alright, this week we studied for our test, and we did our packets; however, last week I didn't know how to do optimization, so I am going to explain that and other stuff that I know.
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).
Steps for optimization
1.Identify primary and secondary equations. The primary will be the one you are maximizing or minimizing, and the secondary will be the other one.
2.Solve secondary equation for one variable, and plug into the primary. (if the primary only has one variable, this step is not necessary.)
3.Take the derivative of the primary, and set it equal to zero; solve for x.
4.Plug into secondary equation to find the other value, check end points if necessary.
Something I don't understand is the tangent line stuff, I do not even know how to start a question that asks about that. And it's all over the packets we just got!
Sunday, October 4, 2009
Ash's 7th post
Wow...is it merely the 7th week of school? It feels as if I should be going into the Christmas Holidays in about 2 weeks......not in the middle of October.
Alright, this week....Studying and Study Guides. But I do remember Mrs. Robinson saying something about us being able to explain something from past weeks right? Good!
How about I get what I don't understand out of the way right now?
1. STILL!! Optimization
2. Tangent Lines [[I get the general gist of it, just...not all of it]]
First Derivative Test (I'm sorry if everyone understands it, but I don't want to look dumb by explaining something I don't get at all wrong)
The terms for the First Derivative Test:
1. Increasing
2. Decreasing
3. Horizontal Tangent
4. Min/Max
Steps:
1. Take the derivative
2. Set it equal to zero
3. Solve for x to get the possible critical points [[can someone clarify this part???? My notebook says it equals critical points AND mins/maxs/horizontal tangents]]
4. Set up intervals with your x value(s)
5. Plug into your first derivative
6. To find an absolute extrema, plug in the values from step 5 into your original function
Can someone check this for me? I don't know if I'm not in my right mind right now [no comments] or if I'm really failing at this? Or if I just do it and not think about it now?
f(x) = 1/2x - sinx
Find the extrema on the interval (0, 2pi)
f'(x) = 1/2 - cosx
1/2 - cosx = 0
-cosx = -1/2
cosx = 1/2
x=cos^-1(1/2)
x=pi/3 and 5pi/3
Because
| +
-------
| +
cos is positive on the first and forth quadrants
300 degrees in radians is 5pi/3
and
60 degrees in radians is pi/3
Also, for the powerpoint tomorrow:
1. If I downloaded fonts from the Internet, will they show up on Mrs. Robinson's computer?
2. Will someone be clicking the next slide button or does it have to be set on a timer?
Thanks!! =]
Alright, this week....Studying and Study Guides. But I do remember Mrs. Robinson saying something about us being able to explain something from past weeks right? Good!
How about I get what I don't understand out of the way right now?
1. STILL!! Optimization
2. Tangent Lines [[I get the general gist of it, just...not all of it]]
First Derivative Test (I'm sorry if everyone understands it, but I don't want to look dumb by explaining something I don't get at all wrong)
The terms for the First Derivative Test:
1. Increasing
2. Decreasing
3. Horizontal Tangent
4. Min/Max
Steps:
1. Take the derivative
2. Set it equal to zero
3. Solve for x to get the possible critical points [[can someone clarify this part???? My notebook says it equals critical points AND mins/maxs/horizontal tangents]]
4. Set up intervals with your x value(s)
5. Plug into your first derivative
6. To find an absolute extrema, plug in the values from step 5 into your original function
Can someone check this for me? I don't know if I'm not in my right mind right now [no comments] or if I'm really failing at this? Or if I just do it and not think about it now?
f(x) = 1/2x - sinx
Find the extrema on the interval (0, 2pi)
f'(x) = 1/2 - cosx
1/2 - cosx = 0
-cosx = -1/2
cosx = 1/2
x=cos^-1(1/2)
x=pi/3 and 5pi/3
Because
| +
-------
| +
cos is positive on the first and forth quadrants
300 degrees in radians is 5pi/3
and
60 degrees in radians is pi/3
Also, for the powerpoint tomorrow:
1. If I downloaded fonts from the Internet, will they show up on Mrs. Robinson's computer?
2. Will someone be clicking the next slide button or does it have to be set on a timer?
Thanks!! =]
Post 7
This week in Calculus we pretty much just reviewed and had a quiz. At the end of last week and the beginning of this week, I was still confused with optimization. By the time of the quiz, I finally caught on to optimization. This week we were also given studyguides to work on right up until our exam. They're pretty big, but we'll get through them.
