Showing posts with label Derivative Functions. Show all posts
Showing posts with label Derivative Functions. Show all posts
Friday, October 19, 2007
BoB 1
Well, this unit was a unit that I had struggled with. It was all alright until the day Mr.K introduced the graphing of a function to find its derivative and second derivative. After doing all the homework and discussing it with other students in the class I felt more reassured, although many times I do still get confused on the second derivative of the parent function. All in all i believed i had studied enough and get the understanding of derivative functions. So good luck to everyone on today's derivative test.
BOB
So here we are bobbing before another test. This unit has been somewhat difficult for me. However with my lovely boyfriend, Manny, I can usually clear things up quite well. If i concentrated a tad bit more i think things would be a lot easier right now. I can always change for the next units. (: It's hard to devote all your time to school when there's so many other things going on. Committees and so on and so forth. I know i should try a little harder next time though..
anyway. Good luck to everyone on the test today!
anyway. Good luck to everyone on the test today!
Thursday, October 18, 2007
Scribulous
Alright a pretest scribe! I guess I have some explaining to do.... about the questions on the pretest!
Question 1)
If f(x) = ln(sqr(x)) then the average rate of change on the interval [3,7] is... then 5 possible answers. Using Roberts method you punch the equation into the calculator and use that handy-dandy slope program and out pops the answer or you can do it manually by hand and come up with A) 0.106
Question 2)
Suppose that the number of bacteria in a certain organism grows over time and the number, N(t), of bacteria (measured in thousands) at any time is given by:
N(t) = t(2 + cos(t))^(4/3) + 3t
At approximately what time ,t, in the first ten days is the colony growing the fastest?
First and foremost I must remind you all that IN THIS CLASS CALCULATORS SHOULD BE IN RADIANS! It is truly *very* annoying to miss a mark because of your calculator...
Now entered into your calculator you turn on that oh so useful derivative-of-whatever-you-have-in-your-Y1-slot graph and you look for the peak! (Then use that maximum function to give you a value)
The answer is.... 5.18!
Question 3)
Limit as x approaches one of:
(2x^2 + x - 3)/(3x^2 -x -2)
is equal to.... ?.
Well first you try putting one in right away and find your dividing by 0!
Algebraic massage required...
These both factor readily and two of the terms reduce. Then 1 can be entered in and you end up with 5/5 or 1... which is the answer!
Question 4)
The graph of the second derivative of a function f is shown at the right. Which of the following statements are true (I realize that there is no graph there but there but there is on this slide)
I) The graph of f has an inflection point at x = -1.
II) The f graph is concave down on the interval (-1,3).
III) The graph of the derivative function f prime is increasing at x = 1.
Then you find which of those statements is true.
I) An inflection point on a second derivative graph would be a zero and there happens to be one where they say that there is an inflection point therefore it is true!
II) The second derivative measures concavity, since it is negative on the interval that they say then the statement is true!
III) The second derivative is a measure of the slope of the first derivative. Since the value is negative the slope of that point is negative so it is decreasing at that point making this statement false.
So the answer is I and II are true!
Question 5)
Suppose a function f is defined so that is has derivatives:
f'(x) = x^2(x-1) and f''(x) = x(3x-2)
Over what intervals is f both increasing and concave up?
Well for it to be increasing the first derivative must be positive. Finding the roots of that we find them to be x = 0 and 1. Then do a sign check by entering a value before, in between and after to find when it is positive. The process is repeated for the second derivative because where that is positive the graph is concave up. We find the overlapping positive areas to be when x > 1!
Question 6)
a) Condider the following table of data.
Estimate f'(5.2) as accurately as possible.
Well f' would be the slope at that point!
Since they are on equal intervals and there are values before and after 5.2 we are able to use the symmetric difference quotient to get a more accurate result. It is found by finding the slope of that point with the points before and after and averaging it which results in you getting -9/4.
b) Write and equation for the slope of the tangeant line at that point.
