This week in calculus we started off working some more optimization problems and then took a quiz on optimization. I understood the concept of optimization and how to work it, i just need help figuring out my different formulas.
After the quiz, we started working on exam study guides in class. I understand the derivative packet the best. I also understand what to do when they ask you for the slope of a function and when they ask for an equation to the tangent line.
For those of you who are still struggling with derivatives, heres an example of one:
8xsinx+5cosx.
the derivative of that would be
8x(cos(x)) + 8sin(x)-5sinx
to simplify that you would come out with:
8xcos(x)+5sin(x)
I also now understand how to figure out different types of limits. You find vertical asymptotes when you set the bottom of a fraction equal to zero and solve. For horizontal asymptotes, you use the three rules when looking at the degree on the top of the fraction and on the bottom. Anything you can cancel from a function is a removable. Another thing i understand better is how to use the first and second derivative test. When using the first derivative test, you take the derivative of the function and set it equal to zero and solve for x. Then you set up your x values into intervals to see which ones are max and mins etc. When using the second derivative test, you take the derivative of a function two times and set equal to zero and solve for x. You put these x values into intervals as well to find if a graph is concave up or down, point of inflection, etc.
The problems i still seem to have is how to figure out what formulas need to be used in optimization problems. After i figure that out, i can finish the problems. Another thing that i am still confused with is when they give you a graph of an original function. Then they want you to calculate what the max and min of the graph. i do not understand how to look at the original function and figure out what the first derivative is supposed to look like. If anyone can help me with these things, it would be big help.. Thanks :)
Saturday, October 3, 2009
Friday, October 2, 2009
Week 7
Calculus Week#7
So this week in Calculus we started off by studying and doing more problems with Optimization. The thing about Calculus in general that I've learned is that everything comes with more and more practice. We took a short quiz on optimization which I thought was pretty easy, but I know a lot of you struggled with it...and I hope that won't happen on the exam or something.
To go over optimization, the way I think about it:
You look at the problem. Read what it says...if it says to find the "maximum area" or the "area of maximum size", it obviously means you are going to be optimizing the area...because it wants the maximum..or minimum.
So, we know our primary equation. Now we look at the equation. Does it have 3 or more variables? If so, you have to find a secondary equation. In the problem it will probably say something like "Find the maximum area of a rectangle that has a perimeter of 36 units". Okay, so perimeter was mentioned...let's make that our secondary.
So now you have two equations and want to maximize one. So, to make the one that you want to maximize in terms of one variable, solve the secondary equation for a variable and then plug in what you get for that variable in the primary.
For instance, say we have
2x+2y=36 and xy=A
Solving the secondary (the first one) for y gives us (36-2x)/2. Simplifying and plugging into the first one for y, you get x(18-x)=A.
Now that we have it in terms of one variable, simplify and take the derivative.
(18x-x2).
The derivative is simply 18-2x.
Now we set the derivative equal to 0 and solve. x=9.
Plugging that back into our secondary, which is the next step, would give us that y=18 as well.
You are done. :-) The maximum area rectangle has dimensions 9x9. Btw, 9+9+9+9 = 36 (the perimeter works out if you want to check yourself)
Also, a hint for the future of rectangles....the maximum area will always be a n x n sized rectangle...that's just how it works.
Good luck on study guides and exam preparation!
So this week in Calculus we started off by studying and doing more problems with Optimization. The thing about Calculus in general that I've learned is that everything comes with more and more practice. We took a short quiz on optimization which I thought was pretty easy, but I know a lot of you struggled with it...and I hope that won't happen on the exam or something.
To go over optimization, the way I think about it:
You look at the problem. Read what it says...if it says to find the "maximum area" or the "area of maximum size", it obviously means you are going to be optimizing the area...because it wants the maximum..or minimum.
So, we know our primary equation. Now we look at the equation. Does it have 3 or more variables? If so, you have to find a secondary equation. In the problem it will probably say something like "Find the maximum area of a rectangle that has a perimeter of 36 units". Okay, so perimeter was mentioned...let's make that our secondary.
