This first week of calculus started off easy when we learned about Rolle's theorum and the Mean Value theorum. When using Rolle's theorum you are given an equation and a point. First you figure out if the function is continuous and differentiable. If it is not continuous or differentiable on the point give, then the problem cannot be worked. If it is continuous and differentiable, then you can move on to the next steps.
Let's say they give you x^2-2x and (0,2) you would first plug in
f(0)=0^2-2(0)=0
f(2)=(2)^2-2(2)=0
The y values have to match, if they do not the problem cannot be worked further.
Now you take the derivative of the function and set it equal to the y value 0
2x-2=0
which gives you x=1. Therefor c=1
In Mean Value theorum, you still check for continuous and differentiability.
x^3 and (0,1)
first you plug into the formula like Rolle's
so f(0)= 0^3=0
f(10= 1^3=1
then you plug into the formula to find the slope between the two points: f(b)-f(a)/b-a
so then you plug in: 1-0/1-0=1
now you take the derivative of the function and set equal to the slope
3x^2=1
which will give you x= +or- 1/3, but c will equal 1/3 because it is positive and (0,1) is positive.
What i did not understand this week is optimization. Once some were worked in class, i started to get the feel on how to do it, but i do not know how to distinguish from primary functions and when to plug in different numbers to different functions. If anyone can help me with this it would be a big help. Thanks :)
Saturday, September 26, 2009
6TH POST
During the sixth week of calculus we learned a few new concepts such as: Rolle's theorem, the mean value theorem, and optimization.
Rolle's theorem starts off with a polynomial and a point. the first step is to plug both points into the f(x)= (equation). if both numbers come out to equal the same number then you take the first derivative. however, if both numbers do not equal the same number then you stop right there. on our last quiz mrs. robinson gave us one like that and asked us if it could be solved and then we had to justify our answer.
Another thing we learned about was the mean value theorem (MVT). the mean value theorem states that given a section of a smooth differentiable curve, there is at least one point on that section at which the derivative (slope) of the curve is equal to the "average" derivative of the section.Let f : [a, b] → R be a continuous function on the closed interval [a, b], and differentiable on the open interval (a, b), where a <>f1(c)=f(b)-f(a)/b-a. EXAMPLE:
Optimization refers to choosing the best element from some set of available alternatives.
In the simplest case, this means solving problems in which one seeks to minimize or maximize a real function by choosing the values of real or interger variables from within an allowed set. More generally, it means finding "best available" values of some objective function given a defined domain, including a variety of different types of objective functions and different types of domains.
In the simplest case, this means solving problems in which one seeks to minimize or maximize a real function by choosing the values of real or interger variables from within an allowed set. More generally, it means finding "best available" values of some objective function given a defined domain, including a variety of different types of objective functions and different types of domains.
I understand rolles theorem but im still having a little trouble with the mean value theorm so if anyone could explain it that would be helpful to me.
Thursday, September 24, 2009
Mrs Robinson...
hey mrs robinson...i'm just letting you know...my mom left the football pool on her desk at work and people were putting their name before she knew so we got nine names...and i can get more..but WE HAVE NINE..is that okay or do i need to pay them back?
Sunday, September 20, 2009
Post 5
So at the beginning of this week we didn't learn many new things. We just added onto old things and worked problems on them. We worked problems that included using the frist derivative test, the second derivative test to find points of inflection, where a graph concaves up or down, and a shortcut way to find maximums and minimums; however, we can only use these shortcuts on multiple choice and when a short answer question asks for it. We were given an assignment to find a career that involves math. The new material we covered was how to find absolute maximums and minimums, which are either the highest point of all high points, which would be the absolute maximum, and the lowest point of all low points, which would be the absolute minimum.
To find the absolute maximums and minimums the steps are as follows:
1. First derivative test
2. Plug critical values into the origonal function to get y-values
3. Plug endpoints into the origional function to get y-values
4. The highest y-value is the absolute maximum
5. The lowest y-value is the absolute minimum
*Remember that absolute maximums or minimums are written as a point, or simply as the y-value
After this, we were pretty much given the whole week to review on different types of problems covering everything from when we started out the school year, all the way down to finding the absolute maximums and minimums.
I understood how to read and graph origional function graphs, first derivative graphs, and second derivative graphs much better after I got together with a few classmates who were willing to help me understand them on Wednesday. Before this study session, I really didn't understand the concept that if the origional function graph was decreasing, that its first derivative would be starting below the x axis and if the orgional function graph was increasing, that its first derivative would be starting above the x axis. I also learned that when there is a zero on an origional function graph, the zero will become the maximum or minimum on the first derivative graph, and vice versa.
I feel very accomplished this week after completing all the things I've completed and I really feel greatful for the people who've helped me complete these things.
The only real question that I have is how to graph a tangent line. I've asked this question before and I've learned it time and time again, but I keep on forgetting. Can someone please explane?
To find the absolute maximums and minimums the steps are as follows:
1. First derivative test
2. Plug critical values into the origonal function to get y-values
3. Plug endpoints into the origional function to get y-values
4. The highest y-value is the absolute maximum
5. The lowest y-value is the absolute minimum
*Remember that absolute maximums or minimums are written as a point, or simply as the y-value
After this, we were pretty much given the whole week to review on different types of problems covering everything from when we started out the school year, all the way down to finding the absolute maximums and minimums.