Well... you are given a point.... you are given a slope... so let's use point slope form shall we? and you end up with y - 8.8 = -2.25 (x - 5.2)
c) Use your answer in part (b) to approximate f'(5.26)
Since the two points are so close together we can just plug that value into the equation we made to get a fairy accurate answer... which ends up being 8.665.
d) What is the sign of f''(5.2)? Explain.
Well this one I got wrong. I drew a graph where this point really looked like a point of inflection and that's what I based my answer on. WRONG! Mr.K ends up saying something along the lines of "Using the algebra to prove a point is far better then the graphical approach and will be expected of you on the exam". Since f' is decreasing to the left and right of this point we can deduce that this point is also decreasing meaning the graph is concave down meaning that f'' is negative at that point! (Though I still think it could be a point of inflection, but I do see the logic in what was said).
Woo that was a workout for the fingers!
The next scribe will be DINO!
Question 1)
If f(x) = ln(sqr(x)) then the average rate of change on the interval [3,7] is... then 5 possible answers. Using Roberts method you punch the equation into the calculator and use that handy-dandy slope program and out pops the answer or you can do it manually by hand and come up with A) 0.106
Question 2)
Suppose that the number of bacteria in a certain organism grows over time and the number, N(t), of bacteria (measured in thousands) at any time is given by:
N(t) = t(2 + cos(t))^(4/3) + 3t
At approximately what time ,t, in the first ten days is the colony growing the fastest?
First and foremost I must remind you all that IN THIS CLASS CALCULATORS SHOULD BE IN RADIANS! It is truly *very* annoying to miss a mark because of your calculator...
Now entered into your calculator you turn on that oh so useful derivative-of-whatever-you-have-in-your-Y1-slot graph and you look for the peak! (Then use that maximum function to give you a value)
The answer is.... 5.18!
Question 3)
Limit as x approaches one of:
(2x^2 + x - 3)/(3x^2 -x -2)
is equal to.... ?.
Well first you try putting one in right away and find your dividing by 0!
Algebraic massage required...
These both factor readily and two of the terms reduce. Then 1 can be entered in and you end up with 5/5 or 1... which is the answer!
Question 4)
The graph of the second derivative of a function f is shown at the right. Which of the following statements are true (I realize that there is no graph there but there but there is on this slide)
I) The graph of f has an inflection point at x = -1.
II) The f graph is concave down on the interval (-1,3).
III) The graph of the derivative function f prime is increasing at x = 1.
Then you find which of those statements is true.
I) An inflection point on a second derivative graph would be a zero and there happens to be one where they say that there is an inflection point therefore it is true!
II) The second derivative measures concavity, since it is negative on the interval that they say then the statement is true!
III) The second derivative is a measure of the slope of the first derivative. Since the value is negative the slope of that point is negative so it is decreasing at that point making this statement false.
So the answer is I and II are true!
Question 5)
Suppose a function f is defined so that is has derivatives:
f'(x) = x^2(x-1) and f''(x) = x(3x-2)
Over what intervals is f both increasing and concave up?
Well for it to be increasing the first derivative must be positive. Finding the roots of that we find them to be x = 0 and 1. Then do a sign check by entering a value before, in between and after to find when it is positive. The process is repeated for the second derivative because where that is positive the graph is concave up. We find the overlapping positive areas to be when x > 1!
Question 6)
a) Condider the following table of data.
Estimate f'(5.2) as accurately as possible.
Well f' would be the slope at that point!
Since they are on equal intervals and there are values before and after 5.2 we are able to use the symmetric difference quotient to get a more accurate result. It is found by finding the slope of that point with the points before and after and averaging it which results in you getting -9/4.
b) Write and equation for the slope of the tangeant line at that point.
