3/17/2011 Review Day Blog

Today on St. Patricks Day i dont know how Mr. cope started off class because I was late, but when i arrived he was checking our 8.6 homework. We then went over our homework discussing problems having to deal with rationalizing the denominator. After we went over our homework problems Mr. Cope then handed out an 8.4 - 8.6 review sheet with practice problems on it. For the rest of class we worked on the problems on the review sheet to get us ready for out Chapter 8 test tomorrow. The basic ideas of each section are:

8.4 :
The basic idea of this section was just being able to find the Nth roots for radicals. The Root of a Power Theorem is n(radical sign) X^m = X^m/n. The conditions are x has to be greater then or equal to zero. N must be an integer greater then or equal to 2 and m must be an integer. Another important point is finding the geometric mean, which is just when you multiply all the N set numbers and then take the nth root of that number and thats your geometric mean. Lastly you have to be able to convert between rational exponents notation and radical notation.

8.5 :
The main ideas of 8.5 was to learn how to break down products with radicals and being able to multiply them. We learned how to simplify using the factor tree when you break the number in the product to its factors.

8.6 :
8.6 was the hardest lesson to learn this week. The main objective of 8.6 was to learn how to rationalize the denominator for quoteints with radicals. You carry out this task by multiplying the denominator by its conjugate and then foiling (similar to imaginary numbers).

Good luck on the test tomorrow everyone.

8-6 Quotients with Radicals

When we came into class on Wednesday, Mr. Cope had watched the video he recorded from the day before. After watching the video, he said he noticed two things: he answers his own questions sometimes, and that if you don't want to participate in class, then this class is fairly easy because you can just sit back and listen. He decided to take a new approach for the day to get everyone involved by using small erase boards and doing problems on them.

In today's lesson we learned about rationalizing the denominator. It is very similar to imaginary numbers if you look back to chapter six. It is the same concept, getting the radical out of the denominator. We started with easier examples:

3
--
√4 from here, you need to multiply the fraction by a WFOO (weird form of one)

3 √4 3 √4 3√4
--- x ---- = ----- = ----
√4 √4 √16 4


For the next examples, you need to multiply the numerator and the denominator by the conjugate, for example:

2 3 - √6 6 - 2√6
---- x -------- = --------
3 + √6 3- √6 3


Some of the harder examples, ones that you need to use information from both 8.5 and 8.6, look like this:

y^2 y^2
---- first, you need to simplify the denominator ------
√y^7 y^3√y


then, you need to get the radical out of the denominator
y^2 √y y ^2√y √y
--------- x ---- = ---------- = -------
y^3 √y √y y^4 y^2

We’re having a quiz on this section along with 8.4 and 8.5 tomorrow so hopefully this helps out!

8.5- Products with Radicals

Before class started today we found out Mr. Cope was video taping us. Then we went over the 8.4 homework answers. Then Mr. Cope gave us six square rooting problems to do. One of them was the √20. There are two methods to solve this. We can either use the method Mr. Cope taught us, or we can use the jailbreak method, otherwise known as the factor tree method. An example of the factor tree method is this:

√20
/ \
4 5
/ \
2 2

The bolded numbers are the leaves that you want to pick. There has to be two leaves that are the same before you can pick them. Since there are two 2’s that’s one numbers we can pick. However there is only one 5 leaf so that one has to stay on the tree. So the final answer is 2√5. This is just a warm up to the lesson ahead. The first example on the noted was this ^4√48p^11. This may look difficult but you need to just use the jailbreak method on √48.
√48
/ \
6 8
/ \ / \
3 2 2 4
/ \
2 2
The little 4 above the square root means that we need to find 4 pairs of a single number of the square root of 48. There are exactly 2 pairs of 2 so that and a 3 left over that we cannot pick from the tree so the answer will look like this so far 2√3. This is still not done because we still need to get rid of the p^11. Now write out 11 p’s.

Ppppppppppp

Now we need to find all the 2 pairs of p. So we will circle(bold) 4 p’s.

Ppppppppppp

There are two 2 pairs of 4 p’s and 3 p’s left over. Then we will write the final answer out like this. 2p^2 ^4√3p^3. This is read as 2p squared times 4 square roots of 3p cubed.
Another sample problem of using products with radicals is ^5√32y^9. First we need to use the factor tree method on √32.

√32
/ \
8 4
/ \ / \
4 2 2 2
/ \
2 2

There has to be 5 of a certain number and luckily there is five 2's. That means so far we have only one 2 in the answer. Next we find out how many y's are in the problem. There are 9 so there is only one group of five y's. There is 4 y's that have to stay so the final answer will be 2y^5√y^4. This is read two y to the fifth times the fifth sqaure root of y to the fourth.

Hopes this helps.

8.4

So today in class, we received our 8.1-8.3 quizzes and went over those. He said the class did well, so that was good!

Then, we went over today's lesson: Radical notation for Nth roots. We finished with enough time to complete our homework assignment (so there is no homework tonight except read this blog- which- probably no one will do:)

Now I will get to the point, and summarize the lesson. Today's point was to link the idea of a number raised to a fraction (25^1/2) to the idea of a number under a radical(√25). This involved reviewing the notation from chapter 7.

x^1/2 meant the positive square root of x.

x^1/n meant the positive nth root of x.

So then we were able to justify this: ^n √ x= x^1/n. Conditions: x must be a positive integer and n must be an integer greater than 2.