I understood how to read and graph origional function graphs, first derivative graphs, and second derivative graphs much better after I got together with a few classmates who were willing to help me understand them on Wednesday. Before this study session, I really didn't understand the concept that if the origional function graph was decreasing, that its first derivative would be starting below the x axis and if the orgional function graph was increasing, that its first derivative would be starting above the x axis. I also learned that when there is a zero on an origional function graph, the zero will become the maximum or minimum on the first derivative graph, and vice versa.
I feel very accomplished this week after completing all the things I've completed and I really feel greatful for the people who've helped me complete these things.
The only real question that I have is how to graph a tangent line. I've asked this question before and I've learned it time and time again, but I keep on forgetting. Can someone please explane?
week 5 post
This week in calculus, we mostly just practiced stuff we already knew so that we could do those study guides and take home tests and stuff... but I still struggled a lot with it. But anyway, one thing that I specifically learned how to do this week was finding an absolute max or min. steps and procedure to do this:
1.First derivative test.
2.Plug critical values into original function to get y values.
3.Plug endpoints into original function to get y values.
4.Highest y value will be absolute max. – Lowest y value will be absolute min.
Note: Absolute max's and min's are written as a point or a simple y value.
Example: Find the absolute max or min of f(x)= 3x^(4)-4x^(3) on [-1,2] (-1) less than or equal to (x) less than or equal to (2)
First, you take the derivative.
F' (x)= 12x^(3)-12x^(2) = 0 which simplifies to (12x^(2))(x-1)=0. So x=1,0
so you have to plug in numbers between these points (-1,0) u (0,1) u (1,2) into the first derivative:
so, f'(-.5)= negative # f'(.5)= negative # f'(1.5)= positive #
so then you plug in your endpoints into the original function:
3(1)^(4)-4(1)^(3)= -1 so, (1,-1)
3(-1)(4)-4(-1)^(3)=7 so, (-1,7)
3(2)^(4)-4(2)^(3)= 3(16)-4(8)=16
so your absolute min is (1,-1), or -1, or -1 at x=1. They all mean the same thing.
And your absolute max is (2,16) or 16, or 16 at x=2. Once again, they all mean the same thing.
NOTE: there is a shortcut to this, but I don't completely understand it.
So my question is, can someone show me the shortcut to finding the absolute max's and min's?
Ok thanks and good night everyone. God help us on tomorrow's test...
1.First derivative test.
2.Plug critical values into original function to get y values.
3.Plug endpoints into original function to get y values.
4.Highest y value will be absolute max. – Lowest y value will be absolute min.
Note: Absolute max's and min's are written as a point or a simple y value.
Example: Find the absolute max or min of f(x)= 3x^(4)-4x^(3) on [-1,2] (-1) less than or equal to (x) less than or equal to (2)
First, you take the derivative.
F' (x)= 12x^(3)-12x^(2) = 0 which simplifies to (12x^(2))(x-1)=0. So x=1,0
so you have to plug in numbers between these points (-1,0) u (0,1) u (1,2) into the first derivative:
so, f'(-.5)= negative # f'(.5)= negative # f'(1.5)= positive #
so then you plug in your endpoints into the original function:
3(1)^(4)-4(1)^(3)= -1 so, (1,-1)
3(-1)(4)-4(-1)^(3)=7 so, (-1,7)
3(2)^(4)-4(2)^(3)= 3(16)-4(8)=16
so your absolute min is (1,-1), or -1, or -1 at x=1. They all mean the same thing.
And your absolute max is (2,16) or 16, or 16 at x=2. Once again, they all mean the same thing.
NOTE: there is a shortcut to this, but I don't completely understand it.
So my question is, can someone show me the shortcut to finding the absolute max's and min's?
Ok thanks and good night everyone. God help us on tomorrow's test...
fifth post
I know a little bit of the first derivitave test.
1) take derivative
2) find critical values by setting equal to zero
3) set up intervals. ( -infinity, x; x to x;......;x to infinity)
4) choose proper values that are equivilent to infinity< n < x or x < n < x (or what ever value is between what ever possible critical values you find.
5) determine weather your answer when plugged backinto the first derivative is positive or negative. If it is positive then it is a min meaning that the slope of the graph at that point is going up. If it is negative then it is a max saying that the graph is going down and the slope is going down.
6) determine your maxs and mins by using your data that you have found.
*****7) the one that i forget to do and the one that requires writing. JUSTIFY. After problem is worked you must justify your answer by explaining what you did in writing that will be able to give enough infortion so that your first derivitive test is clear and easy to follow due to your explanation.
The problem with this test is not remembering the steps, but for me it is to actually work the problem. Most of the time i dont know what it is asking and as the GROUP KNOWS lol that packet was beast. Taking the derivitave to me is the easiest, but after that i get confused and dont understand what part to set equal to zero or how in some cases. Like when it is a fraction with a polynomial at the top and bottom, do i set both numerator and denominator to zero or just top???????