Well... you are given a point.... you are given a slope... so let's use point slope form shall we? and you end up with y - 8.8 = -2.25 (x - 5.2)
c) Use your answer in part (b) to approximate f'(5.26)
Since the two points are so close together we can just plug that value into the equation we made to get a fairy accurate answer... which ends up being 8.665.
d) What is the sign of f''(5.2)? Explain.
Well this one I got wrong. I drew a graph where this point really looked like a point of inflection and that's what I based my answer on. WRONG! Mr.K ends up saying something along the lines of "Using the algebra to prove a point is far better then the graphical approach and will be expected of you on the exam". Since f' is decreasing to the left and right of this point we can deduce that this point is also decreasing meaning the graph is concave down meaning that f'' is negative at that point! (Though I still think it could be a point of inflection, but I do see the logic in what was said).
Woo that was a workout for the fingers!
The next scribe will be DINO!
BOB and the chocolate factory
My first bob for this class and for the year. I think that the overall course hasn't presented a lot of trouble, though I feel like a broken BOB record after saying that too frequently. I have encountered much of this unit prior to the course, so quite a bit of it is a review for me, though there were some questions that required some serious concentration.
These questions put my brain to work, but the more practice I embarked upon regarding these types of questions they inevitably sparked a further comprehension of the overall unit and of all of the unit's constituent areas. I cannot stress the significance of the excercises enough. Well, as for Mr. K, I'm sure he will optimize these questions and utilize them to the maximum extent possible to challenge us on tomorrow's test. But the test itself should not be too difficult, just don't forget to convert your calculator mode from DEGREES to RADIANS, just in case some trig questions present themselves similar to counteract the consequences of today's fiasco (sorry to hear your calc cost you that mark Graeme).
I'll probably just end this bob with a couple quick notes for the class:
When f has a max or a min at x=a, f' will have a root at x=a.
When f has an point of inflection (change in concavity) at x=a, then f'' will have a root at x=a.
When f is increasing, f' is positive. When f is decreasing, f' is negative.
When f is concave up, f'' is positive. When f is concave down, f'' is negative.
When f' has a max or a min at x=a, f'' will have a root at x=a.
Well that concludes my BOB post for today, I tried to keep it as short as possible, I'd just like to wish everyone luck on tomorrow's test and I want to remind everyone to study!
GOOD LUCK! and GOOD NIGHT!
These questions put my brain to work, but the more practice I embarked upon regarding these types of questions they inevitably sparked a further comprehension of the overall unit and of all of the unit's constituent areas. I cannot stress the significance of the excercises enough. Well, as for Mr. K, I'm sure he will optimize these questions and utilize them to the maximum extent possible to challenge us on tomorrow's test. But the test itself should not be too difficult, just don't forget to convert your calculator mode from DEGREES to RADIANS, just in case some trig questions present themselves similar to counteract the consequences of today's fiasco (sorry to hear your calc cost you that mark Graeme).
I'll probably just end this bob with a couple quick notes for the class:
When f has a max or a min at x=a, f' will have a root at x=a.
When f has an point of inflection (change in concavity) at x=a, then f'' will have a root at x=a.
When f is increasing, f' is positive. When f is decreasing, f' is negative.
When f is concave up, f'' is positive. When f is concave down, f'' is negative.
When f' has a max or a min at x=a, f'' will have a root at x=a.
Well that concludes my BOB post for today, I tried to keep it as short as possible, I'd just like to wish everyone luck on tomorrow's test and I want to remind everyone to study!
GOOD LUCK! and GOOD NIGHT!
BOB
This unit about derivatives was really tough for me. It started off relatively easily until the concepts just kept coming and coming, but as the days go by and with the continued help from my peers, i am starting to understand the concepts a little bit more. I just hope that my knowledge will be sufficient for the test tomorrow. Well, thats it. Good luck to everyone and study hard!
BOB
Hello everyone! The time has come again to bob. Well this unit was a tough one for me and I found it really challenging. With so many concepts to soak in, I really felt overwhelmed with this unit. I'm starting to understand more and more each day and hope to know enough for the test. Well good luck to everyone on the test and best wishes. Bye for now and don't forget to study!
