Scribe Post: Day 13

HI everyone! It's Katrin and I will be your scriber for today! :) Sorry for posting this super late!!! I took a nap after work and ended up waking up late!!! =)

Today, Mr. K started talking about Calculus AB and Calculus BC. He said that Calculus AB (the course we're all taking right now) has about nine chapters in the textbook that we're going to take up. Calculus BC is some other math course. Then, he started talking about the "Complete Solution Guide," a book that consists of answers from the text. He said that we can look at it at a regular basis for reference. He placed that on top of the filing cabinet by the window. He then started talking about the TEST!!!

The test consists of 3 parts:

1) blogging on blogging (BOB) <-- 1 mark (i think)
--Have you blogged before this test? will be asked...

2) multiple choice

3) long answers


***Test is going to be by the end of the week!!! Friday, I think!!!! *** NOT ANYMORE!!! Mr. K changed it to Monday!!!***


Mr. K continued and said that the post on the blog will have a series of questions.

IMPORTANT!

Mr. K also noted that the blogging on blogging (short, BOB) MUST BE LABELED when writing a post!!! Label post ---> BOB

and

Make the post before the test!!! (preferably days before the test and do sooner than later!)


We then did our daily questions....

1) If a ploynomial function , f, of degree of 3 has the numbers -6, 4 and 5 as zeros. If f(1)=210, determine the formula for f.

I started off by making the equation f(x) = (x+6) (x-4) (x-5)

----> remember to put opposite signs for -6, 4 and 5 b/c they're zeros!

Mr. K then taught me to make k as the "constant" and plug it in front of the zeros in order to have f(1)=210.

like this,

f(x) = k (x+6) (x-4) (x-5)
210= k (1+6) (1-4) (1-5)
210= k (7) (-3) (-4)
210 = 84k
84....... 84
2.5=k

therefore, we apply f(1)=210 in the equation...

f(x) = k (x+6) (x-4) (x-5)
f(1) = 2.5 (x+6) (x-4) (x-5)
f(1)= 2.5 (1+6) (1-4) (1-5)
f(1)= 2.5 (7) (-3) (-4)
f(1) = 210


2) A rational function g has the line x=2 and x=-2 as vertical asymptotes, the line y=4 as a horizontal asymptote and the numbers 1 and 3 as zeros. Find a formula for g.

clue used 1st: roots --> 1 and 3

g(x)=(x-1)(x-3)

---> goes in the numerator (above line)


clue used 2nd: vertical asymptotes x=2 and x=-2

g(x)=(x-1)(x-3)
..........(x-2)(x+2) <----(x² -4)

---> goes in the denominator (below line)


clue used 3rd: horizontal asymptote y=4

g(x)=4(x-1)(x-3)
..........(x-2)(x+2)

---> you have to look at the degree of the numerator and the degree of the
denominator...

---> if the degree of the numerator is smaller than the degree of the denominator
and if the degrees are the same in both the numerator and the denominator,
then there is a horizontal asymptote...
but...
if the degree of the numerator is greater than the degree of the denominator,
then there is NO horizontal asymptote...

---> highest power of x in numerator and divide by highest power of x in denominator...

..............the result of doing the arrowed (--->) steps is horizontal asymptote=1
BUT...
..............we don't want the horizontal asympote to equal 1, we want it to equal 4 since in the question it said that y=4 for the horiontal asymptote.

therefore, we put 4 as constant... like such:

g(x)=4(x-1)(x-3) which can be written as g(x)=x²-4x+3
........(x-2)(x+2)..................................................x²-4


3) Let f(x) = ax+b,
.....................cx-a
where a, b and c are positive contants. Show that f is one-to-one and f is its own inverse.

f(f^-1(x))=x

therefore you plug in f(x) in the equation itself... like such:

f(f(x))=a (ax+b) + b
_______(cx-a)____
_____c (ax+b) - a
______(cx-a)
f(f(x))=a(ax+b) + b(cx-a)
......... ____cx-a_______
_____c(ax+b) - a(cx-a)
.......................cx-a
f(f(x))=a(ax+b) + b(cx-a) x ____ cx-a______
....................cx-a__________c(ax+b) - a(cx-a)
f(f(x))=a(ax+b) + b(cx-a)
..............(cx+b) - a(cx-a)
f(f(x))=ax² + ab + bcx - ab
.............acx + cb - acx + a²
f(f(x))=a²x + bcx
...............cb+a²
f(f(x))=x(a² + bc)
.............a² + cb
f(f(x))=x

therefore when the final answer is equal to x, then it proves that f is one-to-one and f is its own inverse.


After thoose three questions Mr. K talked about Composition of Functions

f(x)=x - 1.........................g(x)=x - 2
......x + 4.................................. x

domain of f cannot = -4
domain of g cannot = 0

we have to plug in the function g(x) into f(x) so wherever you see x in f(x) you plug in the function of g(x)... like such...

f(x)=x - 1........................g(x)=x - 2
.........x + 4.................................x

f(g(x))=(x-2) - 1
....... _ (x)______
..............(x-2) + 4
................(x)
f(g(x))=(x-2) - (x)
............__x__ x___
..............(x-2) + 4x
.................x.......x
f(g(x))=-2__
..............5x-2 , x cannot = 2/5 and 0


Exclude:

5x-2=0
5x=2
x=2/5

and

x-2 = -4
x
x-2 = -4x
5x-2=0


Well... i don't really get composition of functions so i'm don' t really know how to explain this
much....

Homework: 1.9 odd #'s and study for quiz!!!

This is my scribe =) and I'm very very sorry for posting this super late!!! Well, i hope i covered everything we did in class today! hmmm... if i did anything wrong please feel free to comment since i'm not 100% sure about everything i posted... and oh yeah... the next scribe for tommorow will be CHARLENE!!! hehe i know u know already... but i know i still have to post it anyway! ;)