Something else that is confusing me is:
sinx-cosx
derivative: cos(x)+sin^2(x)
***Do i set this whole thing equal to zero or do i brake up the terms and set cosx equal to zero and six^2x equal to zero???This problem confused the crap out of me.
1) take derivative
2) find critical values by setting equal to zero
3) set up intervals. ( -infinity, x; x to x;......;x to infinity)
4) choose proper values that are equivilent to infinity< n < x or x < n < x (or what ever value is between what ever possible critical values you find.
5) determine weather your answer when plugged backinto the first derivative is positive or negative. If it is positive then it is a min meaning that the slope of the graph at that point is going up. If it is negative then it is a max saying that the graph is going down and the slope is going down.
6) determine your maxs and mins by using your data that you have found.
*****7) the one that i forget to do and the one that requires writing. JUSTIFY. After problem is worked you must justify your answer by explaining what you did in writing that will be able to give enough infortion so that your first derivitive test is clear and easy to follow due to your explanation.
The problem with this test is not remembering the steps, but for me it is to actually work the problem. Most of the time i dont know what it is asking and as the GROUP KNOWS lol that packet was beast. Taking the derivitave to me is the easiest, but after that i get confused and dont understand what part to set equal to zero or how in some cases. Like when it is a fraction with a polynomial at the top and bottom, do i set both numerator and denominator to zero or just top???????
Something else that is confusing me is:
sinx-cosx
derivative: cos(x)+sin^2(x)
***Do i set this whole thing equal to zero or do i brake up the terms and set cosx equal to zero and six^2x equal to zero???This problem confused the crap out of me.
posting... #5
Well this weekend I was pretty bummed because I knew I had a lot of math to do. I was even more bummed when I actually started working because those take home test were uber hard but now that I’m actually finished I feel ok about the test tomorrow. Anyways:
Well this week we particularly stayed on graphs and I feel like I know it but then I get it wrong. The graphs I don’t have trouble with or the ones that are on the study guide.
But I don’t really understand how to get the points of inflection and intervals when you have a problem with lets say sin or cos in it. Working the problem is not the problem; its just when I get to the part when your getting the possible points of inflection that messes me up because I don’t really remember or understand when and how do the part in the problem when you use the quadrants of the graph to find the points of inflection. So if anybody understands what I’m asking and wouldn’t mind explain it to me that would really help. I know my quadrants for like sin, cos, and etc. but its just when you get like 30 and you have to subtract 360 I get all lost.
But other than when I’m dealing with a problem like that I’m pretty sure I grasp the first derivative test and the second derivative test.
About feeling ok on that test tomorrow I LIED!!! :o
GOOD LUCK. :dodgy:
Well this week we particularly stayed on graphs and I feel like I know it but then I get it wrong. The graphs I don’t have trouble with or the ones that are on the study guide.
But I don’t really understand how to get the points of inflection and intervals when you have a problem with lets say sin or cos in it. Working the problem is not the problem; its just when I get to the part when your getting the possible points of inflection that messes me up because I don’t really remember or understand when and how do the part in the problem when you use the quadrants of the graph to find the points of inflection. So if anybody understands what I’m asking and wouldn’t mind explain it to me that would really help. I know my quadrants for like sin, cos, and etc. but its just when you get like 30 and you have to subtract 360 I get all lost.
But other than when I’m dealing with a problem like that I’m pretty sure I grasp the first derivative test and the second derivative test.
About feeling ok on that test tomorrow I LIED!!! :o
GOOD LUCK. :dodgy:
Post Number Five
This week in Calculus we basically just reviewed and got the take home portions of out tests and the study guide. We also learned how to find absolute max's and min's. The one thing i'm actually comfortable with is the first and second derivative tests. You simply just follow the steps we learned in class:
1. Take a derivative
2. Set it equal to 0
3. Solve for x to find max, min, horizontal tangents, extrema (critical points)
4. Set up intervals using step 3.
5. Plug in first derivate
6. To find an absolute max/min plug values from #5 into original function
7. Check endpoints
An example problem of this is f(x)=1/2x - sinx
Find the relative extrema of f(x) on the interval (0,2pi)
f ' (x)= 1/2 - cos(x)
1/2 - cos(x)=0
cos(x)=1/2
x=cosinverse(1/2)
x=pi/3, 5pi/3
You know these are the values because cos is positive in those spots.
After doing this you can set up intervals in order to find the max and mins:
(0,pi/3) u (pi/3,5pi/3) u (5pi/3, 2pi)
f'(pi/4)=-ve which is decreasing
f'(pi)=+ve which is increasing
f'(11pi/6)=-ve which is decreasing.
Over all this means that pi/3 is a min and 5pi/3 is a max.
One thing i'm still not too sure about is looking at graphs and knowing what the first derivative or second derivative graph looks like. If anyone can help with that it'd be greatly appreciated.
We also went over justifications this week and i officially hate that we have to justify every problem by the way. It's so tedious mannnnn.
I've worked a lot of problems over the weekend and think I accomplished a lot from the take home tests. I still have a little trouble on simplifying derivatives but hey i'm getting better. If anyone can also help with explaining what differentiability it would be greatly appreciated also.
