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.
post 6
To start off the week we had a test on the first derivative test and second derivative test. The rest of the week consisted of doing some different things that I understand fully for once. One is the Rolles Theorem. And the other is the Mean Value Theorem. They are similar in a way that the equation has to be both continuous on a closed interval differentiable on an open interval in order to use both of these.
Rolle's
In Rolle's f(a)=f(b) or it can not be done. And there is always at least one answer which is "c". But mainly the way to do Rolle's is to start off making sure the equation is continuous and differentiable. Then make sure f(a)=f(b). Next take the derivative and solve for x. And finally pick the answer that falls between the interval given.
Mean Value
First you find the slope and plug into f(b)-f(a)/b-a. Then take the derivative and set it equal to the slope and solve for x. Finally take the answer between the interval given in the problem.
But I am not clear at all on optimization. Once I get started I can get most of it but it is getting started that is killing me. So if there is any way someone can help me it would be appreciated. And also I had no idea how to even think about starting the last problem on the first derivative and second derivative tests test if someone can help me for future reference.
Rolle's
In Rolle's f(a)=f(b) or it can not be done. And there is always at least one answer which is "c". But mainly the way to do Rolle's is to start off making sure the equation is continuous and differentiable. Then make sure f(a)=f(b). Next take the derivative and solve for x. And finally pick the answer that falls between the interval given.
Mean Value
First you find the slope and plug into f(b)-f(a)/b-a. Then take the derivative and set it equal to the slope and solve for x. Finally take the answer between the interval given in the problem.
But I am not clear at all on optimization. Once I get started I can get most of it but it is getting started that is killing me. So if there is any way someone can help me it would be appreciated. And also I had no idea how to even think about starting the last problem on the first derivative and second derivative tests test if someone can help me for future reference.
Post 6
So this has to be the earliest I've ever posted a blog.
At the beginning of this week we had our second calculus test. The test focused on the first derivative test, second derivative test, and how to find absolute maximums and minimums. Tuesday, we learned Extreme Value Theorem, Rolle's theorem, and Mean Value Theorem. Wednesday we started Optimization, Thursday we had a quiz on the theorems and at the end of the class through Friday we continued to learn and work problems using Optimization.
EVT:
a continuous function on a closed interval [a,b] must have both a mnimmum and a maximum on the interval; however, the maximum and minimum can occur at the endpoints.
Rolle's:
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 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
Optimization is either streaching out an equation or simplifying it
The steps for optimization are:
1. Identify all given quantities and quantities to be determined. If possible, make a sketch.
2. Write a primary equation for the quantity that is to be maximized or minimized.
3. Reduce the primary equation to one having a single independent variable. This may involve the use of secondary equations relating the independent variables of the primary equation.
4. Determine the feasible domain of the primary equation. That is, determine the values for which the stated problem makes sense.
5. Determine thedesired maximum or minimum value
I understand that I have to follow these steps, but I don't know when I'm finished a step. I get confused with when to use a primary or secondary equation and how to get these equations. Can someone please explain?
At the beginning of this week we had our second calculus test. The test focused on the first derivative test, second derivative test, and how to find absolute maximums and minimums. Tuesday, we learned Extreme Value Theorem, Rolle's theorem, and Mean Value Theorem. Wednesday we started Optimization, Thursday we had a quiz on the theorems and at the end of the class through Friday we continued to learn and work problems using Optimization.
EVT:
a continuous function on a closed interval [a,b] must have both a mnimmum and a maximum on the interval; however, the maximum and minimum can occur at the endpoints.
Rolle's:
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 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
Optimization is either streaching out an equation or simplifying it
The steps for optimization are:
1. Identify all given quantities and quantities to be determined. If possible, make a sketch.
2. Write a primary equation for the quantity that is to be maximized or minimized.
3. Reduce the primary equation to one having a single independent variable. This may involve the use of secondary equations relating the independent variables of the primary equation.
4. Determine the feasible domain of the primary equation. That is, determine the values for which the stated problem makes sense.
5. Determine thedesired maximum or minimum value
I understand that I have to follow these steps, but I don't know when I'm finished a step. I get confused with when to use a primary or secondary equation and how to get these equations. Can someone please explain?
Post Number Six
This week in Calculus was pretty overwhelming to me, since i started off the week horribly. We won't go there though. Tuesday we learned a few new different concepts, such as Rolle's Theorem and Mean Value Theorem. Then we learned about optimization. Before understanding Rolles you must first know the extreme value theorem.
