differentiable and the y-values MUST match.
Rolles theorem gives conditions that guarantee the existance of an extrema in the interior of a closed interval.
It states, let f be continuous on a closed interval [a,b] and differentiable on the open interval (a,b). If f(a) = f(b) then there is AT LEAST one number, or c, in the interval (a,b) such that f`(c) = 0.
****Y-VALUES MUST MATCH!****
For the function f(x) = x^2 - 3x + 2, show that f`(x) = 0 on some interval.
x^2 - 3x + 2 = 0
(x-2)(x-1)
x-intercepts = [1,2]
f(1) = f(2) = 0
Continuous: check mate
Differentiable: check mate
f`(x) = 2x - 3
2x = 3
x = 3/2
c = 3/2
Using Rolle's Theorem because the function is continuous and differentiable and the y-values match; after the derivative was taken and set equal to zero, one found the c = 3/2.
The newest thing I don't understand and get lost with in integration is how you can not use product and quotient rule. I do not understand how to break it up..so can anyone explain how to break these things up? Also, substitution confuses me because i don't understand when you use substitution and when you break it up..
HELP PLEASEEE!
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instead of using product rule you just integrate the two things separately. say you were give (x^5)(2x) you would first integrate x^5 then 2x then multiply them together. and whenever you are given a fraction and you would usually use the quotient rule, just raise the bottom to -1 and multiply it to the top. hope i helped :)
ReplyDeletealso if you have something like
ReplyDelete4x^2+3x-5
---------
7
Because you only have one term on the bottom you can separate it to 4/7x^2+3/7x-5/7
If you have more than one term you cannot do this