Today in class we went over the inverse function. We first went over how to find the inverse of a single function. For example: refer to the function f(x)=2x+5

Step 1: replace the f(x) with the letter y;
y=2x+5
Step 2: swap x with y;
x=2y+5
Step 3: solve for y
(x-5)/2=y

We then learned that the definition of an inverse could be expressed as (fof^-1)(x)=x
and (f^-1of)(x)=x. Refer to the function above for this example on verifying functions as an inverse.

f(f^-1(x))=f(x-5)/2
=2(x-5)/2+5
=x-5+5
=x; because we got x as the answer, we therefore verified this function as
an inverse!
Just to double check we will also prove this function as an inverse with the second definition given above.

f^-1(f(x))=f^-1(2x+5)
=(2x+5-5)/2
=2x/2
=x; because we got x as the answer, we therefore verified this function as
an inverse!

The two examples above show how to prove that a function has an inverse algebraically. We can also prove this by graphing. The way to check if a function has an inverse is if the inverse reflects y=x. What this is saying is that if f(2, 7) is on a graph than so is its inverse of f^-1(7, 2)

We then learned that some functions do not have inverses :( These functions include...
-absoloute value
-functions that don't pass the horizontal line test

We then learned a very helpful way to test if a function has a true inverse. You can check if a function is one-to-one to verify if it has an inverse.
-One-to-one means...
if and only if (iff) f(a)=f(b), implies a=b

Examples!
f(x)=x^2
a^2=b^2
a=-b or a=b, because a could equal b OR -b, this is not a one-to-one function

f(x)= 2x+5
f(a)= 2a+5; f(b)=2b+5
2a+5=2b+5
2a=2b
a=b; because we got one solution for a... which should always be b, this is a one-to-one function, so it has an inverse!

Tonight's homework is...
Assignment 8: 1.5 problems 16, 21, 31, 43, 46, 54, 56, 73, 81, 83, 85, 88, 92, 93, 96

Ok im going to make this blog post short and sweet. Today in class we learned about Composition. The easiest way to show you what that is, is by example. So here goes...

.

.

f(x)=x + 2 and g(x)=x^2+3

(f o g) (x)= f(g(x)) this means "f composed with g of x"

.

.

(f o g) (x)= f(g(x)

=f(x^2+3)

= (x^2+3) + 2

=x^2 + 5

.

(g o f)(x)= g(f(x))

=g (x+2)

=(x+2)^2 + 3

= x^2 + 4x + 7

.

These are examples of the reverse way of doing it. Knowing the solution and trying to find g(x) and f(x)

.

(f o g)(x)= x + 3√x+1

f(x)=x^2+3x+1

g(x)=√x

.

or

f(x)=x+3√x+1

g(x)=x

g(x)=x is a trivial solution, don't use them on the test, you will get marked down :(

.

the hw today is assigment 7 section 1.4

(46,47,49) (a & b), 52, 53, 55, 58, 59, 63, 75, 84, 86, 90, 99-101

.

plus a neat vid nothing to do with math

http://www.youtube.com/watch?v=iTnoUB8KXVI

My post probably wont be as crazy long as the last one, but today we talked about combinations of functions. We learned the first four combinations which are arithmetic including the sum, difference, product and quotient.

Let f and g be two functions with overlapping domains. Then, for all x common to both domains, the sum, difference, product, and quotient of f and g are defined as follows:

use f(x) = x + 2 & g(x) = x² + 3 as an example

sum: (f + g)(x) = f(x) + g(x)

(x + 2) + (x² + 3)
= x² + x + 5

difference: (f - g)(x) = f(x) - g(x)

(x + 2) - (x² + 3)
= -x² + x - 1

> note: it is important to remember the parentheses when finding the difference of functions.

product: (f g)(x) = f(x) * g(x)

(x + 2)(x² + 3)
= x³ + 2x² + 3x + 5

quotient: (f /g)(x) = f(x) / g(x)

(x + 2) / (x² + 3)

> note: g(x) ≠ 0

next we will look into compositions of functions, but until then the homework for tonight is assignment 6 -
1.3 # 42, 53, 59, 63
1.4 # 1-25 odd, 26, 106-110 even

also happy birthday to those born on 9.28 -- Spencer Rogers

enjoy this birthday wish from Lauren, and that new kid Lily. http://www.youtube.com/watch?v=8Q45N-oNu-4

Friday's lesson was all about the moving of graphs; translating, stretching, and reflecting. We first reviewed the parent functions of the graphs by going over what they look like. Here are the 6 he showed us:

y=x

y=x^2

y=x^3

y=|x|

y=

y=c (assuming that c=3)

You can manipulate any type of graph using the concept of f as a function of x (or any variables other than f and x). This is represented as f(x), and for the most part, replaces y. So a function such as y=x could also be written as f(x)=x because it is the function of whatever value x is. This is called function notation. We'll first look at translating graphs using the parent function f(x)=x^2.

