Hello everybody! I'm MrSiwWy, and today, as the scribe, I will convey the events and educational endeavours that took place during today's class. Nothing more to say but that I shall continue on with this scribe!
The class began with Mr. K commenting on why we all should have posted our answers for the previous class' assignment so that the rest of the class can comment on each groups answers by tomorrow. Then we proceeded to our lesson plan for the day.
Operations on Functions
We began by conversing upon the utilization of numbers in terms of our progression of mathematical knowledge. Not only can be implement operations on these numbers, but on the functions used to represent and manipulate numbers as well. Functions and numbers can cumulatively be manipulated for any specific use, as we have learned through the wonders of mathematics.
Mr. K continued on with the slides on the smart board by displaying three pairs of functions to us. We were then given time to implement the specific operations within each pair, and were also told to specify the domain for the output function. These function pairs were given as:
f(x) = x2 , g(x) = x + 5.
f(x) = √(1+x) , g(x) = √(1-x).
f(x) = √x , g(x) = ∛x.
The operations we were asked to perform on the above pairs of functions are as follows:
f + g (which can also be written as (f + g)(x)) .
f - g (which can also be written as (f - g)(x)) .
f g (which can also be written as (f*g)(x)) .
f/g (which can also be written as (f/g)(x)) .
Once the class was given sufficient time (of course deemed sufficient by Mr. K) to complete the above, we all went over the answers to each with Mr. K as a class. The answers as well as the actual slides can be seen here (have to wait for Mr. K to post slides). As we analyzed the answers to the given problems, Mr. K had a unique and very intriguing utility to show us on the net. This site of course being http://www.math4mobile.com/.
The site offers a utility that is not only accessible (via an example of how it works and how to use it) on the website, but the tool is available for download to your very own cell phone. The tool transforms your cell phone from the once inept, feeble calculating device into a mathematical modeling and graphing calculator extraordinaire. Mr. K then briefly demonstrated how the graph would portray an example function. He also vaguely noted that in the example function we used (x2 +x + 5), altering the magnitude of the coefficient for the term containing x1 would cause an "oblique translation", or rather a diagonal shift. Though, we did not establish the significance of this concept, and we did not go further into detail concerning the matter so it might not be important as of yet.
After the short math4mobile presentation, Mr. K then transitioned the class to continue with our lesson by presenting us a question. The question proposed that we determine the turning point of the function f within quadrant II for:
f(x) = x3 - x2 - 3x + 7.
To determine this, it's simply achievable by graphing the given function on a calculator and then finding the local max for the function, seeing as if you examine the graph, the point at which the function will turn (change from increasing to decreasing, or vice versa) found in quadrant two is a maximum point. The point was determined to be (-0.72, 8.27).
As Mr. K was dissecting the given graph up on the smart board, he then digressed and began explaining that specific functions can represent other functions on specific intervals. If you imagine viewing the given function, or any cubic of a similar shape, from one turning point to another, you might be able to visualize many functions that behave similarly on the same domain, such as sin(x). Polynomials are often used to approximate many functions on specific domains, seeing as polynomial functions are often more simplistic and are relatively less complicated to conduct calculations on. After we absorbed the useful tidbit conveyed to us, we continued our lesson with yet another question.
This question, required us to analyze the function and determine not only the domain, but the range and the left and right-hand asymptotes of the following function:
f(x) = (x2 - x)/(x2 - 1).
We were given time to perform such duties, and we determined the following:
domain: {x l xeR;x≠±1} or (-∞,-1)U(1,∞).
range: {y l yeR;y≠1} or (-∞,1)U(1,∞).
left-hand asymptote: 1.
right-hand asymptote: 1.
As Mr. K explained why each answer was true, we somewhat diverged to a quite elaborate exploration of horizontal asymptotes. We in turn learned once again how to determine the horizontal asymptotes for a function. To determine such, we must analyze the degree's of both the numerator and the denominator. Here are some rules which Mr. K projected to us to signify the relevance of finding these degree's:
- If the degree of the numerator and the denominator are equal, then to determine the horizontal asymptote, you must simply divide the leading coefficients within the function. This meaning that if the degree's are equal, divide the leading coefficient of the numerator by the leading coefficient of the denominator to determine the horizontal asymptote of the function.
- If the degree of the numerator is greater than that of the denominator, then there is no horizontal asymptote for the function.