So now you have two equations and want to maximize one. So, to make the one that you want to maximize in terms of one variable, solve the secondary equation for a variable and then plug in what you get for that variable in the primary.
For instance, say we have
2x+2y=36 and xy=A
Solving the secondary (the first one) for y gives us (36-2x)/2. Simplifying and plugging into the first one for y, you get x(18-x)=A.
Now that we have it in terms of one variable, simplify and take the derivative.
(18x-x2).
The derivative is simply 18-2x.
Now we set the derivative equal to 0 and solve. x=9.
Plugging that back into our secondary, which is the next step, would give us that y=18 as well.
You are done. :-) The maximum area rectangle has dimensions 9x9. Btw, 9+9+9+9 = 36 (the perimeter works out if you want to check yourself)
Also, a hint for the future of rectangles....the maximum area will always be a n x n sized rectangle...that's just how it works.
Good luck on study guides and exam preparation!
Sunday, September 27, 2009
week 6 post
This week we learned a lot of stuff, like, the Extreme Value Theorem, Rolles Theorem, and Mean Value Theorem.
EVT: the EVT states that a continuous function on a close interval [a,b], must have both a minimum and a maximum on the interval. However, the max and min can occur at the endpoints.
Rolles theorem gives the conditions that guarantee the existance of an extrema in the interior of a closed inteval.
Rolles: 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, “c” in (a,b) such the f '(c)=0
MVT: If f is continuous on the closed interval [a,b] and differentiable on the interval (a,b) then there exists a number c in (a,b) such that f '(c)= (f(b)-f(a))/(b-a).
Example: f(x)-x^(2)-3x+2, show that f ' (x)=0 on some interval. (Hint: if they don't give an interval, assume it's x-intercepts)
so first you factor and find that x=1 and x=2
so then, you check and see that f(1)=f(2)=0 so it's continuous and it's differentiable.
So after you've found that, you take the derivative and set it equal to zero.
F '(x)=2x-3
2x-3=0
2x=3
x=3/2
c=3/2. And that's it. You've found c.
----------------
Another example
f(x)=x^(4)-2x^(2) on [-2,2] find all values of c for which f '(c)=0
continuous: yes
differentiable: yes. So let's get started.
First, you have to plug in to see if f(-2) is equal to f(2)
so, f(-2)=(-2)^(4) -2 (-2)^(2)= 8
f(2)= (2)^(4)-2(2)^(2)=8
so that's good. They equal each other. By the way, i'm not sure if i'm typing up all my parentheses correctly, so sorry if I mess that up.
So then you take the derivative and set = to zero
f '(x)=4x^(3)-4x=0
4x(x^(2)-1)=0
x= 0, 1, -1
c= -1, 0, 1. and you're finished again.
one thing i don't understand is optimization. one reason for this is because i have like no notes on it... i'm not exactly sure why though. i have a bunch of examples in my notebook, but i can't find any notes. so i pretty much don't understand it at all. thanks.
EVT: the EVT states that a continuous function on a close interval [a,b], must have both a minimum and a maximum on the interval. However, the max and min can occur at the endpoints.
Rolles theorem gives the conditions that guarantee the existance of an extrema in the interior of a closed inteval.
Rolles: 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, “c” in (a,b) such the f '(c)=0
MVT: If f is continuous on the closed interval [a,b] and differentiable on the interval (a,b) then there exists a number c in (a,b) such that f '(c)= (f(b)-f(a))/(b-a).
Example: f(x)-x^(2)-3x+2, show that f ' (x)=0 on some interval. (Hint: if they don't give an interval, assume it's x-intercepts)
so first you factor and find that x=1 and x=2
so then, you check and see that f(1)=f(2)=0 so it's continuous and it's differentiable.
So after you've found that, you take the derivative and set it equal to zero.
F '(x)=2x-3
2x-3=0
2x=3
x=3/2
c=3/2. And that's it. You've found c.
