Combinations of Functions

Combinations of Functions:

All of these are Arithmetic combinations of functions

Let f(x) =2x+3
let g(x) =

Creating new functions - examples

(f+g)(x) =f(x)+g(x)

=2x+3+

(f-g)(x) =f(x)-g(x)

= (2x+3)-( )

= -+2x+3

(fg)(x) =f(x)*g(x)

= (2x+3)( )

= 2+3

(f/g)(x)= ------> don’t forget to find the domain after dividing!

Remember-denominator cannot be zero!

= (2x+3)/( )

1.2

Finding Domain and Range of a Function

• Algebraic Solution

Note: Because the expression is under a radical it must end up being ≥ 0 otherwise it would be imaginary.

Step 1. Set the equation as  ≥ 0

             ≥ 0

Step 2. Square both sides to get rid of the radical. It should end up being 

            x-4 ≥ 0

Step 3. Isolate the variable

          x-4+4≥ 0+4

                  x ≥ 4

x is all real numbers greater than or equal to zero [4,∞)

• Graphical Solution

f(x)=

The graph for this function looks like this

To find the domain and range simply look at the real line

As you an see the range is all real numbers greater than or equal to zero & the range is all real numbers greater than or equal to 4

Even and Odd Functions

Function 1

Find if f(x) = –3x² + 4 even, odd, or neither.

Plug in f(-x) and solve

f(–x) = –3(–x)² + 4 

      = -3(-x)(-x) +4

         = –3(x²) + 4          = –3x² + 4

The function is the same as it started out as therefore it's an even function, the graph would end up looking like this with the y axis as the line of symmetry

Function 2

Is f(x) = 2x³ – 4x even, odd, or neither?

Same as last time, plug in f(–x)

f(–x) = 2(–x)³ – 4(–x) 

         = 2(–x³) + 4x          = –2x³ + 4x

As you can see the result is the exact opposite of the original, which means that the function is odd. When graphed it would look like this with the point of origin as the line of symmetry 

Function 3

Is f(x) = 2x³ – 3x² – 4x + 4 odd, even or neither?

Plug in f(–x)

f(–x) = 2(–x)³ – 3(–x)² – 4(–x) + 4          = 2(–x³) – 3(x²)  + 4x + 4          = –2x³ –3x² + 4x + 4

It's not the same or opposite of the original equation which means it is neither and a graph for it would not follow the symmetry on the y axis or the point of origin.

Domain and Range of Functions, Difference Quotient

Domain = all possible x-values
Range = all possible y-values

Domain

When finding the domain, check for:

• negative square roots


• denominators that equal zero
  

Examples

1.) In equations with fractions, the domain is determined only by the denominator. It is important to make sure that it will not equal zero.

Start by setting the equation equal to zero and solving.

These are the values that will make the denominator equal zero, so the domain of the function is all numbers except 2 and -1.

2.) In equations with a square root, it is important to make sure that we are not finding the square root of a negative number.

To start, set the equation as greater than or equal to zero and solve.

This is the lowest x-value that can be used before the equation becomes negative. Therefore, the domain of the equation is all values equal to or less than three halves.

Range

To find the range, graph the equation to see all the different y-values.

In this example, we can see that the lowest y-value on the line is -1, and that the lines extend upwards indefinitely.

Therefore, the range is

Difference Quotient

http://www.youtube.com/watch?v=1O5NEI8UuHM

Ch1- Functions

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 doman ( or set of inputs) of the function f, and the set B contains the range ( or set of outputs).

Functions can be expressed Verbally, Numerically, Graphically, and Algebraically
Use the vertical line test to check if there is a function on a graph. If the line passes 2 points then there are two of the same x values, therefore it is not a function.


Function Notation-
y is a function of x
y = f (x) ↔ The point (x, y) is on the graph f

Example- f (x)= 3x+10
f (2)= 3(2)+10
f (2)=16

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.