Oh and shoutout to s to the sixth :)
1. Take a derivative
2. Set it equal to 0
3. Solve for x to find max, min, horizontal tangents, extrema (critical points)
4. Set up intervals using step 3.
5. Plug in first derivate
6. To find an absolute max/min plug values from #5 into original function
7. Check endpoints
An example problem of this is f(x)=1/2x - sinx
Find the relative extrema of f(x) on the interval (0,2pi)
f ' (x)= 1/2 - cos(x)
1/2 - cos(x)=0
cos(x)=1/2
x=cosinverse(1/2)
x=pi/3, 5pi/3
You know these are the values because cos is positive in those spots.
After doing this you can set up intervals in order to find the max and mins:
(0,pi/3) u (pi/3,5pi/3) u (5pi/3, 2pi)
f'(pi/4)=-ve which is decreasing
f'(pi)=+ve which is increasing
f'(11pi/6)=-ve which is decreasing.
Over all this means that pi/3 is a min and 5pi/3 is a max.
One thing i'm still not too sure about is looking at graphs and knowing what the first derivative or second derivative graph looks like. If anyone can help with that it'd be greatly appreciated.
We also went over justifications this week and i officially hate that we have to justify every problem by the way. It's so tedious mannnnn.
I've worked a lot of problems over the weekend and think I accomplished a lot from the take home tests. I still have a little trouble on simplifying derivatives but hey i'm getting better. If anyone can also help with explaining what differentiability it would be greatly appreciated also.
Oh and shoutout to s to the sixth :)
post 5
After this week in calculus i realized that it's not getting much easier. We mainly worked with the first and second derivative tests again this week and justifications.
A justification is when you pretty much explain everything that you did in the problem. You first explain that you took the first derivative test then you explain all the internal steps, like possible points of inflection, actual points of inflection, concavity, max/min, and things along those lines.
We also learned how to find the absolute max or min in a problem.
1. Take the first derivative
2. Find your critical values
3. Plug in for x to find y-values
4. Find your endpoints
5. Plug in for x to find y-values
6. Highest y-value is your absolute max, lowest y-value is your absolute min
We also learned about differntiability. This is one of the things I did not understand. Another thing I did not understand was horizontal tangent. I tried going over these two things at the study group today, but it still doesn't completely make sense to me. if anyone knows an easy and simple explanation, that would be nice :)
I think calc will start to get easier as the weeks continue because it is only our fifth week. What probably confused me the most this week were those extremely hard take home tests and study guides for the test. They kind of made me nervous for the test tomorrow, but i've been studying all weekend so I think it will turn out okay.
A justification is when you pretty much explain everything that you did in the problem. You first explain that you took the first derivative test then you explain all the internal steps, like possible points of inflection, actual points of inflection, concavity, max/min, and things along those lines.
We also learned how to find the absolute max or min in a problem.
1. Take the first derivative
2. Find your critical values
3. Plug in for x to find y-values
4. Find your endpoints
5. Plug in for x to find y-values
6. Highest y-value is your absolute max, lowest y-value is your absolute min
We also learned about differntiability. This is one of the things I did not understand. Another thing I did not understand was horizontal tangent. I tried going over these two things at the study group today, but it still doesn't completely make sense to me. if anyone knows an easy and simple explanation, that would be nice :)
I think calc will start to get easier as the weeks continue because it is only our fifth week. What probably confused me the most this week were those extremely hard take home tests and study guides for the test. They kind of made me nervous for the test tomorrow, but i've been studying all weekend so I think it will turn out okay.
Post #5
With a test Monday, I'm kind of scared. I understand the First Derivative Test and Second Derivative Test, but most of the vocabulary is screwing me up. Some words are simple and commonly seen, while others are rare to find, so some do not recognize them as well as others.
These are the terms that I know so far:
1. Points of Inflection-relates to the Second Derivative Test
2. Extrema, Critical numbers, and intercepts- relates to the original graph (found using the First Derivative Test)
3. Intervals-relates to both the First Derivative Test and Second Derivative Test (depends on which intervals need to be found)
4. f'-first derivative
5. f"-second derivative
There is one question on 3.3 Homework Quiz that really throws me off. It is: "A differentiable function f has a critical number at x=2 and no other critical numbers. f'(1)=(-3) and f'(3)=5. Identify the relative extrema of f at x=2." I'm just utterly confused with what to do. I do know that differentiable means a derivative, probably the first derivative, but after that I'm lost. Can anyone help me with this problem?
A little reminder if people still can't remember both tests:
First Derivative Test:
1. Take the derivative of the original problem.
2. Set the first derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the first derivative for x.
6. Solve.
Second Derivative Test:
1. Take the derivative of the first derivative.
2. Set the second derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the second derivative for x.
6. Solve.
These are the terms that I know so far:
1. Points of Inflection-relates to the Second Derivative Test
2. Extrema, Critical numbers, and intercepts- relates to the original graph (found using the First Derivative Test)
3. Intervals-relates to both the First Derivative Test and Second Derivative Test (depends on which intervals need to be found)
4. f'-first derivative
5. f"-second derivative
There is one question on 3.3 Homework Quiz that really throws me off. It is: "A differentiable function f has a critical number at x=2 and no other critical numbers. f'(1)=(-3) and f'(3)=5. Identify the relative extrema of f at x=2." I'm just utterly confused with what to do. I do know that differentiable means a derivative, probably the first derivative, but after that I'm lost. Can anyone help me with this problem?