- If the degree of the numerator is less than that of the denominator, then the asymptote can be found at y=0. The due to the fact that is that when a horizontal asymptote occurs, we can then use the fact that the x value for that asymptote will be approaching infinity. If we then substitute infinity into the given equation, the resultant will yield a final value of 0.
And as dismissal became imminent, we briefly went over what classifies functions as either even or odd. Here are some quick reminders concerning both.
For an even function, the function must be symmetrical about the y-axis. This meaning that the function lying left of the y-axis, can be flipped and will give exactly what is present to the right of the y-axis.
Algebraically, a function can be determined to be even if:
f(x) = f(-x)
ex. f(x) = x2 + 2
*To check if the function is even, then substitute -x into each x in the equation. If the new equation is equivalent to the original, the function is said to be even.
f(-x) = (-x)2 + 2
f(-x) = x2 + 2
For an odd function, the function must be symmetrical about the origin. This meaning that the function can be flipped over the x-axis then the y-axis, or rotated 180o and the graph will stay the exact same.
Algebraically, a function can be determined to be odd if:
-f(x) = f(-x)
ex. f(x) = x3
*To check if the function is odd, then substitute the -x into each x in the equation to get f(-x). Then, multiply the original function by -1 to determine -f(x). If both functions are equal to each other, then the function is odd.
f(-x) = (-x)3
f(-x) = -x3
-f(x) = -(x3)
-f(x) = -x3
Since -f(x) = f(-x), the function can be deemed to be odd.
Now, that was all that we did today in class! Don't forget to complete your homework (a question that I will be posting subsequent to this very post) as well as the exercise 1.8 questions 1, 3, 5, 6, 11, 19 and 21. Have a great night everyone! Hope this post helped anyone who missed class (though I don't think anyone did). Oh yes, and the next scribe will be... surprise surprise...
The class began with Mr. K commenting on why we all should have posted our answers for the previous class' assignment so that the rest of the class can comment on each groups answers by tomorrow. Then we proceeded to our lesson plan for the day.
Operations on Functions
We began by conversing upon the utilization of numbers in terms of our progression of mathematical knowledge. Not only can be implement operations on these numbers, but on the functions used to represent and manipulate numbers as well. Functions and numbers can cumulatively be manipulated for any specific use, as we have learned through the wonders of mathematics.
Mr. K continued on with the slides on the smart board by displaying three pairs of functions to us. We were then given time to implement the specific operations within each pair, and were also told to specify the domain for the output function. These function pairs were given as:
f(x) = x2 , g(x) = x + 5.
f(x) = √(1+x) , g(x) = √(1-x).
f(x) = √x , g(x) = ∛x.
The operations we were asked to perform on the above pairs of functions are as follows:
f + g (which can also be written as (f + g)(x)) .
f - g (which can also be written as (f - g)(x)) .
f g (which can also be written as (f*g)(x)) .
f/g (which can also be written as (f/g)(x)) .
Once the class was given sufficient time (of course deemed sufficient by Mr. K) to complete the above, we all went over the answers to each with Mr. K as a class. The answers as well as the actual slides can be seen here (have to wait for Mr. K to post slides). As we analyzed the answers to the given problems, Mr. K had a unique and very intriguing utility to show us on the net. This site of course being http://www.math4mobile.com/.
The site offers a utility that is not only accessible (via an example of how it works and how to use it) on the website, but the tool is available for download to your very own cell phone. The tool transforms your cell phone from the once inept, feeble calculating device into a mathematical modeling and graphing calculator extraordinaire. Mr. K then briefly demonstrated how the graph would portray an example function. He also vaguely noted that in the example function we used (x2 +x + 5), altering the magnitude of the coefficient for the term containing x1 would cause an "oblique translation", or rather a diagonal shift. Though, we did not establish the significance of this concept, and we did not go further into detail concerning the matter so it might not be important as of yet.
After the short math4mobile presentation, Mr. K then transitioned the class to continue with our lesson by presenting us a question. The question proposed that we determine the turning point of the function f within quadrant II for:
f(x) = x3 - x2 - 3x + 7.
To determine this, it's simply achievable by graphing the given function on a calculator and then finding the local max for the function, seeing as if you examine the graph, the point at which the function will turn (change from increasing to decreasing, or vice versa) found in quadrant two is a maximum point. The point was determined to be (-0.72, 8.27).