So first I didn't understand optimization and where the equasions came from, or even how to decide which equation was to be maximized. I kept on confusing myself and going in circles with the problems. After we were given the list for optimization "in English" I understood it a lot better. It's really simple when you know what you're looking for
The steps for optimization are as follows:
1. Identify your primary and secondary equations. Primary will be the one the problem is asking you to minimize or maximize. I've also noticed that the secondary will usually be set equal to a number.
2. After finding secondary, solve it for one variable if there are two.
3. Once you have this variable, plug it into your primary equation for the variable you solved the secondary equation for
4. Take the derivative of the equation you just formulated and set it equal to zero (this zero will become one of your answers)
5. Once you come out with your zeros, plug them into your secondary equation and solve for the variable you have left (this will give you your second answer)
So far looking through the packets, I've only gotten to the limits. I don't understand how to look at a graph and find what the limit is. Can someone please explain?
So first I didn't understand optimization and where the equasions came from, or even how to decide which equation was to be maximized. I kept on confusing myself and going in circles with the problems. After we were given the list for optimization "in English" I understood it a lot better. It's really simple when you know what you're looking for
The steps for optimization are as follows:
1. Identify your primary and secondary equations. Primary will be the one the problem is asking you to minimize or maximize. I've also noticed that the secondary will usually be set equal to a number.
2. After finding secondary, solve it for one variable if there are two.
3. Once you have this variable, plug it into your primary equation for the variable you solved the secondary equation for
4. Take the derivative of the equation you just formulated and set it equal to zero (this zero will become one of your answers)
5. Once you come out with your zeros, plug them into your secondary equation and solve for the variable you have left (this will give you your second answer)
So far looking through the packets, I've only gotten to the limits. I don't understand how to look at a graph and find what the limit is. Can someone please explain?
post 7
most of the week that just passed was studying for the quiz on wednesday and on wednesday we got the study guides for the first nine weeks exam. the study guides include optimization, first and second der tests, and tangent lines.
A Tangent Line is a line which touches a curve at one and only one point. The slope-intercept formula for a line is y = mx + b,where m is the slope of the line and b is the y-intercept.
The point-slope formula for a line is y – y1 = m (x – x1).This formula uses a point on the line, denoted by (x1, y1),and the slope of the line, denoted by m, to calculate the slope-intercept formula for the line.
The first derivative is an equation for the slope of a tangent line to a curve at an indicated point.
to find max and mins you use first der test then plug the critical values into original functions to get y values then plug endpoints in to original function to get y values the highest y value is the absolute max and the lowest y value is the absolute min.
i know both theorems we learned a while ago. Mean value theorem and rolle's theorem.
im somewat confused on optimization and knowing which eqn is the secondary and which one is thee primary. other than that im good
A Tangent Line is a line which touches a curve at one and only one point. The slope-intercept formula for a line is y = mx + b,where m is the slope of the line and b is the y-intercept.
The point-slope formula for a line is y – y1 = m (x – x1).This formula uses a point on the line, denoted by (x1, y1),and the slope of the line, denoted by m, to calculate the slope-intercept formula for the line.
The first derivative is an equation for the slope of a tangent line to a curve at an indicated point.
to find max and mins you use first der test then plug the critical values into original functions to get y values then plug endpoints in to original function to get y values the highest y value is the absolute max and the lowest y value is the absolute min.
i know both theorems we learned a while ago. Mean value theorem and rolle's theorem.
im somewat confused on optimization and knowing which eqn is the secondary and which one is thee primary. other than that im good
this week was mostly going over optimization in preparation for the quiz on wednesday..friday we got our study guides for the exam.
Some of the stuff on the study guides are tangent lines and using the first/second derivative test.
A Tangent Line is a line which locally touchesa curve at one and only one point.• The slope-intercept formula for a line is y = mx + b,where m is the slope of the line and b is the y-intercept.
• The point-slope formula for a line is y – y1 = m (x – x1).This formula uses a point on the line, denoted by (x1, y1),and the slope of the line, denoted by m, tocalculate the slope-intercept formula for the line.
