Section 1.4 Combinations of Functions

The domain of an Arithmetic combination of functions f and g consists of all real numbers that are common to the domains of f and g. In the case of the quotient f(x)/g(x) - g(x) ≠ 0.

1. Sum: (f + g)(x) = f(x) + g(x)

2. Difference: (f – g)(x) = f(x) – g(x)

3. Product: (fg)(x) = f(x) ∙ g(x)

4. Quotient: (f/g)(x) = f(x)/g(x), g(x) ≠ 0

ex. (f + g)(x) = f(x) + g(x)

= (2x + 1) + (x² + 2x – 1)

= x² + 4x

Composition of Functions

(f ○ g)(x) = f(g(x))

ex. f(x) = x + 2 g(x) = 4 – x²

(f ○ g)(x) = f(g(x))

= f(4 – x²)

= (4 – x²) + 2

= –x² + 6

Section 1.5- Inverses

Finding the inverse of functions are based on composition

ex. Find inverse of F(x)

F(x)=2x-3

F^-1(x)= x+3/2

(FoF^-1)(x)= F(F^-1(x))

= F (x+3/2)

= 2(x+3/2)

= x+3-3

= x

OR

(F^-1 o F)(x)= F^-1 (F(x))

= F^-1 (2x-3)

= 2x-3+3/2

= 2x/2

= x

*You should get x regardless of which way you solve it

- two functions, f and g, are inverses of each other if and only if....

(F o G) (x)= (G o F)(x)= x

*graphs are reflections by Y=X

One-to-one function- passes horizontal line test and has inverse that is a function

-A function is a one-to-one function if and only if F(a)=F(b)----> a=b

Section 1.4 - Arithmetic Operations

This section is about how to add, subtract, multiply, and divide functions.

Addition, subtraction, multiplication, and division are all called arithmetic functions.

1. Adding Functions

When adding functions, you have to add them as two separate quantities

ex. f(x) = (x + 1)
g(x) = (x - 1)

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

2. Subtracting Functions

The same principal applies to subtracting functions.

ex. f(x) = (2x - 5)
g(x) = (1 - x)

(f - g)(x) = (2x - 5) - (1 - x)
= 3x - 6

3. Multiplying Functions

Multiplying functions requires the use of FOIL.

ex. f(x) = (3x - 2)
g(x) = (5x + 7)

(fg)(x) = (3x - 2)(5x + 7)
=

4. Dividing Functions

If you are asked to divide two functions, you will also have to state the domain.

ex. f(x) =

g(x) =

(f/g)(x) =

Domain:

Section 1.3 - Shifting, Reflecting, and Stretching Graphs

Section 1.3 covers the shifting, reflecting, and stretching of graphs.

These parent functions provide the basis for a strong understanding of the section:

(This graph uses c=4, though
c is an unknown value until

specified)

Shifting, Reflecting, and Stretching Graphs

1. Adding or subtracting values from the y-coordinate shifts the graph vertically

Adding c shifts the graph up the value of c

Subtracting c shifts the graph down the value of c

Ex:

The graph is shifted up 2 units

2. Adding or subtracting values from the x-coordinate shifts the graph horizontally

Adding c shifts the graph left the value of c

Subtracting c shifts the graph right the value of c

Ex:

The graph shifts 5 units right

3. Multiplying the y-coordinate stretches or compresses the graph vertically

y=c(f(x))

If the value of c is greater than one, the graph is stretched vertically
If the value of c is less than one, the graph is compressed vertically

Ex: If the absolute value of x in multiplied by 4, the graph will be vertically stretched by a factor of 4.

4. Multiplying the x-coordinate stretches or compresses the graph horizontally

y=f(c(x))
If the value of c is greater than one, the graph is compressed horizontally

If the value of c is less than one, the graph is stretched horizontally
Ex: If the square root of x is multiplied by 3, the graph will be horizontally compressed by a factor of 3.

5. Multipying the y-coordinate by a negative number reflects the graph in the x-axis
y=-f(x)
The x values do not change
The y values are opposite of what they were before

Ex:
y=x+2
y=-(x+2)

y=-(x+2) = y=-f(x)

The y values are now opposite of what they were before while the x values do not change

6. Multiplying the x-coordinate by a negative number reflects the graph in the y-axis
y=f(-x)
The x values are opposite of what they were before
The y values do not change
Ex:

y=5x-2

y=5(-1)-2

y=5(-1)-2 = y=f(-x)

The x values are now opposite of what they were before while the y values do not change

Section 1.2

A graph has symmetry with respect to the y-axis if whenever (x,y) is on the graph, so is the point (-x,y). A graph has symmetry with respect to the origin if whenever (x,y) is on the graph, so is the point (-x,-y). A graph has symmetry with respect to the x-axis if whenever (x,y) is on the graph, so is the point (x,-y).

A function whose graph is symmetric with respect to the y-axis is an EVEN function.
A function f is EVEN if, for each x in the domain of f, f(-x) = f(x)

A function whose graph is symmetric with respect to the origin is an ODD function.
A function f is ODD if, for each x in the domain of f, f(-x) = -f(x)

Here is a practice problem that can be found on page 95 in the book.
a. g(x) = x^3 - x
Is this function even, odd, or neither?
g(x) = x^3 - x =
g(-x) = (-x)^3 - (-x) =
-x^3 + x =
-(x^3 - x) =
-g(x)
This function is odd because f(-x) = -f(x)

Thursday, September 15th 2011

Today we learned two main concepts:

1. Piecewise-Defined Functions: Functions in which not all points on the graph would be able to be obtained from only one equation. With the different equations for the same function would come an inequality that would limit the domain, so that only if the X value was true to that specific inequality, then that X value could be plugged into that equation to solve for the range.
2. Evaluating Difference Quotient: Taking whatever value is given for f(x) and plugging that value (s) into this equation.

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.