A little reminder if people still can't remember both tests:
First Derivative Test:
1. Take the derivative of the original problem.
2. Set the first derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the first derivative for x.
6. Solve.
Second Derivative Test:
1. Take the derivative of the first derivative.
2. Set the second derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the second derivative for x.
6. Solve.
Post #5
So I finally finished my take home tests now it is time to write about everything I had trouble with.
I find I still have a major problem with simplifying. I tend to cancel things that do not cancel or if I have cos x + sin x and I have to set it equal to zero I separate the two and set each equal to zero which I know I can't do but I still accidentally did it three times on the take home tests, that only works if you're multiplying by the way. I also am having trouble going from and equation to a graph especially numbers 1 and 2 on 3.6. I understand how to find the relative extrema, points of inflection, intercepts and asymptotes, but I do not know how to actually form a graph from the information I find.
The thing I am most comfortable with and I think I fully understand is first and second derivative test but since a lot of people explained that I am going to take a stab at explaining the relationship between the original graph and the first derivative graph because I feel I have gotten better at it over the course of this week.
If you are giving an original graph of f(x) and are asked to find the critical numbers, they are just the maxs and mins of the graph.
If you are asked to find the zeros, it is everywhere the graph crosses the x-axis.
If asked to find the zeros of f'(x) it is the maxs and mins of the original function (the same as the critical numbers) because when you sketch the derivative you move the maxs and mins to the x-axis.
If the original graph is increasing, the f' will be above the axis but if the original is decreasing f' will be below the axis.
This may seem easy but I think we learned this about two weeks ago and I finally just caught on for the most part this week. It really helped me with trying to do the study guide so I hope it helps yall too.
Ash's 5th post
After a very interesting day involving newborn guppy fish, I yawn and crawl in bed...almost asleep...until I realize...I FORGOT MY CALCULUS BLOG! =0 *freak out*
Okay...soooo, what I understood most this week was on the study guide...the one with all of the answers...yes, very sad, I know.
Anyway, numbers 124 and 125, I FINALLY understand after someone explained them ever so slowly. Since these require graphs, I guess I'll make some on paint ^^
Number 124 on the Multiple Choice Study Guide With Answers
Note: Sorry for all of the links, but it was much easier for me to see things step by step. Also, it's not perfect, please just pretend it is. The lines are only the even numbers.
The graph of F is shown below. On what interval is F' an increasing function?
http://i153.photobucket.com/albums/s225/Inutaisho_lover/th_Original.jpg
a) (2, ∞)
b) (0, ∞)
c) (-2, ∞)
d) (-4, ∞)
e) (1, ∞)
1) Find your zeros and mark them. (Red dots on graph) http://i153.photobucket.com/albums/s225/Inutaisho_lover/th_1.jpg
2) Mark your minimums and maximums. (Green dots on graph) http://i153.photobucket.com/albums/s225/Inutaisho_lover/th_2-1.jpg
3) Bring your minimums and maximums down to y=0. (Pink arrows on graph) http://i153.photobucket.com/albums/s225/Inutaisho_lover/th_3.jpg
4) The graph is that of a cubic and you know the derivative of a cubic is a parabola, right? (Because (this is an example, NOT the equation of the graph) f(x)=3x^3 and f'(x)=6x^2) This step was to figure that out.
Now here comes the really hard part for me to explain because I don't know HOW to explain what I do.
5) When going left to right, the first line is increasing right? The derivative of that is decreasing. So, draw a line decreasing THROUGH y=0 AT x=-4. ( Blue line on graph) http://i153.photobucket.com/albums/s225/Inutaisho_lover/th_5.jpg
6) For your minimum, I was told to just estimate it. (Light blue curve on graph) http://i153.photobucket.com/albums/s225/Inutaisho_lover/th_6-1.jpg
7) Now, since the next line (when going left to right) is decreasing, the derivative is to increase. (Dark blue line on graph...also, it's very crooked, please ignore that) http://i153.photobucket.com/albums/s225/Inutaisho_lover/th_7-1.jpg
8) Because the derivative of a cubic is a parabola, you're finished graphing!
9) Just look at your graph to see what is increasing, which happens to be c) (-2, ∞)
Hope that helps! Sorry for the horrible graphs and many links.
Ohh, as for my question, can someone explain how to find the equation of a graph. I completely forgot this and I can't find my Advanced Math notebook (if you saw my room, you'd know why). Also, the power point project is due the 28th right?
Okay...soooo, what I understood most this week was on the study guide...the one with all of the answers...yes, very sad, I know.
Anyway, numbers 124 and 125, I FINALLY understand after someone explained them ever so slowly. Since these require graphs, I guess I'll make some on paint ^^
Number 124 on the Multiple Choice Study Guide With Answers
Note: Sorry for all of the links, but it was much easier for me to see things step by step. Also, it's not perfect, please just pretend it is. The lines are only the even numbers.