• The first derivative is an equation for the slope of a tangentline to a curve at an indicated point.The equation for the slope of the tangent line tof(x) = x2 is f '(x), the derivative of f(x).f(x) = x2f '(x) = 2x (1)Therefore, at x = 2, the slope of the tangent line is f '(2).f '(2) = 2(2)= 4
Now , you know the slope of the tangent line, which is 4.All that you need now is a point on the tangent line to beable to formulate the equation. To find that point, simply plugthe coordinate of the shared point into the original equation, this gives you (2,4)The only step left is to use the point (2, 4) and slope, 4,in the point-slope formula for a line. Therefore: Y-4=4(x-2)
TO FIND MAX AND MINS
1. first derivative test
2. plug the critical values into origional function to get y-values.
3. plug endpoints in to origional function to get y-values.
4. highest y-value is absolute max.
5. lowest y-value is absolute min.
-absolute maxs or mins or written as a point or simply as the y-value.
For example:find the absolute max or min of f(x)=3x^4-4x^ on [-1,2].f1(x)=12x^3-12x^2=012x^2(x-1)=0x=1,0(-1, 0)U(0,1)U(1,2)f1(-.5)=-ve f1(.5)=-ve f1(1.5)=+ve
min @x=13(1)^4-4(1)^3=-1 (1,-1)3(-1)^4-4(-1)^3=-7 (-1,-7)3(2)^4-4(2)^3 (2,16)-32=16
abs min:(1,-1)or -1 -1 at x=1
i still don't fully understand optimization as a whole, anyone want to explain from the beginning?
Some of the stuff on the study guides are tangent lines and using the first/second derivative test.
A Tangent Line is a line which locally touchesa curve at one and only one point.• The slope-intercept formula for a line is y = mx + b,where m is the slope of the line and b is the y-intercept.
• The point-slope formula for a line is y – y1 = m (x – x1).This formula uses a point on the line, denoted by (x1, y1),and the slope of the line, denoted by m, tocalculate the slope-intercept formula for the line.
• The first derivative is an equation for the slope of a tangentline to a curve at an indicated point.The equation for the slope of the tangent line tof(x) = x2 is f '(x), the derivative of f(x).f(x) = x2f '(x) = 2x (1)Therefore, at x = 2, the slope of the tangent line is f '(2).f '(2) = 2(2)= 4
Now , you know the slope of the tangent line, which is 4.All that you need now is a point on the tangent line to beable to formulate the equation. To find that point, simply plugthe coordinate of the shared point into the original equation, this gives you (2,4)The only step left is to use the point (2, 4) and slope, 4,in the point-slope formula for a line. Therefore: Y-4=4(x-2)
TO FIND MAX AND MINS
1. first derivative test
2. plug the critical values into origional function to get y-values.
3. plug endpoints in to origional function to get y-values.
4. highest y-value is absolute max.
5. lowest y-value is absolute min.
-absolute maxs or mins or written as a point or simply as the y-value.
For example:find the absolute max or min of f(x)=3x^4-4x^ on [-1,2].f1(x)=12x^3-12x^2=012x^2(x-1)=0x=1,0(-1, 0)U(0,1)U(1,2)f1(-.5)=-ve f1(.5)=-ve f1(1.5)=+ve
min @x=13(1)^4-4(1)^3=-1 (1,-1)3(-1)^4-4(-1)^3=-7 (-1,-7)3(2)^4-4(2)^3 (2,16)-32=16
abs min:(1,-1)or -1 -1 at x=1
i still don't fully understand optimization as a whole, anyone want to explain from the beginning?
Week 7
Wow, I cannot believe we are already in the seventh week of school. It seems like just yesterday we were walking into the first day of class scared of how hard calculus may be. Anyways, this week we reviewed optimization on Monday and Tuesday. On Wednesday we had a quiz on optimization and recieved our review packets for exams. On Thursday we had a field trip, and on Friday we worked on our packets.
So far I have understood most of the things in our packets.
I understand how to take derivatives.
Example:
9xsin+4cox
you would use product and rule and the addition property to solve this problem.
= [9x(cosx) + sinx(9)] + 4(-sinx)
= 9xcosx + 5sinx
I also understand that slope is the same thing as derivative
I have recently discoved that a higher-order derivative is just that you keep taking derivatives until you are told to stop.
Example:
f ''(x) = 2x^(7/5), find f^(iv)(x)
first you would take the derivative of f ''(x) and get (14/5)x^(2/5). Then you would take the derivative of that to get (28/25)x^(-3/5).