----------------
Another example
f(x)=x^(4)-2x^(2) on [-2,2] find all values of c for which f '(c)=0
continuous: yes
differentiable: yes. So let's get started.
First, you have to plug in to see if f(-2) is equal to f(2)
so, f(-2)=(-2)^(4) -2 (-2)^(2)= 8
f(2)= (2)^(4)-2(2)^(2)=8
so that's good. They equal each other. By the way, i'm not sure if i'm typing up all my parentheses correctly, so sorry if I mess that up.
So then you take the derivative and set = to zero
f '(x)=4x^(3)-4x=0
4x(x^(2)-1)=0
x= 0, 1, -1
c= -1, 0, 1. and you're finished again.
one thing i don't understand is optimization. one reason for this is because i have like no notes on it... i'm not exactly sure why though. i have a bunch of examples in my notebook, but i can't find any notes. so i pretty much don't understand it at all. thanks.
Post #6
This week in calculus started off pretty easy. We learned Rolle’s Theorem and Mean Value Theorem. To use Rolle’s Theorem, you the equation giving has to be continuous and differentiable on the interval giving and f(a)=f(b). If it meets these requirements, you then have to take the derivative and set it equal to zero. Then solve for x and that is your c (which is your maxs and mins on that interval).
An example of Rolle’s Theorem is x^2-2x, [0,2]
Since the equation is a parabola, it is both continuous and differentiable.
F(0) = 0^2-2(0) = 0
F(2)= 2^2-2(2) = 0
Since f(0)=f(2)=0 , Rolle’s theorem can be used.
Derivative is 2x-2
2x-2 = 0
X= 1 so c=1
For Mean Value Theorem, you have to check to see if it is continuous and differentiable on the interval given such that f’(c)= f(b)-f(a)/b-a. Once you plug into that formula, you take the derivative and set it equal to the number you got then solve for x.
Example: x^2 , [-2,1]
It is both continuous and differential therefore mean value theorem can be used.
F(-2) = -2^2= 4
F(1) = 1^2 = 1
Plug into f(b)-f(a)/b-a = 4-1/ 4-1 = 1
Take the derivative: 2x
Set the derivative = to 1
2x=1
X= ½
C=1/2
The thing I did not understand this week was optimization. I have always been horrible with word problems. I get confused with the words and also for optimization, I never know what formula to plug into or how to get the primary formula ( if it’s not the one given) in the first place.
An example of Rolle’s Theorem is x^2-2x, [0,2]
Since the equation is a parabola, it is both continuous and differentiable.
F(0) = 0^2-2(0) = 0
F(2)= 2^2-2(2) = 0
Since f(0)=f(2)=0 , Rolle’s theorem can be used.
Derivative is 2x-2
2x-2 = 0
X= 1 so c=1
For Mean Value Theorem, you have to check to see if it is continuous and differentiable on the interval given such that f’(c)= f(b)-f(a)/b-a. Once you plug into that formula, you take the derivative and set it equal to the number you got then solve for x.
Example: x^2 , [-2,1]
It is both continuous and differential therefore mean value theorem can be used.
F(-2) = -2^2= 4
F(1) = 1^2 = 1
Plug into f(b)-f(a)/b-a = 4-1/ 4-1 = 1
Take the derivative: 2x
Set the derivative = to 1
2x=1
X= ½
C=1/2
The thing I did not understand this week was optimization. I have always been horrible with word problems. I get confused with the words and also for optimization, I never know what formula to plug into or how to get the primary formula ( if it’s not the one given) in the first place.
Posting...#6
Hello Calc
These past few days we have tested a lot learned a lot and died on the inside a lot. We learned three new main ideas / topics / formulas. We learned Rolle’s theorem, mean value theorem and optimization.
What I understand is Rolle’s theorem and the Mean Value theorem. So ill just explain the mean value theorem.
Explanation: Mean Value is if (f) is continuous on the closed interval (a,b) and differentiable on the open interval (a,b)then and only then a number c exist in (a,b)
F’(c) =f(b)-f(a)/b-a now take derivative and set equal to (a) and (b)’s slope.