The Extreme Value theorem simply states that there must contain both a max and min on a closed interval of a continuous function. That then brings you to Rolle's:
The Extreme Value theorem simply states that there must contain both a max and min on a closed interval of a continuous function. That then brings you to Rolle's:
- 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 that f'(c)=0.
- Remember, if you are not given an interval assume the x intercepts.
An example problem of Rolle's is f(x)= x^4 - 2x^2 on [-2, 2] find all values c for which f'(c)=0.
First you must check if the function is continuous and differentiable: yes and yes
Now check if f(-2) = f(2)
f(-2)=(-2)^4 - 2(-2)^2=8
f(2)= (2)^4 - 2(2)^2=8
f(-2) and f(2) both equal 8, so you can move on to the next step.
Take the derivative and set it equal to zero, then solve for x.
f'=4x^3 - 4x=0
4x(x^2 - 1) = 0
x=0, 1, -1
All of these numbers are in the interval [-2,2], so your answer is c=-1, 0, 1.
I caught on to Rolle's theorem and Mean Value Theorem pretty quickly and was excited that i finally understood something. Until of course we learned optimization. I understood optimization in class when we were working the problems together but i can't seem to do them by myself. I don't know what to do first to the problem and how to find the primary and secondary equation. If anyone can help me with optimization i'd really appreciate it, or once again i'll receive another grade to not be spoken about on the quiz Tuesday.
Let's hope for a good seventh week of calculus :)
Optomization
Several people are asking if there are steps to optimization. We went over them on the first day we did optimization. I even copied them into your packets so you wouldn't have to copy them down. Refer to the beginning of your packet and you will find the steps.
post 6
i understood all the beginning of the week but when we got to optimization i was a little confused. i understand Rolle's Theorem which gives the conditions that guarantee the existence of an extema in the interior of a closed interval. and i understand mean value theorem which just states that a continuous function on a closed interval must have both a max or min on the intervavl.You find out if it is continuous and differentiable between the interval. If not, you cannot use this theorem. If no interval just use the x-intercepts. plug in both interval points. These must equal each or you cannot use this theorem. Then take the derivative, set it equal to 0 and solve for x, and you get your critical values. (make sure the points fall between the interval)
I understand Mean Value Theorem which just says 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^1(c)=f(b)-f(a)/b-a.
All you do is make sure it is continuous and differentiable between the interval. (if not you cannot use this theorem) Then find the slope using f^1(c)=f(b)-f(a)/b-a. Then take the derivative and set it equal to the slope and solve for x. This gives the critical points. (make sure the points fall between the interval)
im not too sure about optimization and some of the stuff that goes w/ it. if anyone wants to help w/ that holla at me :)
I understand Mean Value Theorem which just says 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^1(c)=f(b)-f(a)/b-a.
All you do is make sure it is continuous and differentiable between the interval. (if not you cannot use this theorem) Then find the slope using f^1(c)=f(b)-f(a)/b-a. Then take the derivative and set it equal to the slope and solve for x. This gives the critical points. (make sure the points fall between the interval)
im not too sure about optimization and some of the stuff that goes w/ it. if anyone wants to help w/ that holla at me :)
Post #6
Alrighty, the beginning of this week was great because I could actually do the homework, but then we got to optimization.
I really understand Rolle's Theorem, but first you have to know the Extreme Value Theorem which just states that a continuous function on a closed interval must have both a max or min on the interval. (They can occur at the endpoints.) Anyway, Rolle's Theorem gives the conditions that guarantee the existence of an extema in the interior of a closed interval.
You basically see if it continuous and differentiable between the interval. If not, you cannot use this theorem. (If no interval just use x-intercepts.) Then plug in both interval points. These must equal each or you cannot use this theorem. Then take the derivative, set it equal to 0 and solve for x, and you get your critical values. (make sure the points fall between the interval)
Example: f(x)=x^4-2x^2 on [-2,2] Find all values c for which f^1(c)=0
Is it continuous and 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
Also, I understand Mean Value Theorem which just says 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^1(c)=f(b)-f(a)/b-a.
All you do is make sure it is continuous and differentiable between the interval. (if not you cannot use this theorem) Then find the slope using f^1(c)=f(b)-f(a)/b-a. Then take the derivative and set it equal to the slope and solve for x. This gives the critical points. (make sure the points fall between the interval)
Example: f(x)=5-4/x Find all values c on (1,4) such that f^1(c)=f(b)-f(a)/b-a.