Translation

Vertical Translation:

If you want to move the graph f(x)=x^2 up 1 unit, you would +1 on the end of the equation,

and because f(x)=x^2, you can replace x^2 with f(x) when using function notation, the process

would look something like this

Parent function Translated Equation

f(x)=x^2 y=f(x)+1

Translated Graph

To translate to the down simply subtract instead of adding (ex. y=f(x)-1). This means that the equation for vertical translation is y=f(x)+c

To move the graph to the right or left, you would use the same principal of function notation, only instead of changing the y value (f(x)) you are going to change the x value. This is easily done by inserting the value you want to move the graph by, right next to the x. There is only one change in the format of the two equations, and that is when manipulating a graph horizontally (in this case to the left or right), the sign is the opposite of what you would think. So if you're trying to move f(x)=x^2 to the right 2 units, the equation would be y=f(x-2) NOT y=f(x+2). The resulting graph would look like this:

To move the parent function to the left 2, the equation would be y=f(x+2), meaning that the equation for moving a graph horizontally is y=f(x-c)

Stretching

To make the graph narrower, you would multiply its parent function (f(x)) by however much narrower you would want to make it. For example, if you want to make it 2 times narrower, the equation would be y=2f(x). This is because you are making the parent function larger, resulting in a graph with the same x values, but with y values that move higher up on the graph. For example, to make an equation with the parent function of f(x)=x^2 2 times narrower, the equation would look like this: y=2f(x), and would yield this graph:

(the red being the parent function of f(x)=x^2 and the blue being y=2f(x))

To make a graph wider, you would multiply its x value by the amount you want enlarge it. Same basic concept as the making it narrower, so the equation if you want to make it 2 times wider would be y=f(2x). Taking all of this information into account, the equation for making graphs narrower by increasing its parent function is y=cf(x), and the equation to make a graph wider by increasing its x value is y=f(cx).

Reflecting

Reflecting a graph is the simplest one of the three. To reflect an image on the graph is basically like you are putting a mirror up to the image and it shows up on the other side. To reflect a graph vertically (meaning flip it upside down), you simply multiply the parent function (the f(x)) by negative 1. An equation with the parent function f(x)=x^2 upside down would be y=-f(x) and its graph would look like this

To reflect it to the right or left, you would multiply its x value by negative 1, making the equation look like this: y=f(-x). This is called horizontal reflection, and the equation for vertical reflection is y=-f(x). You can manipulate graphs by translating (shifting), stretching, or reflecting them. You may see one or all in a given equation, and can all be broken down as long as you remember the basic equations. A couple reminders; vertical changes effect the f(x), or y-coordinates, and horizontal changes effect the x, or x-coordinates. And finally, when horizontally changing, the things that happen inside the parenthesis (such as f(x-c)) will have an opposite effect than what you would expect. Goodnight and good luck.

September 21

Today we learned more about even and odd functions, specifically what the graphs of them look like.


To review, an odd function is when f(-x) = -f(x).  If the point (x,y) is on the graph so will the point (-x,-y).  A good example of an odd function is f(x) = x³.  Graphically, odd functions are symmetrical about the origin. If you were to draw a line from one point on the graph to the origin, and then continue that line for the same distance you would hit the opposite of the original point.  Above is an example of a graph of an odd function.













An even function is when f(-x) = f(x).  If the point (x,y) is on the graph, so will the point (-x,y).  A good example of an even function is f(x) = x².  If you were to plot an even function, it would be symmetrical about the y-axis.  An example of a graph of an even function:

Friday’s lesson was more about functions.

To review the problems we did in class:
1: If f(x) = x^2-2x find f(x-2)
F(x-2) = (x-2)^2 - 2(x-2)
= x^2- 4x + 4 - 2x + 2
= x^2 - 6x + 8

2: Domain of g(x) = (x^3-8)/√(x^2-9)
√(x^2-9) > 0
x = ± 3
D: (-∞, -3) U (3, ∞)

3: If a =2b^2 + 5 and b= c - 3, show a as a function of c
a = 2(c-3)^2 + 5
= 2(c^2 - 6c + 9) + 5
= 2c^2 - 12c + 18 +5
= 2c^2 - 12c +23

Later we learned the meaning of this statement:
A function F is increasing on an interval if, for any x and x_2 in the interval x < x_2 implies
f(x) < f (x_2).