Also, if it tells you to find a derivative a certain point, then take the derivative and plug the point in for x.
We copied down the Intermediate Value Theorem on Friday. I do not know how to solve a problem with it.
Example:
Find the value of c guaranteed by the Intermediate Value Theorem.
f(x) = x^2 - 2x - 3, [4,8], f(c) = 12.
Can anybody help me with that? Thanks.
So far I have understood most of the things in our packets.
I understand how to take derivatives.
Example:
9xsin+4cox
you would use product and rule and the addition property to solve this problem.
= [9x(cosx) + sinx(9)] + 4(-sinx)
= 9xcosx + 5sinx
I also understand that slope is the same thing as derivative
I have recently discoved that a higher-order derivative is just that you keep taking derivatives until you are told to stop.
Example:
f ''(x) = 2x^(7/5), find f^(iv)(x)
first you would take the derivative of f ''(x) and get (14/5)x^(2/5). Then you would take the derivative of that to get (28/25)x^(-3/5).
Also, if it tells you to find a derivative a certain point, then take the derivative and plug the point in for x.
We copied down the Intermediate Value Theorem on Friday. I do not know how to solve a problem with it.
Example:
Find the value of c guaranteed by the Intermediate Value Theorem.
f(x) = x^2 - 2x - 3, [4,8], f(c) = 12.
Can anybody help me with that? Thanks.
post 7
this week in calculus we reviewed and are getting ready for our exam by doing our beastfull study guide.
Things on the study guide that i understand include the first and second derivitive test and optimization.
First Derivative test-determine if it is continious and differentiable on the interval. If it is then you take the first derivitve and set equal to zero. Once set equal to zero solve for x and set your x into intervals. Then from there you plug in and find your max and mins.
Second derivative Test- determine if it is continious and defferentiable, if not then it doesnt work. take first derivative then take the second derivative and set the second equal to zero. Solve for x. Set up your x intervals and then plug in from within those intervals into the equation. Then you find your concavity, concave up and concave down.
Optimization - determine primary equation and then determine secondary and solve the secondary for an equation. Once solved plug it into the primary equation and then solve for that variable and then take the derivative and set equal to zero. Once set equal to zero you then plug back into the primary equatin and get your answers.
I still dont fully understand how to look at a graph and determine the derivative and the lim. The graph stuff messes me up.
Things on the study guide that i understand include the first and second derivitive test and optimization.
First Derivative test-determine if it is continious and differentiable on the interval. If it is then you take the first derivitve and set equal to zero. Once set equal to zero solve for x and set your x into intervals. Then from there you plug in and find your max and mins.
Second derivative Test- determine if it is continious and defferentiable, if not then it doesnt work. take first derivative then take the second derivative and set the second equal to zero. Solve for x. Set up your x intervals and then plug in from within those intervals into the equation. Then you find your concavity, concave up and concave down.
Optimization - determine primary equation and then determine secondary and solve the secondary for an equation. Once solved plug it into the primary equation and then solve for that variable and then take the derivative and set equal to zero. Once set equal to zero you then plug back into the primary equatin and get your answers.
I still dont fully understand how to look at a graph and determine the derivative and the lim. The graph stuff messes me up.
post #7
This week we didn't really learn anything new. On Monday and Tuesday we reviewed optimization. Then on Wednesday we took a quiz on it, then after the quiz we got our study guide for the exam which we will work on until the exam day. Thursday we went on a field trip, and Friday we worked on our packets.
lets go over the rules for finding a limit that approaches inifinty :)
1. if the top degree is equal to the bottom, divide leading coefficients.
2. if top degree is greater than bottom degree, it's infinity
3. if the top top degree is less than bottom degree, it's zero.
Taking derivatives are also all over the calc packet. like product rule, quotient rule, and all the other 36 rules we had to copy.
product rule:
f(x)= x^3cos(3x+4)
f'(x)=3x^2)(cos(3x+4)+(x^3)(-sin(3x+4)(3)
then simplify
one thing i still don't understand is optimization. if they don't give you any problems, how do you know what the primary and secondary equations are! im so confused. i also need help with tangent lines
lets go over the rules for finding a limit that approaches inifinty :)
1. if the top degree is equal to the bottom, divide leading coefficients.
2. if top degree is greater than bottom degree, it's infinity
3. if the top top degree is less than bottom degree, it's zero.