Wednesday, October 17, 2007
ßØß
Well my first bob of the year... I must say I have been finding the base concepts of this unit fairly easy. But my problem is with the recognition of those base concepts. They are as straight forward as can be in the text but I know that Mr.K is never that straight forward. I just have to work on identification of what is given in the question and use the knowledge that I'm comfortable with to get the questions. If I can do that this test will be a cake walk.
Gute Nacht!
GreyM
Gute Nacht!
GreyM
Blogging on Blogging..
Hey! It's Tim-Math-Y and its time to b.o.b. again. Well this new unit was definitive.. different, as it included a new idea, Derivatives. At first this unit really had my mind twisting up on itself trying to visualize a derivative of a function, then holding that image, attempting to visualize the second derivative. I will admit two things. That wasn't a good idea to imagine similtaneously and second, that I'm still a little confused about the matter.
However, thinking back of the leaked characteristics of the power law, which I also admit that I don't know how it works.. yet, it helps me remember that each derivative is a degree down from its parent (cubic>quadratic>linear etc.), and thus aids me in picturing the graphs.
Now with the knowledge of derivatives, it brings fourth questions that we deal with in physics, such as comparing two variables: time and distance, time and velocity, time and acceleration etc. I find that I don't have many problems with these types of Qs because I understand the concepts to a more complete level.
So I can say that I'm semi-prepared for derivative functions and their like. Other than that, I still don't really understand the exact purpose and use of finding limits. Limits still greatly puzzles me as, logically, it just isn't registering for me. Besides that, Continuity, I understand and I'm glad I can say that atleast...
Wow, my bob is long? =\ Good luck on the test on friday guys!
However, thinking back of the leaked characteristics of the power law, which I also admit that I don't know how it works.. yet, it helps me remember that each derivative is a degree down from its parent (cubic>quadratic>linear etc.), and thus aids me in picturing the graphs.
Now with the knowledge of derivatives, it brings fourth questions that we deal with in physics, such as comparing two variables: time and distance, time and velocity, time and acceleration etc. I find that I don't have many problems with these types of Qs because I understand the concepts to a more complete level.
So I can say that I'm semi-prepared for derivative functions and their like. Other than that, I still don't really understand the exact purpose and use of finding limits. Limits still greatly puzzles me as, logically, it just isn't registering for me. Besides that, Continuity, I understand and I'm glad I can say that atleast...
Wow, my bob is long? =\ Good luck on the test on friday guys!
Calculus BOB saluclaC
Well, BOBin' for the first time in Calculus... haha we got away with one last week.
Hmmm... well for anyone who thought Calculus should have been a little harder while doing the first unit, HERE YOU GO! Holy, honnestly, I did not expect it to be this big of a jump from the most familiar things to a brand new concept of Derivatives and Limits. WOW!!! Anyway, now I think I'm finally starting to get going with this stuff in the sense that I know EXACTLY WHAT I'M DOING!!! At the beginning of the unit, with two different teachers teaching, I found it a little difficult to stay on track. However, lately I've started to understand it a little more. One thing that I know I will probably have a little difficulty with is the limits portion... I never really knew exactly what was happening with them. Oh well, I'll see how they are in the review tonight. Other than that I think I should be set, although I'm sure I'll make a couple mistakes... nobody's perfect (cough*Chris*cough) LOL!!! just kidding....
Oh! Almost forgot! REMEMBER:
Keep pushing =)
Hmmm... well for anyone who thought Calculus should have been a little harder while doing the first unit, HERE YOU GO! Holy, honnestly, I did not expect it to be this big of a jump from the most familiar things to a brand new concept of Derivatives and Limits. WOW!!! Anyway, now I think I'm finally starting to get going with this stuff in the sense that I know EXACTLY WHAT I'M DOING!!! At the beginning of the unit, with two different teachers teaching, I found it a little difficult to stay on track. However, lately I've started to understand it a little more. One thing that I know I will probably have a little difficulty with is the limits portion... I never really knew exactly what was happening with them. Oh well, I'll see how they are in the review tonight. Other than that I think I should be set, although I'm sure I'll make a couple mistakes... nobody's perfect (cough*Chris*cough) LOL!!! just kidding....