So for example: F(x) =x^4-2x^2 on [-2,2] find f’(c ) =0
Is it continuous: YESIs it differentiable:YES
F(-2)=(-2)^4-2(-2)^2=8___f(2)=(2)^4-2(2)^2=8____f(-2)=f(2)=8
4x^3-4x=0-----4x(x^2 -1)=0 -----x=0,1,-1-----c= 0,1,-1
And now your done.
But what I really have problems with is optimization… What is optimization? How to do optimization? Where to start optimization? I don’t really understand it I look in my notebook look at the problems we did but I still don’t get it so if anyone can help me understand optimization please do cause confused.
That’s my biggest problem so give me ya’z wordz or wisdomz because I need all the help I can get and we had homework this weekend and I didn’t know we had homework so I didn’t do my homework so now I don’t know what will happen.
These past few days we have tested a lot learned a lot and died on the inside a lot. We learned three new main ideas / topics / formulas. We learned Rolle’s theorem, mean value theorem and optimization.
What I understand is Rolle’s theorem and the Mean Value theorem. So ill just explain the mean value theorem.
Explanation: Mean Value is if (f) is continuous on the closed interval (a,b) and differentiable on the open interval (a,b)then and only then a number c exist in (a,b)
F’(c) =f(b)-f(a)/b-a now take derivative and set equal to (a) and (b)’s slope.
So for example: F(x) =x^4-2x^2 on [-2,2] find f’(c ) =0
Is it continuous: YESIs it differentiable:YES
F(-2)=(-2)^4-2(-2)^2=8___f(2)=(2)^4-2(2)^2=8____f(-2)=f(2)=8
4x^3-4x=0-----4x(x^2 -1)=0 -----x=0,1,-1-----c= 0,1,-1
And now your done.
But what I really have problems with is optimization… What is optimization? How to do optimization? Where to start optimization? I don’t really understand it I look in my notebook look at the problems we did but I still don’t get it so if anyone can help me understand optimization please do cause confused.
That’s my biggest problem so give me ya’z wordz or wisdomz because I need all the help I can get and we had homework this weekend and I didn’t know we had homework so I didn’t do my homework so now I don’t know what will happen.
week six
we Learned alot! it is hectic!
Rolles thereom - this thereom states that for a said closed interval that is continous and differentiable at that interval when the function of a is equal to the functin of b then there is a given number C which equals zero in that interval.
f(a)=f(b), then f(c)=0
Optimization- a process that "stretches" an equation. first you determine all quantities of the equation, then you write a new equation, reduce it and determine the domain of the equation, then from there you determine the max's and min's of the equation. optimization is usefull when dealing with area and volume.
Mean value thereom - somewhat like rolles. says that if a function is continous and differentiable then, f^(c) = f(b)-f(a)/b-a. saying that when its continous and differentiable on the interval then the value c exist on that interval.
The main thing that i am stuggling with is optimization. like i can list the steps but when it comes to working a problem with it I get stuck and dont even know what to do. Rolles and Mean is not as difficult to me because it reminds me of the first derivative test for some reason. I just need more work and help with optimization.
Rolles thereom - this thereom states that for a said closed interval that is continous and differentiable at that interval when the function of a is equal to the functin of b then there is a given number C which equals zero in that interval.
f(a)=f(b), then f(c)=0
Optimization- a process that "stretches" an equation. first you determine all quantities of the equation, then you write a new equation, reduce it and determine the domain of the equation, then from there you determine the max's and min's of the equation. optimization is usefull when dealing with area and volume.
Mean value thereom - somewhat like rolles. says that if a function is continous and differentiable then, f^(c) = f(b)-f(a)/b-a. saying that when its continous and differentiable on the interval then the value c exist on that interval.
The main thing that i am stuggling with is optimization. like i can list the steps but when it comes to working a problem with it I get stuck and dont even know what to do. Rolles and Mean is not as difficult to me because it reminds me of the first derivative test for some reason. I just need more work and help with optimization.