Taking derivatives are also all over the calc packet. like product rule, quotient rule, and all the other 36 rules we had to copy.
product rule:
f(x)= x^3cos(3x+4)
f'(x)=3x^2)(cos(3x+4)+(x^3)(-sin(3x+4)(3)
then simplify
one thing i still don't understand is optimization. if they don't give you any problems, how do you know what the primary and secondary equations are! im so confused. i also need help with tangent lines
Post 7
Another week down and first nine weeks exams are creeping up on us. This means it is study guide time.
In the study guide there are limits. The limit rules are: 1) if the highest exponent is the same on the top and bottom then the limit is the top coefficient over the bottom coefficient of the highest exponents. 2) If the highest exponent is on the top then the limit is infinity. 3) But if the highest exponent is on the bottom then the limit is 0.
Some examples are:
lim->infinity (3x^2-2x)/(2x^2+4x-1) Then the limit is 3/2 because it follows rule 1.
lim->infinity (8x^4+6)/(3x-9) Then the limit is infinity because it follows rule 2.
lim->infinity (x-7)/(5x^3+4x+2) Then the limit is 0 because it follows rule 3.
Also the study guide reviews taking derivivatives. That means using the product rule, quotient rule, and the regular way of taking the derivative of an equation.
Product rule is copy the first times the derivative of the last plus copy the first times the derivative of the second.
Quotient rule is copy the bottom times derivative of the top minus copy the top times the derivative of the bottom all divided by the bottom squared.
But there are some things I do not remember how to do or never did. I do not remember how to do the tangent line things. And I still do not fully understand optimization. Help with these 2 things would be great so that I will be ready for the exam.
In the study guide there are limits. The limit rules are: 1) if the highest exponent is the same on the top and bottom then the limit is the top coefficient over the bottom coefficient of the highest exponents. 2) If the highest exponent is on the top then the limit is infinity. 3) But if the highest exponent is on the bottom then the limit is 0.
Some examples are:
lim->infinity (3x^2-2x)/(2x^2+4x-1) Then the limit is 3/2 because it follows rule 1.
lim->infinity (8x^4+6)/(3x-9) Then the limit is infinity because it follows rule 2.
lim->infinity (x-7)/(5x^3+4x+2) Then the limit is 0 because it follows rule 3.
Also the study guide reviews taking derivivatives. That means using the product rule, quotient rule, and the regular way of taking the derivative of an equation.
Product rule is copy the first times the derivative of the last plus copy the first times the derivative of the second.
Quotient rule is copy the bottom times derivative of the top minus copy the top times the derivative of the bottom all divided by the bottom squared.
But there are some things I do not remember how to do or never did. I do not remember how to do the tangent line things. And I still do not fully understand optimization. Help with these 2 things would be great so that I will be ready for the exam.
Post #7
This week in Calculus we did optimization and reviewed. So, that’s what I’m going to do..explain things I now understand from the beginning of the year, and say I still don’t understand how you know what is the primary and secondary functions for optimization.
So, FIRST DERIVATIVE TEST: let c be a critical point of a function, f, that is continuous on an open interval containing c. If it is differentiable on the interval, except possibly at c, then f[c] can be classified as either:relative min [negative to positive]relative max [positive to negative]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
SECOND DERIVATIVE TEST: concavity & points of refection. 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.
Also, I understand the LIMIT RULES…Remember, you find vertical asymptotes when you set the bottom of a fraction equal to zero, then solve. After you factor the top and bottom of the fraction, if there is anything you can cancel from a function, it is a removable. Another thing i understand better is how to use the first and second derivative test.
I need help with tangent lines and optimization when figuring out which formula is what…can anyone help?
So, FIRST DERIVATIVE TEST: let c be a critical point of a function, f, that is continuous on an open interval containing c. If it is differentiable on the interval, except possibly at c, then f[c] can be classified as either:relative min [negative to positive]relative max [positive to negative]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
SECOND DERIVATIVE TEST: concavity & points of refection. 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.
Also, I understand the LIMIT RULES…Remember, you find vertical asymptotes when you set the bottom of a fraction equal to zero, then solve. After you factor the top and bottom of the fraction, if there is anything you can cancel from a function, it is a removable. Another thing i understand better is how to use the first and second derivative test.
I need help with tangent lines and optimization when figuring out which formula is what…can anyone help?
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