This means than if a function is increasing in value on the graph of that function then the output of x must be lesser than the output of x_2 if x is in fact less than x_2.

For example: f(x) = x + 2, to show that this function is increasing we plug in two values.
f(3) = 3+2 f(4) = 4 + 2
f(3) = 5 f(4) = 6

Since 3 < 4 and f(3) < f(4), f(x) = x + 2 is an increasing function.

9/16/2010

Hey. So today's lesson was building on what we learned yesterday... Functions.

A function f from a set A to a set B is a relation that assigns to each element x in the set A exactly one element y in the set B. The set A is the domain (or set of inputs) of the functions f, and the set B contains the range (or set of outputs)

Functions are represented in 4 ways
       1) Verbally by a sentence that describes how the input variable is related to the output value
       2) Numerically by a table or a list of ordered pairs that matches input values with output values
       3) Graphically by points on a graph in a coordinate plane in which the input values are represented by horizontal axis and the output values are represented by the vertical axis
       4) Algebraically by an equation in two variables

X represents the independent variable and Y represents the dependent value.

There was homework D:    but it wasnt soooo bad :D

it was section 1.1 #57-69 odd, 79, 81, 87-90, 91(b-d), 94, 103-105
          section 1.2 #1-17 odd

Hope this helped :D

Hi thar

So guys.... I fun day of class huh? well most of it was review on functions. Your book is actually a very nice resource if you miss a class period. if you start on page 74 in your book and move all the way through to example 3 then you will have covered everything thing we did in class.
As i said this was just a review really. we talked about the definition of a function. and that is on pg 74. because this is my first post im not entirely sure what I should be posting but this appears to be in order. im not going to summarize the book for you though. and the hw is...

pg 82
1-6, 18-24, 26, 31-33, 37, 43, 46-56 even

Chapter P

Sections P.4 and P.5 were review of topics, mostly from Algebra 2, with which you should be familiar.  Section P.4 focused on solving equations of many types.  The ones we spent the most time with were equations involving fractions and radical expressions.

Perhaps the easiest way to solve an equation with rational expressions (i.e., fractions) is to multiply both sides of the equation by the least common multiple of all the denominators.   Multiplying through will eliminate all the denominators, so there will be no fractions remaining.  From there, the equation will usually be either quadratic or linear and should be relatively straightforward to solve.  Remember to check that none of your solutions make any denominator in the original equation equal zero.

When a variable is under a radical, both sides of the equation will need to be squared.  Before doing this, isolate the radical term.  (When there are two radical terms, it is typically easier to separate them before squaring both sides for the first time.)  After squaring both sides of the equation, you will usually be left with a quadratic or linear equation to solve.  Remember to check for extraneous solutions!  When you square both sides of an equation, you open the door to extraneous solutions, so you have to check them by plugging them into the original equation.

P.5 dealt primarily with absolute value and inequalities.

Definition of absolute value:
|x| = a     if and only if     x = a  or  - x = a

for inequalitites:
|x| < a     if and only if     x < a  or  - x < a

Example:

Once the absolute value expression is isolated on one side of the inequality sign, split the problem into two separate inequalities (based on the definition above).

Using interval notation, we would represent our solution as [2,3].

Polynomial inequalities are a big part of P.5. Solving them is more complicated than solving equations.
Before the official solving begins, you must have a zero on one side of the inequality.
The first to solve a polynomial inequality is to find the zeros of the polynomial.
Use these zero to set up intervals, and then pick a test value in each interval. 
Plug each test value into the polynomial and see if its value is positive or negative.
Your solution will include all the intervals that matched your inequality (+ for >0 and - for <0).
This process makes more sense when you consider the graph of your inequality.  The inequality is >0 when its graph is above the x-axis and <0 when its graph is below the x-axis.  Setting up the test intervals is a numeric way of making that determination.

Example:
The first order of business is to find the zeros of the polynomial.

Next, plot the zeros on a number line to establish the test intervals, then choose a test value from each interval.  Plug that value into the polynomial and determine if the polynomial's value is postive or negative.

We were interested in where the polynomial was <0, so we choose the interval where f (x) was negative. 

Using interval notation, we would express our solution as (3,5).

Below is the graph of the polynomial.  Looking at that, it is clear that the interval where the graph is below the x-axis (<0) is from 3 to 5.

Welcome

Hello everybody. Welcome to our class's blog. This is where you all will be posting your take on what we're learning in class and keep the conversation going.

Please create a Blogger account and email me <tw04bps@birmingham.k12.mi.us> your username, so I can keep track of and give you credit for your contributions.