Oh! Almost forgot! REMEMBER:
Keep pushing =)
13-0-13
Well...it is time for me to bob again. It's the first of my last this year. Well it seems to me that my muddiest point was the whole unit. It has really confused me a lot. I always had to get someone else to explain things to me. (Thanks Chris, Graeme, Craig, Robert) haha. I understand that a derivative is the slope of the rate of change. I get how first and second derivatives work. Continuity kind of confuses me. I think it's when I read a question that I get all mixed up because I don't quite understand it but if I read it over again and try to understand what it's asking me then I will be able to get it. In class I seem to get things more than when I'm just doing a question by myself...it's weird how that works and I think I'm not the only one that happens to. I really like the feeling of when I understand something, it's like yay! lightbulb! That only happens sometimes though. Overall, I found this unit quite challenging. I will try my best to decipher what is in front of me and try to solve it carefully.
Good luck to everyone on the test and pretest! =)
Good luck to everyone on the test and pretest! =)
The Scribe that didn't know he was the scribe until 3:33...
Haha, hmmm.... well to make a long story short, I looked at the blog briefly last night because I thought there was homework that needed to be posted (there wasn't) and while I was there, I completely forgot to check who the next scribe was. (evidently it was me). I learned from Aichelle a couple minutes after the end of class that I was, fortunately the class wasn't very complicated and no new concpets were really introduced. So, here we go with probably the easiest scribe post I've had to do in a while.
Well, this class was a "workshop" class, so we got into our groups and took a look at the first question which seemed very familiar. It has to do with the graph of the height of a thrown ball over time (Grade 11 Pre-Cal question). However he added the velocity of the ball which is the rate of change (a.k.a. derivative) of the function. Then we were asked a series of questions about the scenario. For question and answer, click the "first question link".
Question Two basically asked us to find the slopes of the tangent at each point and list them in increasing order.
The third question was a little interesting because it gave us a told us that an original function was transformed by with a vertical shift. Then it asked us about how the derivative of the transformed function related to that of the original function. Well, when a function is shifted vertically, the vertical and horizontal scale remains the same (same size), as does the shape and it's position along the x-axis (x coordinates). The only thing that differs between it and it's original function is the y coordinates. Now, we know that a derivative is a graph of the slope of the tangents of the parent function vs. the points along the x-axis. So, since the shape, size, and x coordinates of the transformed function do not change, the slope of tangent at each point remain the same meaning the derivatives are the same!!!.
When Mr. K. clicked the next slide button on the SmartBoard Question Four came on the screen. This dealt with the derivatives of even and odd functions. Basically, the derivative of a point [ie. (4,2)] on a even function will result in a derivative that is equal, but opposite at the point that is opposite about the y-axis [(-4,2)]. When dealing with an odd function, the derivative at one point [ie. (1,3)] will be the exact same as the derivative of the point opposite about the origin [(-4,-2)].
Finally, we finished with two questions that were connected. Question Five was the last one we worked with, but has the same rule as Question Six which is to be completed and posted in the comments box this evening. For an explanation of how to do these two, click here.
Well, that concludes my scribe post for the night. I shall post my answer to the Homework question in the comment box now that at least some people have tried to answer it.
Don't Forget:
-PRETEST TOMORROW!!!
-TEST ON FRIDAY!!!
-B.O.B. BEFORE THE TEST (excluding Aichelle as she has already done so)
LAST BUT NOT LEAST, THE NEXT SCRIBE WILL BEEEEEEE:
...
...