Is it continuous? there is an infinite @ x=0 (which is not in the interval so you can continue with this theorem) Is it differentiable? not @ x=0 (which is also not on the interval)
f(4)=5-4/4=4 f(1)=5-4/1= 1 4-1/4-1 = 3/3=1
f^1(x)=4/x^2
4/x^2=1 x= + or - 2 (-2 is not on the interval) c=2
The only real problem is I'm having trouble with optimization. Is there any steps to follow? I just get confused on what equation to use and what to use from the problem. Each problem is sort of different which I guess is what is making me have trouble. And does anyone know when our next quiz is?
I really understand Rolle's Theorem, but first you have to know the Extreme Value Theorem which just states that a continuous function on a closed interval must have both a max or min on the interval. (They can occur at the endpoints.) Anyway, Rolle's Theorem gives the conditions that guarantee the existence of an extema in the interior of a closed interval.
You basically see if it continuous and differentiable between the interval. If not, you cannot use this theorem. (If no interval just use x-intercepts.) Then plug in both interval points. These must equal each or you cannot use this theorem. Then take the derivative, set it equal to 0 and solve for x, and you get your critical values. (make sure the points fall between the interval)
Example: f(x)=x^4-2x^2 on [-2,2] Find all values c for which f^1(c)=0
Is it continuous and 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
Also, I understand Mean Value Theorem which just says 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^1(c)=f(b)-f(a)/b-a.
All you do is make sure it is continuous and differentiable between the interval. (if not you cannot use this theorem) Then find the slope using f^1(c)=f(b)-f(a)/b-a. Then take the derivative and set it equal to the slope and solve for x. This gives the critical points. (make sure the points fall between the interval)
Example: f(x)=5-4/x Find all values c on (1,4) such that f^1(c)=f(b)-f(a)/b-a.
Is it continuous? there is an infinite @ x=0 (which is not in the interval so you can continue with this theorem) Is it differentiable? not @ x=0 (which is also not on the interval)
f(4)=5-4/4=4 f(1)=5-4/1= 1 4-1/4-1 = 3/3=1
f^1(x)=4/x^2
4/x^2=1 x= + or - 2 (-2 is not on the interval) c=2
The only real problem is I'm having trouble with optimization. Is there any steps to follow? I just get confused on what equation to use and what to use from the problem. Each problem is sort of different which I guess is what is making me have trouble. And does anyone know when our next quiz is?
Week #6
So this week in Calculus we learned three new concepts: Rolle's Theorem, Mean Value Theorem, and Optimization.
The first two are just theorems describing maxs and mins...
Rolle's Theorem: states that a differentiable and continuous function, which attains equal values at two points, must have a point somewhere between them where the slope of the tangent line to the graph of the function is zero.
So, to use this...you are given an interval...let's say [-4,4], and a function...let's say f(x)=x^2. To use this, you have to check that first it is differentiable...which it is. Second, it's continuous...which it is. Next, you do f(-4) and f(4). If those two values are equal, which for this example it is, then you can say that rolle's theorem applies. You then take the derivative, which is 2x in our case, and set it equal to 0. We find that @ x=0 there is a min or a max, which IS between -4 and 4. That's Rolle's Theorem.
The next theorem is Mean Value Theorem: states, that given a section of a curve, there is at least one point on that section at which the derivative of the curve is equal (parallel) to the "average" derivative of the section.
To do this, you have to check that it is differentiable and continous. Then you plug into (f(b)-f(a))/(b-a). Set the derivative of the function equal to whatever that comes out to be. Solve for x.
The other thing we learned is Optimization which...at first was very hard to grasp but now that I did a few problems, extremely easy.
Things to do for Optimization that might help you:
1) I like to circle the equation I'm trying to maximize or minimize...this just reminds me of what I'm working with plugging back into at the end of the problem.
2) Next, identify if you have more than one variable in the equation...if so, you need to isolate it to one.
3) If you need to, find a secondary equation and solve it in terms of the variable you want to isolate your primary equation with....this is a LOT like substitution...actually, that's all it is.
4) After getting the equation in terms of one variable, simplify and then take the derivative. Set the derivative = to 0 and solve. This is taking the max's and min's of the function...
5) Look back at your circled equation...did you want to minimize or maximize it? Plug in the values you found when setting the derivative = to 0 into this circled equation...if you are looking for minimums, take the smallest value. If you are looking for maximums, you take the largest value.
Optimization is pretty cool. I finally see the actual application of Calculus in the real world. :-) (Well, one of them.)
-John
The first two are just theorems describing maxs and mins...
Rolle's Theorem: states that a differentiable and continuous function, which attains equal values at two points, must have a point somewhere between them where the slope of the tangent line to the graph of the function is zero.