The graph of F is shown below. On what interval is F' an increasing function?
http://i153.photobucket.com/albums/s225/Inutaisho_lover/th_Original.jpg
a) (2, ∞)
b) (0, ∞)
c) (-2, ∞)
d) (-4, ∞)
e) (1, ∞)
1) Find your zeros and mark them. (Red dots on graph) http://i153.photobucket.com/albums/s225/Inutaisho_lover/th_1.jpg
2) Mark your minimums and maximums. (Green dots on graph) http://i153.photobucket.com/albums/s225/Inutaisho_lover/th_2-1.jpg
3) Bring your minimums and maximums down to y=0. (Pink arrows on graph) http://i153.photobucket.com/albums/s225/Inutaisho_lover/th_3.jpg
4) The graph is that of a cubic and you know the derivative of a cubic is a parabola, right? (Because (this is an example, NOT the equation of the graph) f(x)=3x^3 and f'(x)=6x^2) This step was to figure that out.
Now here comes the really hard part for me to explain because I don't know HOW to explain what I do.
5) When going left to right, the first line is increasing right? The derivative of that is decreasing. So, draw a line decreasing THROUGH y=0 AT x=-4. ( Blue line on graph) http://i153.photobucket.com/albums/s225/Inutaisho_lover/th_5.jpg
6) For your minimum, I was told to just estimate it. (Light blue curve on graph) http://i153.photobucket.com/albums/s225/Inutaisho_lover/th_6-1.jpg
7) Now, since the next line (when going left to right) is decreasing, the derivative is to increase. (Dark blue line on graph...also, it's very crooked, please ignore that) http://i153.photobucket.com/albums/s225/Inutaisho_lover/th_7-1.jpg
8) Because the derivative of a cubic is a parabola, you're finished graphing!
9) Just look at your graph to see what is increasing, which happens to be c) (-2, ∞)
Hope that helps! Sorry for the horrible graphs and many links.
Ohh, as for my question, can someone explain how to find the equation of a graph. I completely forgot this and I can't find my Advanced Math notebook (if you saw my room, you'd know why). Also, the power point project is due the 28th right?
Post #5
Alright I'm so lost, and we have a test tomorrow. I have been all over Vacherie and Laplace trying to learn this.
The thing that I actually understand are the rules for the First and Second Derivative Test. But the problem is carrying out the problem and sometimes I don't understand what it is even asking for.
Okay, I'm most comfortable with the second derivative test rules:
1. take the derivative
2. then take the second derivative
3. set = to 0 (this gives you possible points of inflection)
4. set up intervals
5. plug in (tells you concave up/down, and finds the point of inflection)
Example problem: x^3-6x^2+12x
1. 3x^2-12x+12
2. 6x-12
3. 6x-12=0 x=2--->possible pts. of inflection
4. (-infinity,2) u (2,infinity)
5. 6(0)-12= -ve 6(3)-12= +ve
concave down concave up
x=2 is the point of inflection
It confuses me that I understand the rules, but I can't do a single problem. Especially problems that are like the ones on 3.6 homework quiz. The major thing I can't even start are problems like: find the points of inflection of f(x)=cosx-sinx on [0, 2pi] or something like: find any critical numbers for the function f(theta)=sin(theta)-cos^2(theta) with 0 (like number 2 on 3.1 homework quiz) Can someone help me on this?
The thing that I actually understand are the rules for the First and Second Derivative Test. But the problem is carrying out the problem and sometimes I don't understand what it is even asking for.
Okay, I'm most comfortable with the second derivative test rules:
1. take the derivative
2. then take the second derivative
3. set = to 0 (this gives you possible points of inflection)
4. set up intervals
5. plug in (tells you concave up/down, and finds the point of inflection)
Example problem: x^3-6x^2+12x
1. 3x^2-12x+12
2. 6x-12
3. 6x-12=0 x=2--->possible pts. of inflection
4. (-infinity,2) u (2,infinity)
5. 6(0)-12= -ve 6(3)-12= +ve
concave down concave up
x=2 is the point of inflection
It confuses me that I understand the rules, but I can't do a single problem. Especially problems that are like the ones on 3.6 homework quiz. The major thing I can't even start are problems like: find the points of inflection of f(x)=cosx-sinx on [0, 2pi] or something like: find any critical numbers for the function f(theta)=sin(theta)-cos^2(theta) with 0
post 5
after the fifth week of calculus im pretty confused on some things. i know the first and second der tests pretty well. i know how to get the maxes and the mins which has to do w/ the first der test and i know the concave up and down which has to do w/ the second der test. we also have to justify all the answers we get. the justifying thing is what is kind of hard to do. i'm not really goodd at explaining myself in math talk haha the first and second der tests are what im pretty goodd at.
Find the absolute MAX/MIN of f(x) = 3x^4 - 2x^4 on [0,3] :
First, take the derivative. 12x^3 - 8x^3
Set equal to zero, find critical values and endpoints, plug in for x to find y values, smallest = min ; largest = max
you need to put the max and min in point form.
im not too sure of drawing the graphs when the information is given. i knonw the points of discontinuities and what they look like on teh graph. i know how to find them when given an eqn as well. i dont know differtiability too much so if anyone wants to help comment me
here is an example problem of max andd min.