...
once again I pick GREY M!!!
good night =D
Well, this class was a "workshop" class, so we got into our groups and took a look at the first question which seemed very familiar. It has to do with the graph of the height of a thrown ball over time (Grade 11 Pre-Cal question). However he added the velocity of the ball which is the rate of change (a.k.a. derivative) of the function. Then we were asked a series of questions about the scenario. For question and answer, click the "first question link".
Question Two basically asked us to find the slopes of the tangent at each point and list them in increasing order.
The third question was a little interesting because it gave us a told us that an original function was transformed by with a vertical shift. Then it asked us about how the derivative of the transformed function related to that of the original function. Well, when a function is shifted vertically, the vertical and horizontal scale remains the same (same size), as does the shape and it's position along the x-axis (x coordinates). The only thing that differs between it and it's original function is the y coordinates. Now, we know that a derivative is a graph of the slope of the tangents of the parent function vs. the points along the x-axis. So, since the shape, size, and x coordinates of the transformed function do not change, the slope of tangent at each point remain the same meaning the derivatives are the same!!!.
When Mr. K. clicked the next slide button on the SmartBoard Question Four came on the screen. This dealt with the derivatives of even and odd functions. Basically, the derivative of a point [ie. (4,2)] on a even function will result in a derivative that is equal, but opposite at the point that is opposite about the y-axis [(-4,2)]. When dealing with an odd function, the derivative at one point [ie. (1,3)] will be the exact same as the derivative of the point opposite about the origin [(-4,-2)].
Finally, we finished with two questions that were connected. Question Five was the last one we worked with, but has the same rule as Question Six which is to be completed and posted in the comments box this evening. For an explanation of how to do these two, click here.
Well, that concludes my scribe post for the night. I shall post my answer to the Homework question in the comment box now that at least some people have tried to answer it.
Don't Forget:
-PRETEST TOMORROW!!!
-TEST ON FRIDAY!!!
-B.O.B. BEFORE THE TEST (excluding Aichelle as she has already done so)
LAST BUT NOT LEAST, THE NEXT SCRIBE WILL BEEEEEEE:
...
...
...
once again I pick GREY M!!!
good night =D
Tuesday, October 16, 2007
Scribe version 3: Continuing our discussion of continuity
Well, we started off today's class by forming different groups and working on two problem solving questions about derivatives and continuity.
For the first part of the first question G`(t) represents the exact definition of a derivative so in this case we can see that G(x+h)=log((9+Δx)+1). Therefore G(x)=log (x+1). It can be seen on the second slide.
For the second part of the question since x=9 ,therefore a=9 just by looking at how g`(t) represent the exact definition of a derivative.
Now, for the second question it will be a nonremovable discontinuity. WHY? It is a nonremovable discontinuity because a limit has to exist in order for this to be removable. It can be seen on the third slide.
For the second part of the question the answer would have to be NO, because if you buy more than 3 bags of grain from this company the price jumps up. As it can be seen on the graph of the piece wise function that after exceeding three bags of grain the behavior of the function changes for being linear into being quadratic.
Then after doing those two questions we spent the rest of our time, taking a quiz at visual calculus.
Thats it for today. Our homework for tonight is to start doing the supplementary questions for chapter 2. The next scribe will be......(drum rolls)...CRAIG. (sorry but i have to pick someone)
For the first part of the first question G`(t) represents the exact definition of a derivative so in this case we can see that G(x+h)=log((9+Δx)+1). Therefore G(x)=log (x+1). It can be seen on the second slide.
For the second part of the question since x=9 ,therefore a=9 just by looking at how g`(t) represent the exact definition of a derivative.
Now, for the second question it will be a nonremovable discontinuity. WHY? It is a nonremovable discontinuity because a limit has to exist in order for this to be removable. It can be seen on the third slide.
For the second part of the question the answer would have to be NO, because if you buy more than 3 bags of grain from this company the price jumps up. As it can be seen on the graph of the piece wise function that after exceeding three bags of grain the behavior of the function changes for being linear into being quadratic.