Ash's 6th post
This week I actually understood something on my own....and explained it to people! My gosh I felt accomplished. Also, I almost hyperventilated when I saw my quiz grade. I was excited, I'm sorry! =P Anyway, Rolle's Theorem.
Let's look at the definition
Rolle's Theorem gives the conditions that guarantee the existence of an extrema in the interior of a closed interval.
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 1 number c in (a,b) such that f'(c)=0
Okay, what that basically says is
that if the function is continuous on the interval and it's differentiable on the interval, when you plug in the endpoints of your interval (or x intercepts) and they equal each other, when solved for x in the first derivative, there will be numbers inside your interval.
I have no idea if that made sense to you or not...here's an example problem:
f(x)=x^4-2x^2 on [-2,2] find all values c for which f'(x)=0
1) Is it continuous?
Yes
2) Is it differentiable?
Yes
3) Plug in your first end point.
f(-2)=(-2)^4-2(-2)^2 = 8
4) Plug in your second end point
f(2)=(2)^4-2(2)^2 = 8
5) Are they equal?
f(2)=f(-2)=8
Yes
6) Take the first derivative
f'(x)=4x^3-4x
7) Solve for x
x= -1, 0, 1
8) Which points are on your interval?
Yes so c=-1, 0, 1
9) Justify
Rolle's Theorem is applied to this function because it is continuous and differentiable and f(-2)=f(2)=8. When Rolle's Theorem is applied, the first derivative is taken and c is found to equal -1, 0, and 1.
Now, can someone please help me with optimization? I have NO clue what I'm doing at all for this. I don't even know how to start it! =0
Let's look at the definition
Rolle's Theorem gives the conditions that guarantee the existence of an extrema in the interior of a closed interval.
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 1 number c in (a,b) such that f'(c)=0
Okay, what that basically says is
that if the function is continuous on the interval and it's differentiable on the interval, when you plug in the endpoints of your interval (or x intercepts) and they equal each other, when solved for x in the first derivative, there will be numbers inside your interval.
I have no idea if that made sense to you or not...here's an example problem:
f(x)=x^4-2x^2 on [-2,2] find all values c for which f'(x)=0
1) Is it continuous?
Yes
2) Is it differentiable?
Yes
3) Plug in your first end point.
f(-2)=(-2)^4-2(-2)^2 = 8
4) Plug in your second end point
f(2)=(2)^4-2(2)^2 = 8
5) Are they equal?
f(2)=f(-2)=8
Yes
6) Take the first derivative
f'(x)=4x^3-4x
7) Solve for x
x= -1, 0, 1
8) Which points are on your interval?
Yes so c=-1, 0, 1
9) Justify
Rolle's Theorem is applied to this function because it is continuous and differentiable and f(-2)=f(2)=8. When Rolle's Theorem is applied, the first derivative is taken and c is found to equal -1, 0, and 1.
Now, can someone please help me with optimization? I have NO clue what I'm doing at all for this. I don't even know how to start it! =0
Week Six
This week was very hectic and we learned a lot. We had a test on monday, which didn't really go well for me. On Tuesday we learned about the Extreme Value Theorem, the Rolle's Theorem, and the Mean Value Theorem. On Wednesday, we learned about Optimization. On Thursday and Friday, we did example problems with Optimization.
Things I now understand:
1.) Everything I got wrong on my test, thanks to doing corrections.
2.) Extreme Value Theorem
3.) Rolle's Theorem
F(x) must be continuous on a closed intervale. It must be differentiable on the open interval. And f(a) must equal f(b). If all this is true, then there is at least one number 'c' within (a,b) such that f'(c) = 0.
Example:
You are given: f(x) = x^4 -2x^2 on [-2,2]. Find all values c such that f '(c) = 0.
Since it is continous (polynomial) and differentiable (no corners/discontinuities), you now check that f(a) = f(b).
f(-2) = f(2) = 8
Since all this works, you take the derivative of f(x).