So, to use this...you are given an interval...let's say [-4,4], and a function...let's say f(x)=x^2. To use this, you have to check that first it is differentiable...which it is. Second, it's continuous...which it is. Next, you do f(-4) and f(4). If those two values are equal, which for this example it is, then you can say that rolle's theorem applies. You then take the derivative, which is 2x in our case, and set it equal to 0. We find that @ x=0 there is a min or a max, which IS between -4 and 4. That's Rolle's Theorem.
The next theorem is Mean Value Theorem: states, that given a section of a curve, there is at least one point on that section at which the derivative of the curve is equal (parallel) to the "average" derivative of the section.
To do this, you have to check that it is differentiable and continous. Then you plug into (f(b)-f(a))/(b-a). Set the derivative of the function equal to whatever that comes out to be. Solve for x.
The other thing we learned is Optimization which...at first was very hard to grasp but now that I did a few problems, extremely easy.
Things to do for Optimization that might help you:
1) I like to circle the equation I'm trying to maximize or minimize...this just reminds me of what I'm working with plugging back into at the end of the problem.
2) Next, identify if you have more than one variable in the equation...if so, you need to isolate it to one.
3) If you need to, find a secondary equation and solve it in terms of the variable you want to isolate your primary equation with....this is a LOT like substitution...actually, that's all it is.
4) After getting the equation in terms of one variable, simplify and then take the derivative. Set the derivative = to 0 and solve. This is taking the max's and min's of the function...
5) Look back at your circled equation...did you want to minimize or maximize it? Plug in the values you found when setting the derivative = to 0 into this circled equation...if you are looking for minimums, take the smallest value. If you are looking for maximums, you take the largest value.
Optimization is pretty cool. I finally see the actual application of Calculus in the real world. :-) (Well, one of them.)
-John
Post #6
This week in Calculus, the thing I understand most is Rolle's Theorem. Later on in the week we learned optimization. This is the thing I do not understand..while we are working it, I can follow along, but i cannot do it by myself, because I don't understand the steps..if there are any?
The first thing we learned this weed was Rolle's theorem. It states: "Let f be continuous on a 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).
To put this in simpler terms it means there must be a point, c, where the instantaneous rate of change equals the average rate of change. Rolles Theorem gives conditions that promise the existance of an extrema in the interior of the closed interval.
An example of this would be:
f(x) = x^2 - 3x + 2 Show that f'(x) = 0 on some interval.
The first thing you should know, is if they do not give you an interval, you should ALWAYS assume the x-interval.
First, to find the interval, solve for x :
x^2 -3x + 2 = 0
(x-2)(x-1)
x-interval --> [1,2]
Then, begin to check and see if Rolle's theorem can be used. The standards are that the function must be continuous and differentiable (on the interval given or found), and the y-values MUST be equal.
Plug in the numbers in the x-interval to find out if the y-values are equal.
f(1)=f(2)=0
This part of the test is successful.
Next, you must check if the function is continuous and differentiable. It is, so then you can proceed by taking the derivative and setting it equal to zero.
f'(x) = 2x-3 = 0
2x = 3
x = 3/2
c = 3/2
After checking to make sure that the c-value is in between the values in the closed interval, you justify...of course.
Using Rolle's Theorem, because the function is continuous and differentiable, and the y-values are equal, when the derivative was taken and set equal to zero, i found that c is equal to 3/2.
Optimization is the only thing I do not understand..if anyone can explain to me the steps, it would probably save me from failure on the quiz tuesday..thankss.
The first thing we learned this weed was Rolle's theorem. It states: "Let f be continuous on a 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).
To put this in simpler terms it means there must be a point, c, where the instantaneous rate of change equals the average rate of change. Rolles Theorem gives conditions that promise the existance of an extrema in the interior of the closed interval.
An example of this would be:
f(x) = x^2 - 3x + 2 Show that f'(x) = 0 on some interval.
The first thing you should know, is if they do not give you an interval, you should ALWAYS assume the x-interval.
First, to find the interval, solve for x :
x^2 -3x + 2 = 0
(x-2)(x-1)
x-interval --> [1,2]
Then, begin to check and see if Rolle's theorem can be used. The standards are that the function must be continuous and differentiable (on the interval given or found), and the y-values MUST be equal.
Plug in the numbers in the x-interval to find out if the y-values are equal.
f(1)=f(2)=0
This part of the test is successful.