Then after doing those two questions we spent the rest of our time, taking a quiz at visual calculus.
Thats it for today. Our homework for tonight is to start doing the supplementary questions for chapter 2. The next scribe will be......(drum rolls)...CRAIG. (sorry but i have to pick someone)
Monday, October 15, 2007
Derivative Assignment
Well here is my derivative assignment finally being posted up on the blog. 
For the answers to my derivative question, I will create a comment on this very post detailing the answers for my derivative question.
For the answers to my derivative question, I will create a comment on this very post detailing the answers for my derivative question.
Scribe post # 3
[filling in for Mark, he will be tomorrow's scribe]
In today's class we talked about the assignment we had on Friday, which was to create a problem identifying which 3 out of the 4 functions were related. Next, we talked about the hits we are getting from around the world. The map can be seen on the right, if you scroll down. After that, we searched for a wonderful picture to start of the slides with. It's the picture with the swirly smarties! Yay! Smarties = Yummy!
Mr. K. then brought up the term continuous functions. He asked us what we thought it meant. After a short discussion (and spelling analysis) about continuous functions we came up with an informal definition. We said that a continuous function is a function that we can draw without lifting the pencil off the paper at any point. If you cannot draw the graph without lifting the pencil then it is not continuous. We also concluded that all polynomial functions are continuous functions (X², X³...). Functions like 1/X are not continuous. Absolute value functions can be continuous depending on the kind of function. When a function contains a hole/corner/cusp it is not a continuous function. If a function is differentiable then it is continuous. However, if the function is continuous it does not mean it is differentiable.
We also learned how to Factor a Difference of Cubes and Factor By Grouping.
***Factoring By Grouping should be learned in grade ten, although it is not in the curriculum. Since a number of us didn't learn it, we were taught/re-taught it today.
See slide 10.
On slide eleven we talked about the Intermediate Value Theorem and Corollary. A corollary means it does not need to be proven; follows naturally. On slide twelve, we talked about the Extreme Value Theorem. We said that if a function has a solution it has a root and if it is continuous it has to have a min and a max aka extreme values.
Homework: 2.8 ODDS up to 13 + 2
Tomorrow's scribe will be Mark.
***sorry if I did anything wrong...please tell me if I did so I can fix it.
In today's class we talked about the assignment we had on Friday, which was to create a problem identifying which 3 out of the 4 functions were related. Next, we talked about the hits we are getting from around the world. The map can be seen on the right, if you scroll down. After that, we searched for a wonderful picture to start of the slides with. It's the picture with the swirly smarties! Yay! Smarties = Yummy!
Mr. K. then brought up the term continuous functions. He asked us what we thought it meant. After a short discussion (and spelling analysis) about continuous functions we came up with an informal definition. We said that a continuous function is a function that we can draw without lifting the pencil off the paper at any point. If you cannot draw the graph without lifting the pencil then it is not continuous. We also concluded that all polynomial functions are continuous functions (X², X³...). Functions like 1/X are not continuous. Absolute value functions can be continuous depending on the kind of function. When a function contains a hole/corner/cusp it is not a continuous function. If a function is differentiable then it is continuous. However, if the function is continuous it does not mean it is differentiable.
We also learned how to Factor a Difference of Cubes and Factor By Grouping.
***Factoring By Grouping should be learned in grade ten, although it is not in the curriculum. Since a number of us didn't learn it, we were taught/re-taught it today.
See slide 10.
On slide eleven we talked about the Intermediate Value Theorem and Corollary. A corollary means it does not need to be proven; follows naturally. On slide twelve, we talked about the Extreme Value Theorem. We said that if a function has a solution it has a root and if it is continuous it has to have a min and a max aka extreme values.
Homework: 2.8 ODDS up to 13 + 2
Tomorrow's scribe will be Mark.
***sorry if I did anything wrong...please tell me if I did so I can fix it.
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