F'(x) = 4x^3 -4x
You then set f '(x) = 0
You get that x = 0, 1, -1
Since all these are within the interval:
c= -1, 0, 1
4.) Mean Value Theorem
Same as Rolle's such that: must be continous and differentiable.
Different because: f '(c) = [f(b) - f(a)] / (b-a).
Something I do not understand is Optimization.
I understand most of the steps, but usually get stuck when trying to find the primary and secondary equations and which is one is which.
And example would be Example 5 on the handout Ms. Robinson gave us.
I am still kicking myself in the butt over not studying hard enough for the test on Monday, but I guess I know now.
Things I now understand:
1.) Everything I got wrong on my test, thanks to doing corrections.
2.) Extreme Value Theorem
3.) Rolle's Theorem
F(x) must be continuous on a closed intervale. It must be differentiable on the open interval. And f(a) must equal f(b). If all this is true, then there is at least one number 'c' within (a,b) such that f'(c) = 0.
Example:
You are given: f(x) = x^4 -2x^2 on [-2,2]. Find all values c such that f '(c) = 0.
Since it is continous (polynomial) and differentiable (no corners/discontinuities), you now check that f(a) = f(b).
f(-2) = f(2) = 8
Since all this works, you take the derivative of f(x).
F'(x) = 4x^3 -4x
You then set f '(x) = 0
You get that x = 0, 1, -1
Since all these are within the interval:
c= -1, 0, 1
4.) Mean Value Theorem
Same as Rolle's such that: must be continous and differentiable.
Different because: f '(c) = [f(b) - f(a)] / (b-a).
Something I do not understand is Optimization.
I understand most of the steps, but usually get stuck when trying to find the primary and secondary equations and which is one is which.
And example would be Example 5 on the handout Ms. Robinson gave us.
I am still kicking myself in the butt over not studying hard enough for the test on Monday, but I guess I know now.
WEEK SIXX!
week six was definately stressful.
.. in the beginning of the week we learned the different theorems.
EVT: a continuous function on a closed interval [a,b] must have both a mnimmum and a maximum on the interval
Rolle's: Let F be continuous on the closed interval [a,b] and differentiable on the open interval (a,b). If f(a)=f(b), then there is at least one number "c" in (a,b) such that f(c)=0.
MVT: If F is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number "c" in (a,b) such that f '(c)=f(b)-f(a)/b-a
and later in the week we recieved packets on optimization..
OPTIMIZATION:
A rancher wants to fence in an area of 3000000 square feet in a rectangular
field and then divide it in half with a fence down the middle parallel to one
side. What is the shortest length of fence that the rancher can use?
There is a top side side and a bottom side (that's two sides).
There are 3 sides the run vertically - one on either side and one in the middle.
Let's call the top and the bottom x.
Let's call the vertical ones y.
The total distance is 2x + 3y.
We know that the area, 3,000,000, is defined by xy.
This says that y = 3000000/x
The equation for length is then 2x + 9000000/x.
There derivative of this function is 2 - 9000000/x², which we set to 0 and solve for x.
That gives us x² = 9000000/2 = 4,500,000.
The answer for x would be √(4.5) thousand.
The answer for y would be 3M/x.
.. i can easly explan both optimization and the theorems, but when it comes to actually applying them i get stuck. i think i have more difficulty with optimization because i get lost in the steps and forget what exactly the questions asking me to find, the only thing i can identify and the primary and secondary functions.. after that i quit. anyone have any ideas on how to make sense of what i'm supposed to do after that?
.. in the beginning of the week we learned the different theorems.
EVT: a continuous function on a closed interval [a,b] must have both a mnimmum and a maximum on the interval
Rolle's: Let F be continuous on the closed interval [a,b] and differentiable on the open interval (a,b). If f(a)=f(b), then there is at least one number "c" in (a,b) such that f(c)=0.
MVT: If F is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number "c" in (a,b) such that f '(c)=f(b)-f(a)/b-a
and later in the week we recieved packets on optimization..