Find the absolute MAX/MIN of f(x) = 3x^4 - 2x^4 on [0,3] :
First, take the derivative. 12x^3 - 8x^3
Set equal to zero, find critical values and endpoints, plug in for x to find y values, smallest = min ; largest = max
you need to put the max and min in point form.
im not too sure of drawing the graphs when the information is given. i knonw the points of discontinuities and what they look like on teh graph. i know how to find them when given an eqn as well. i dont know differtiability too much so if anyone wants to help comment me
post 5
Another week down in calculus and it is not getting any easier. Mainly we remained working with the first and second derivative tests. The only thing that was new was justifications. Justifications are where you explain what you did to get the answer you got. The main things that are used are the definitions such as concave up, concave down, increasing, decreasing, etc. An example of a justification is of the derivative of sinx+cosx. You get cosx-sinx as your derivative. The justification is that the derivative formulas were used to get the derivative. Also we kept working with the second derivative test. This test deals with concave up, concave down, and points of inflection. But this can only be done after the first derivative test has been done. And negative slope, positive slope, horizontal tangent, and zero of derivative are used in the first derivative test. Another thing is we started have to look at old notes and limits were a part of that. And limits are not complicated at all especially if the highest exponents are the same on the top and bottom because the limit of that is the top coefficient over the bottom coefficient.
But I do not understand how to differentiate. I read the notes and try to understand it but I just can not grasp it fully. Also I am fuzzy on exactly how to get the horizontal tangent of an equation. I mean I kind of remember I just do not remember everything. But if someone can help me before tomorrow that would be great but if not i will try to figure it out.
But I do not understand how to differentiate. I read the notes and try to understand it but I just can not grasp it fully. Also I am fuzzy on exactly how to get the horizontal tangent of an equation. I mean I kind of remember I just do not remember everything. But if someone can help me before tomorrow that would be great but if not i will try to figure it out.
Week 5
This week we prepared for the test on Monday.
Something I understood was the Second derivative test.
You must first take the derivative, then the derivative of the derivative. You go through your first derivative test, but it is really second derivatieve test.
Second Derivative test: take the derivatives.
2. set equal to zero
3. solve for x - this will give you possible points of inflection
4. set up intervals with your possible points of inflection
5. plug into second derivative
6. to see if it is a point of inflection, see if the intervals change concavitiy.
I still do not get the max & min shortcut. I don't know the difference between it and the regular way to find it.
Don't get differentiability. I don't understand how to pick one out or whether it matters or not. Help?
Something I understood was the Second derivative test.
You must first take the derivative, then the derivative of the derivative. You go through your first derivative test, but it is really second derivatieve test.
Second Derivative test: take the derivatives.
2. set equal to zero
3. solve for x - this will give you possible points of inflection
4. set up intervals with your possible points of inflection
5. plug into second derivative
6. to see if it is a point of inflection, see if the intervals change concavitiy.
I still do not get the max & min shortcut. I don't know the difference between it and the regular way to find it.
Don't get differentiability. I don't understand how to pick one out or whether it matters or not. Help?
Post #5
This week we basically reviewed for the test on Monday. After getting together with a couple of people for a study group, I understand some concepts that I didn't know before. So overall, I think I'm somewhat prepared for the test. I can do the First and Second Derivative Tests with my eyes closed. And then we learned how to find absolute max and mins which is even easier than the Tests. For example:
Find the absolute extrema(max and mins) of f(x) 1/(x^2 + 1) on [-1, 1].
After taking the derivative by the quotient rule you get:
2x/(x^2 + 1)^2
Set the top equal to zero and it gives you x = 0. Then set up your intervals with you given interval.
(-1, 0) U (0, 1)
Plug in your critical values and your intervals into the original function to get your y values. So you will have (-1, 1/2), (0, 1), (1, 1/2).
Since 0 gives you the biggest value, there is an absolute max at x=0. And from what I understand, -1 and 1 are not mins because there are two of the same values. Someone might want to clarify that for me.
Justification: By using the 1st Derivative Test, I set the derivative equal to zero and solved for x. I then plugged my critical values and endpoints into the original function and found an absolute max at x=0.
Couls someone please explain the graphs to me? Like when you're given a graph of the second der. and asked to find the graph of the original. I know what it looks like and can eliminate a couple of choices, but I usually cannot decide between the two or three left. That whole study packet on graphs kind of confused me, and I could really use some help. Also, for some reason I have difficulties graphing a function after finding the extrema and such, but that could just be because I fail at graphs. Thanks :)
Find the absolute extrema(max and mins) of f(x) 1/(x^2 + 1) on [-1, 1].
After taking the derivative by the quotient rule you get:
2x/(x^2 + 1)^2
Set the top equal to zero and it gives you x = 0. Then set up your intervals with you given interval.
(-1, 0) U (0, 1)
Plug in your critical values and your intervals into the original function to get your y values. So you will have (-1, 1/2), (0, 1), (1, 1/2).
Since 0 gives you the biggest value, there is an absolute max at x=0. And from what I understand, -1 and 1 are not mins because there are two of the same values. Someone might want to clarify that for me.