As Mr. K was dissecting the given graph up on the smart board, he then digressed and began explaining that specific functions can represent other functions on specific intervals. If you imagine viewing the given function, or any cubic of a similar shape, from one turning point to another, you might be able to visualize many functions that behave similarly on the same domain, such as sin(x). Polynomials are often used to approximate many functions on specific domains, seeing as polynomial functions are often more simplistic and are relatively less complicated to conduct calculations on. After we absorbed the useful tidbit conveyed to us, we continued our lesson with yet another question.
This question, required us to analyze the function and determine not only the domain, but the range and the left and right-hand asymptotes of the following function:
f(x) = (x2 - x)/(x2 - 1).
We were given time to perform such duties, and we determined the following:
domain: {x l xeR;x≠±1} or (-∞,-1)U(1,∞).
range: {y l yeR;y≠1} or (-∞,1)U(1,∞).
left-hand asymptote: 1.
right-hand asymptote: 1.
As Mr. K explained why each answer was true, we somewhat diverged to a quite elaborate exploration of horizontal asymptotes. We in turn learned once again how to determine the horizontal asymptotes for a function. To determine such, we must analyze the degree's of both the numerator and the denominator. Here are some rules which Mr. K projected to us to signify the relevance of finding these degree's:
- If the degree of the numerator and the denominator are equal, then to determine the horizontal asymptote, you must simply divide the leading coefficients within the function. This meaning that if the degree's are equal, divide the leading coefficient of the numerator by the leading coefficient of the denominator to determine the horizontal asymptote of the function.
- If the degree of the numerator is greater than that of the denominator, then there is no horizontal asymptote for the function.
- If the degree of the numerator is less than that of the denominator, then the asymptote can be found at y=0. The due to the fact that is that when a horizontal asymptote occurs, we can then use the fact that the x value for that asymptote will be approaching infinity. If we then substitute infinity into the given equation, the resultant will yield a final value of 0.
And as dismissal became imminent, we briefly went over what classifies functions as either even or odd. Here are some quick reminders concerning both.
For an even function, the function must be symmetrical about the y-axis. This meaning that the function lying left of the y-axis, can be flipped and will give exactly what is present to the right of the y-axis.
Algebraically, a function can be determined to be even if:
f(x) = f(-x)
ex. f(x) = x2 + 2
*To check if the function is even, then substitute -x into each x in the equation. If the new equation is equivalent to the original, the function is said to be even.
f(-x) = (-x)2 + 2
f(-x) = x2 + 2
For an odd function, the function must be symmetrical about the origin. This meaning that the function can be flipped over the x-axis then the y-axis, or rotated 180o and the graph will stay the exact same.
Algebraically, a function can be determined to be odd if:
-f(x) = f(-x)
ex. f(x) = x3
*To check if the function is odd, then substitute the -x into each x in the equation to get f(-x). Then, multiply the original function by -1 to determine -f(x). If both functions are equal to each other, then the function is odd.
f(-x) = (-x)3
f(-x) = -x3
-f(x) = -(x3)
-f(x) = -x3
Since -f(x) = f(-x), the function can be deemed to be odd.
Now, that was all that we did today in class! Don't forget to complete your homework (a question that I will be posting subsequent to this very post) as well as the exercise 1.8 questions 1, 3, 5, 6, 11, 19 and 21. Have a great night everyone! Hope this post helped anyone who missed class (though I don't think anyone did). Oh yes, and the next scribe will be... surprise surprise...
DINO!!!
4 comments:
Excellent Scribe post! Bravo.. /clap
Very good Chris, used it to do a question in the homework as soon as it was up. Don't think you missed anything in that class.
Hi Mr. Siwwy,
I'm amazed at the depth and detail of your post. As I read through it, I found myself sitting among you in the classroom - not an easy feat.Great job!
Just something to think about: how do you think technology affects your learning (re: calculus), and does it really make a difference? If so, in what way(s)?
Again, thanks for the good scribe post.
Take care,
C.
haha geez Chris..lol by the third paragraph I was like too many bigs words..lol=Þ haha I wish my vocabulary was like that..but unfortunately...it's not haha by excellent post Chris! I enjoyed it! the explanations rocked! and the colours were sweet....ROCK ON!
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