OPTIMIZATION:
A rancher wants to fence in an area of 3000000 square feet in a rectangular
field and then divide it in half with a fence down the middle parallel to one
side. What is the shortest length of fence that the rancher can use?
There is a top side side and a bottom side (that's two sides).
There are 3 sides the run vertically - one on either side and one in the middle.
Let's call the top and the bottom x.
Let's call the vertical ones y.
The total distance is 2x + 3y.
We know that the area, 3,000,000, is defined by xy.
This says that y = 3000000/x
The equation for length is then 2x + 9000000/x.
There derivative of this function is 2 - 9000000/x², which we set to 0 and solve for x.
That gives us x² = 9000000/2 = 4,500,000.
The answer for x would be √(4.5) thousand.
The answer for y would be 3M/x.
.. i can easly explan both optimization and the theorems, but when it comes to actually applying them i get stuck. i think i have more difficulty with optimization because i get lost in the steps and forget what exactly the questions asking me to find, the only thing i can identify and the primary and secondary functions.. after that i quit. anyone have any ideas on how to make sense of what i'm supposed to do after that?
post number 6
So let’s go through all the days of the week. Monday we had a test from the previous week. Tuesday we learned three theorems. Wednesday we learned Optimization. Thursday we had a quiz on what we learned Tuesday. Friday we continued with optimization.
WHAT I UNDERSTOOD THIS WEEK:
The 3 theorems that we learned on Tuesday were Mean Value Theorem, Rolle’s Theorem, and Extreme Value Theorem.
MEAN VALUE THEOREM-
Must be continuous and differentiable on closed interval [a,b]. Then there must be a number ‘C’ on the interval such that f’(c)=f(b)-f(a)/b-a, or in simpler terms, take the derivative of ‘C’ and set it equal to the slope of [a,b].
ROLLE’S THEOREM-
Must be continuous and differentiable on closed interval [a,b]. If f(a)=f(b), then there must be a number ‘C’ on [a,b] where f(c)=0.
EXTREME VALUE THEOREM-
Must be continuous on closed interval [a,b] and must contain a max and min.
I kind of put these out of order, but the EVT brings you to the Rolle’s Theorem. The MVT is just by itself.
Optimization is stretching out an equation.
We were given some steps to follow in our packets that she handed out with the notes for optimization, but for some reason I just don’t understand it at all.
STEPS-
1. Identify all quantities
2. Write an equation
3. Reduce equation
4. Determine domain of equation
5. Determine max/min values
I might not understand just because I didn’t have much practice, but I really don’t know. There’s a test on Tuesday and I just need some help, how to even start the problem all the way through to the end.
WHAT I UNDERSTOOD THIS WEEK:
The 3 theorems that we learned on Tuesday were Mean Value Theorem, Rolle’s Theorem, and Extreme Value Theorem.
MEAN VALUE THEOREM-
Must be continuous and differentiable on closed interval [a,b]. Then there must be a number ‘C’ on the interval such that f’(c)=f(b)-f(a)/b-a, or in simpler terms, take the derivative of ‘C’ and set it equal to the slope of [a,b].
ROLLE’S THEOREM-
Must be continuous and differentiable on closed interval [a,b]. If f(a)=f(b), then there must be a number ‘C’ on [a,b] where f(c)=0.
EXTREME VALUE THEOREM-
Must be continuous on closed interval [a,b] and must contain a max and min.
I kind of put these out of order, but the EVT brings you to the Rolle’s Theorem. The MVT is just by itself.
WHAT I DIDN’T UNDERSTAND THIS WEEK:
We were given some steps to follow in our packets that she handed out with the notes for optimization, but for some reason I just don’t understand it at all.
STEPS-
1. Identify all quantities
2. Write an equation
3. Reduce equation
4. Determine domain of equation
5. Determine max/min values
I might not understand just because I didn’t have much practice, but I really don’t know. There’s a test on Tuesday and I just need some help, how to even start the problem all the way through to the end.
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