Next, you must check if the function is continuous and differentiable. It is, so then you can proceed by taking the derivative and setting it equal to zero.
f'(x) = 2x-3 = 0
2x = 3
x = 3/2
c = 3/2
After checking to make sure that the c-value is in between the values in the closed interval, you justify...of course.
Using Rolle's Theorem, because the function is continuous and differentiable, and the y-values are equal, when the derivative was taken and set equal to zero, i found that c is equal to 3/2.
Optimization is the only thing I do not understand..if anyone can explain to me the steps, it would probably save me from failure on the quiz tuesday..thankss.
Post #6
The sixth week of calculus began with a test on chapter three. Unfortunately all of my hours of work didn't help me out so well. Tuesday we started learning something new.
First was Rolle's Theorem:
In order to be able to do Rolle's Theorem you have to have a continuous equation and it has to be differentual. When you plug in the points that are given to you, it has to equal to the same number that way f(a)=f(b)=0 [or any other number]
If all of this is passed, then you can start using the theorem:
find the first derivative, set it equal to zero and solve. The number[s] you get for x are the possible answers. If any of the numbers are not in range of the points given then those numbers cannot be your answer. The numbers that are in the range are your answer so you say c = ____ <--those numbers.
FOR EXAMPLE:
f(x)=x^2-2x [0,2]
it is continuous and differentiable.
plug in: f(0)=0
f(2)=0
take the first derivative: 2x-2=0
now solve: x= -1
since -1 isn't between 0 and 2 it is not an answer. Therefore there is no answers!
Then we learned Mean Value Theorem:
In order to be able to do Mean Value Theorem you have to have a continuous equation and it has to be differentual. When you have the points in the problem [a,b] you plug those to the equation. When you solve and find numbers you take [the "b" answer] - [the "a" answer] at the top of the equation, and from the original points given you take [b - a] for the bottom of the equation. When you solve and get a number that's the number that you set the first derivative to. [Take the derivative of the equation given and then set equal to that number] The number[s] you get for the x are the possible answers. If any of the numbers are not in range of the points given then those numbers cannot be you anwer. The numbers that are in the range are your answer so you say c = ____ <--those numbers.
FOR EXAMPLE:
f(x)=x^2 [-2,1]
it is continuous and differentiable.
plug in: f(-2)=4
f(1) =1
so you plug into the equation...1-4 = -3
----- ----
1+2 = 1
since that equals to -3 when you take the first derivative [2x] you will set it equal to -3.
giving you x= -3/2 [which is between [-2,1] so c = -3/2
Next we learned about Optomization. To be honest, i still don't get the concept of it so if someone would be so kind and explain that to me .. it would be greatly appreciated!
First was Rolle's Theorem:
In order to be able to do Rolle's Theorem you have to have a continuous equation and it has to be differentual. When you plug in the points that are given to you, it has to equal to the same number that way f(a)=f(b)=0 [or any other number]
If all of this is passed, then you can start using the theorem:
find the first derivative, set it equal to zero and solve. The number[s] you get for x are the possible answers. If any of the numbers are not in range of the points given then those numbers cannot be your answer. The numbers that are in the range are your answer so you say c = ____ <--those numbers.
FOR EXAMPLE:
f(x)=x^2-2x [0,2]
it is continuous and differentiable.
plug in: f(0)=0
f(2)=0
take the first derivative: 2x-2=0
now solve: x= -1
since -1 isn't between 0 and 2 it is not an answer. Therefore there is no answers!
Then we learned Mean Value Theorem:
In order to be able to do Mean Value Theorem you have to have a continuous equation and it has to be differentual. When you have the points in the problem [a,b] you plug those to the equation. When you solve and find numbers you take [the "b" answer] - [the "a" answer] at the top of the equation, and from the original points given you take [b - a] for the bottom of the equation. When you solve and get a number that's the number that you set the first derivative to. [Take the derivative of the equation given and then set equal to that number] The number[s] you get for the x are the possible answers. If any of the numbers are not in range of the points given then those numbers cannot be you anwer. The numbers that are in the range are your answer so you say c = ____ <--those numbers.
FOR EXAMPLE:
f(x)=x^2 [-2,1]
it is continuous and differentiable.
plug in: f(-2)=4
f(1) =1
so you plug into the equation...1-4 = -3
----- ----
1+2 = 1
since that equals to -3 when you take the first derivative [2x] you will set it equal to -3.
giving you x= -3/2 [which is between [-2,1] so c = -3/2
Next we learned about Optomization. To be honest, i still don't get the concept of it so if someone would be so kind and explain that to me .. it would be greatly appreciated!
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