So here is an example which will probably make more sense then the above.

^3 √512... THINK: ?^3=512... ANSWER: 8

After this we did a little more complicated problems but we just had to apply our knowledge of properties we learned back in chapter 7.

After practicing this, we came to the Root of a Power Theorem:

^n √x^m = X^m/n. Conditions: x must be a positive integer, n must be an integer greater than 2 and m must be an integer. After understanding this we applied it in this problem: ^3 √X^12... THINK: (X^12)^1/3... ANSWER: X^4

After this, we went to the geometric mean. This was a very easy concept. If we have data with n values, we multiply all of the numbers then take the ^n square root of that product. I will not waste space by giving a lengthy example so I will make it super easy: data is : 1,3,4,6... product is 72. then take the 4th root of 72. Answer is 8. 48.. easy right? :)

Then came the concept of roots of roots. √√x hmm... all you do here is however many radical signs there are, you just multiply them. In this case there are 3. So that's like doing 1/2*1/2*1/2 which is 1/8. so x^1/8. But now is the A-HA moment where we can use what we learned today. We now can call X^1/8 the ^8√x

example: √L^4...THINK: (L^4)^1/2... So 4/2=2... L^2.

So that is the lesson in a nutshell! The notes are very meaty and helpful if the above summarization didn't work for you. Also, Mr. Cope stressed that his door is wide open if anyone is struggling.

03/13/11 Thursday & Friday 8.1-8.3 Review

I finally was able to get into the blog today, props to Cope. On thursday we started off doing the daily review of our homework and asking questions about it. After about twenty-five minutes of homework review, we started to review for the quiz on friday over 8.1-8.3. Cope gave us a review packet that we soon started to work on with partners.

8.1 Review Packet

This section mainly focused on the composition of functions. Earlier in the year we learned f(x) notation. for example we learned that when we see an f(x) notation, whatever is in the () gets inputed into the function. however, this came back again in 8.1, although it is more complicated than that. the more complicated part is when we are given a function and then we have to define g(x) and f(x) at the same time.

-ie.) function: g(x)=2x and f(x)=2x+10

then we are dealt with f(g(4)). the first thing we do is plug in 4 for x in the g(x) function.

g(x)=2(4)

x=8

next we plug the solution to that function into the next one f(x).

f(8)= 2(8)+10

f(8)=26

we then have our answer f(g(4))=26

composite- g~f of two function f and g is the function that maps x onto g(f(x)), and whose domain is the set of all values in the domain of f for which f(x) is in the domain of g.

8.2 Review Packet

This section mainly focused on the relations of inverses. The three biggest steps to take from this section are:

1.) Switch x and y and solve for y, when you are told to find the inverse of an f(x) function

2.) The inverse of a graph is found by reflecting it over the line y=x.

3.) The domain of g is the range of f. the range of g is the domain of f. (opposites)\

example 1

let f(x)=3x+5. find the inverse of f.

first step switch x and y. x=3y+5

second step slove for y. x-5=3y

x-5=y

3

example 2

the line y=x is a diagonal line that crosses the origin through the first i and iii quadrants. to find the inverse of a graph all you do is reflect it over the line y=x.

example 3

when f=(3,6),(2,4),(3,7),(1,5)

the domain of f is (3,2,1)

the range of f is (6,4,7,5)

the domain of the inverse of f is (6,4,7,5)

the range of the inverse of f is (3,2,1)

we know that f is not a function because 3 is a repeating x value

8.3 Review Packet

this section mainly focused on inverse functions. inverse of function are wriiten as f^-1. when the inverse of a function is graphed we draw horizontal lines through the graph, and if the lines intersect the graph more than once, it is not a function. to find the inverse to a function the steps we take are:

1.) insert y in place of f(x)

2.) switch x and y

3.)solve for y

example find the inverse of f(x)= 3x-2

step one-----> y=3x-2

step two----->x=3y-2

step three--->x+2=3y

x+2=y

3

On friday we came to class ready to take our quizzes on 8.1-8.3. first we went over our homework from the book by volunteering to put the problem on the board throughout the room, then we asked questions on the problems. Cope handed out the quizzes and we started to take them using the TI-83 calculators because the cas was not allowed. sorry for how long it took for me to get logged in correctly, but i hope this helped.PEACE OUT.

Remember non CAS Calculator

Remember to bring a non-CAS calculator to class tomorrow for our quiz.  A TI-83 or 84 is fine or a simple 4-function calculator.  I am available before school if you need help with the concepts.

3/9/11 8.3 Properties of Inverse Functions

Today in class, we did the same old stuff. Checking over our answers and asking Cope for some explanation on how to do some problems from 8.2. Then, he told us why he was talking oddly. Simple explanation, he talked more than he usually does between teaching and tutoring. We didn't get into one of our usual philosophical conversations we usually do. Then we got onto the lesson with 8.3. We learned more about the secrets of inverses of functions. We learned that plugging in a function with its inverse would produce what you put in in the first place.

Here is an example if you had trouble understanding what I am saying:
g(x) is the inverse of f(x). So, f(g(x)=x and g(f(x))=x. This also can work with almost any number you put in. So, functions undo each other. This example is also known as the INVERSE FUNCTION THEOREM.

Finally, we learned that the graph of x^6 can be a function but only if x is greater than or equal to zero. Also, x^7 is a function either way.

Last but not least, FREE EARL!