Hello! On our blog, I am known as
Tim-math-y, and I will be your scribe for today's lessons.
Introduction:We started off the class with a brief discussion on our
del.icio.us accounts and homework. We are to find, with effort, atleast one site that we can learn from and that can be leveled as a quality find. Then we 'tag' it with:
cal45sw08, so that it will be added to our blog's bucket. Remember that it may not only aid in developing our learning outside the classroom but also, may prove to be great resources for others reading our blog.
Sweeping that discussion aside, we started off our
pre-test on the unit of
limits! The pre-test consisted of 5 questions in total. For those who do not know the procedures of a pre-test, it is an effective practice worth marks where a short test is written. After a set test-writing duration, we are placed into even groups where we share our answers to compile the best solutions onto one paper, as a team. This individual test paper is handed in before Mr. K reveals and explains the correct solutions.
The Pre-test:As mentioned earlier, this Pre-test consisted of 5 questions: 2 multiple choice questions, 2 short answer questions (where work was required to show), and 1 long answer question.
The first question included an
error that stumped everyone. The
x^4 in the numerator was supposed to be
x^2. Because of this unintentional error, this question was
ommitted, as far as marks go.
However, this question could still be solved by exploring the function. This is shown on the slide. First, we notice that there is a
vertical asymptote at x = 4 (Remember that when a question is asking for a limit, it is essentially looking for a horizontal asymptote).
By creating a
number line, one will find that as 'x' approaches 4 from the
negative side, the function goes to
positive infinity. One would also find that as 'x' approaches 4 from the
positive side, the function goes to
negative infinity. Because of this occurrence,
the limit, as 'x' approaches 4 from the positive and negative side DOES NOT EXIST.
This question is simply a give-away, as many may describe it.
As 'x' approaches the value of 1 from the positive and negative side, the value is 1.
A number of groups faltered on this question simply because of the nature of previous 'short answer' questions. In the past, short answer questions were marked based on the final answer only, with a chance to earn partial marks for work shown. However, this question stated: Evaluate using the
Limit Theorems, upon which many did not. This question is extremely simple yet painful. As long as you know your limit theorems, you should be fine. Listed are the
limit theorems from
[visual calculus]:
These are the main limit theorems we are required to know.
To start off this question, we chose to solve for the
horizontal asymptote first. By
dividing each term by the variable with the highest degree, we found that the
horizontal asymptote y=0, when the value of
'x' approaches infinity (any number divided by infinity is extremely close to zero, therefore in this method, terms are reduced substantially).
Next we solved for the
vertical asymptotes. This is done by
factoring the denominator and
solving for restrictions (the denominator can not equal to zero). We found the vertical asymptotes to be @
x=-9, 0.
Finally, to help visualize the
graph and sketch it, a simple method of finding out the positions around the asymptotes is by creating a
number line:
- As 'x' approaches -9 from the negative side, the limit is +infinity
- As 'x' approaches -9 from the positive side, the limit is -infinity
- As 'x' approaches 0 from the negative side, the limit is -infinity
- As 'x' approaches 0 from the positive side, the limit is +infinity
Thus, the graph can be sketched.
In the final question, we started off by running the
piece-wise function through the
three steps of continuity testing.
- Does f(a) exist?
- Does the limit as 'x' approaches 'a' exist?
- Does f(a) = L?
If not, the function is discontinuous.
- f(a) = f(2)
f(2) = 2 - the limit as 'x' approaches 2 is 5
- f(2) does not equal L: 2 does not equal 5
Therefore, this piece-wise function is not continuous. By discovering this, we found that this function is a
removable discontuity.
Finally, we had to sketch this piece-wise function. The graph maintains the
shape of (x+3). However, it has a
hole at x = 2 because there was a
reduction in the factors of (x-2).
Remember: when there is a reduction in factors, there is a hole at that point rather than an asymptote. Because the function of f(x) has a value of 2 @ x = 2, there is a black dot at that location.
The Conclusion:Well that was our pre-test! To sum things up, there were multiple things that should be remembered.
- A limit as 'x' approaches a value from both sides must meet at the same point, otherwise, the limit does not exist
- Remember how to solve using the painful work of writing out all of the evaluation steps using the limit theorems
- A number line really helps in determining the shape of the function
- Remember the three steps to testing continuity
- When factors reduce a restriction in the denominator, there is a hole at that value of 'x' rather than a vertical asymptote
I hope this scribe helped any of the readers! There will be a test on
wednesday so
DON'T DON'T DON'T DON'T DON'T FORGET TO "BOB" !
Good luck everyone on the test! Do not forget to study either! =) Have a great night everyone.
OoOoooOOo! And the scribe for the next class will be: (Give me a sec while I find the scribe list)
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John D. !!!!!