Justification: By using the 1st Derivative Test, I set the derivative equal to zero and solved for x. I then plugged my critical values and endpoints into the original function and found an absolute max at x=0.
Couls someone please explain the graphs to me? Like when you're given a graph of the second der. and asked to find the graph of the original. I know what it looks like and can eliminate a couple of choices, but I usually cannot decide between the two or three left. That whole study packet on graphs kind of confused me, and I could really use some help. Also, for some reason I have difficulties graphing a function after finding the extrema and such, but that could just be because I fail at graphs. Thanks :)
post numero five
During the fifth week of calculus we did a lot of practice on justificiations. We had to do a lot of the same things we did last week, like first and second derivative test, but also we had to justify almost every problem. Justifications are not very hard, it's just hard to remember that you have to write down EVERYTHING you did in order.
We also learned how to find max and min.
We also learned how to look at a graph and figure out what is your absolute max and min.
example problem:
Find the absolute MAX/MIN of f(x) = 2x^3 - 4x^4 on [0,5] :
First, take the derivative. 6x^2 - 16x^3
Set equal to zero, find critical values and endpoints, plug in for x to find y values, smallest = min ; largest = max
Dont forget to write your max & min in point form!
We also learned about differentiability, but that is probably one of the things that confused me the most this week.
Any help? Please and Thank you!
5th post
whew... well this week was a difficult week for me in calculus but luckily we had a study group yesterday at PJ's and i understand some of the concepts better than what i did before. First of all, absolute max and mins are very easy. First you take the derivative of the original function and set it equal to zero. Then you find your critical points and set up intervals. Then you plug in your critical points and endpoints into the original function. Your biggest y value will be your absolute max and your lowest y value will be your absolute min. Another concept i understand better is finding the slope of a line with a given function and x value. You take the derivative of the function and then plug in the x value to get the slope... super easy :).
What i'm not so sure about is how to draw different graphs when the problem gives you different information. If the function gives you a critical point say 5 and they say no extrema or stuff like that, i do not know how to draw that. If someone can explain how to do that it would help a lot. Another thing i'm still not sure about is looking at the graph and figuring out what the derivative form of the original function is supposed to look like or vice versa. So for example if they give me a graph and show a line going through a number on the x axis and they want to know what the first derivative of that graph is supposed to look like, i do not understand what the graph is supposed to look like and if there are certain things that i should be looking at to get the answer. If anyone can help me understand what to look for it would be a great help. Thanks :)
What i'm not so sure about is how to draw different graphs when the problem gives you different information. If the function gives you a critical point say 5 and they say no extrema or stuff like that, i do not know how to draw that. If someone can explain how to do that it would help a lot. Another thing i'm still not sure about is looking at the graph and figuring out what the derivative form of the original function is supposed to look like or vice versa. So for example if they give me a graph and show a line going through a number on the x axis and they want to know what the first derivative of that graph is supposed to look like, i do not understand what the graph is supposed to look like and if there are certain things that i should be looking at to get the answer. If anyone can help me understand what to look for it would be a great help. Thanks :)
Post #4
The thing i understand the most in calculus this week is absolute maximums and minimums. To do this, you simply follow simple steps:
1. first derivative test
which consists of:
Take Derivative
Set Equal to Zero
Set Up intervals
2. plug the critical values into origional function to get y-values
3. plug endpoints in to origional function to get y-values
4. your highest y-value is absolute max
5. your lowest y-value is absolute min.
**absolute maxs or mins or written as a point or simply as the y-value.
These problems are extremely simple to solve because it only involves simple algebra. Just remember, when the graph becomes discontinuous. Jumps only occur when you have a peicewise function. You find removables by factoring the top and bottom of your function..if something cancells out then its a removable at x=whatever that number is. Then after you cancel, if you have anything left at the bottom you have an infintite(asymptote) at that number. These are reallly easy to do, as long as you remember when the discontinuities occur. Remember these simple formulas, and your set for life. :)
The thing I do not understand this week is still the graphs. I do not understand how to take a graph and determine how to take the second derivative, or the first, or original function. I know what the graph will look like, but not where it is on the graph. Please help me because i'm pretty sure it will be on the test tomorrrowwwww.
1. first derivative test
which consists of:
Take Derivative
Set Equal to Zero
Set Up intervals
2. plug the critical values into origional function to get y-values
3. plug endpoints in to origional function to get y-values
4. your highest y-value is absolute max
5. your lowest y-value is absolute min.
**absolute maxs or mins or written as a point or simply as the y-value.
These problems are extremely simple to solve because it only involves simple algebra. Just remember, when the graph becomes discontinuous. Jumps only occur when you have a peicewise function. You find removables by factoring the top and bottom of your function..if something cancells out then its a removable at x=whatever that number is. Then after you cancel, if you have anything left at the bottom you have an infintite(asymptote) at that number. These are reallly easy to do, as long as you remember when the discontinuities occur. Remember these simple formulas, and your set for life. :)
The thing I do not understand this week is still the graphs. I do not understand how to take a graph and determine how to take the second derivative, or the first, or original function. I know what the graph will look like, but not where it is on the graph. Please help me because i'm pretty sure it will be on the test tomorrrowwwww.
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