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11.2 Sets and Compound Inequalities
11.3 Absolute Value Equations
11.4 Absolute Value Inequalities
11.2 Sets and Compound Inequalties
• Set – a collection or group of items. – We usually name sets with capital letters
• Elements – the objects within each set– If we are talking about general terms within a set
we usually use lower case letters to represent them
– When we list the group of elements we enclose them in { } and separate each element by a comma. This method is referred as ROSTER NOTATION.
This class as a set
• A = {Amanda, Kacy, Niki, Megan, Jon, Johnny, Shawn, James}
• To show that an element is contained in a set or that an object belongs to a set we use the symbol
• Amanda A, would be read “Amanda is an element of set A.”
• Kyle A, would be read “Kyle is not an element of set A.”
Sets and Subsets• A subset is a collection of
elements that create another set but all elements come from a larger group.
• We must have a knowledge of the overall group in order to determine a subset.
• If we know that A= {2, 4, 6, 8, 10}
• And we know that B = {4, 6, 8} then it is said that B is a subset of A. The notation/symbols used to talk about subsets is ⊂. We would write B⊂A, read “B is a subset of A”
• Subsets exist only if every element in the smaller set exists in the larger set. “every set is actually a subset of itself…not something this text gets into”
• If we can count the number of items in a set we say that the set is FINITE, and the number associated with the set is called the set’s CARDINALITY. If you cannot count the items within the set then the set is said to be an INFINITE set.
• If there is nothing in a set it is called an EMPTY or NULL SET. This is denoted by the symbol ∅ “no braces”. The empty set is considered an element of every set.
• Some sets that we should be familiar with.• Natural Numbers (counting numbers) – N = {1,2,3,4,5,…}
• Whole Numbers (same as N but start with 0)– N0= {0,1,2,3,4,5,…}
• Integers (Whole numbers and their opposites)– Z= {…-4,-3,-2,-1,0,1,2,3,4,…}
• Rational Numbers (any number written as a/b, b cannot equal 0)
• Irrational Numbers (any number that does not terminate or repeat in regards to decimal)
• Real Numbers denoted as R. Is the overall set that all of the above belong to.
• We can do OPERATIONS with sets. • The two operations that we use are union “ ” and ∪
intersection “∩”. • Union means that we will be taking 2 or more sets
and creating a new set, UNION refers to the new set including all elements of the other sets. – We will say A B, which means the new set will be comprised ∪
of all elements that are either in A or in B. For an element to be in A B then it has to be in A or in B.∪
– The key word is OR
• The other operation is an intersection “∩”• Here we will be taking 2 or more sets and
creating a new set that is comprised of only the elements that are in (shared by) the other sets. So for an element to be in the new set A ∩B then it must be in A and it must be in B. The key word is AND
Examples
• Let A = {2, 4, 6} , B={0, 1, 3, 5} and C = {1,2,3,…}
• a.) A B = {∪• b.) A ∩ B = {• c.) A ∩ C = {• d.) B C = {∪
A B2 4 5 4 3
8 11 12 33 22
A = {
B = {
A U B = {
A ∩ B = {
• Graphing compound inequalities. • When the inequalities use the word or idea of
“and” our final solution will be where the two individual graphs overlap.
• When the inequalities use the word or idea of “or” we will graph both inequalities and our final solution will be EVERYTHING that is graphed.
Graph
{ | 3 2}x x or x
Now represent the answer in interval notation using the correct union or intersection symbol.
Graph
{ | 1 4}x x or x
Graph
{ | 1 4}x x and x
Graph
{ | 2 3}x x and x
Graph
{ | 3 6}x x and x
Solving compound inequalites
• If the word “or” is used you will have two separate inequalities to solve, simply solve both and graph both on the same number line.
4 8 12 4 8 12t or t
• Usually when the word “and” is to be used in the inequality you will not see it. The inequality will be written as three parts.
• With problems like this you treat them just like normal equations however we want to isolate the variable in the center. What you do to one part you have to do to all 3 parts.
5 2 15 3x
11.3 Absolute Value Equations
• In general practices the absolute value of a number is referencing the distance a number is from 0. Distance always has to be positive.
• Thus when asking to solve |x|=5 for x. • We are asking the question for what number
does x have to be so that it is a distance of 5 from zero. In this case x could be 5 or -5.
-This idea of x having two results creates the steps for us in order to solve absolute value equations
• We simply take the equation given and get rid of the |abs| bars and solve the equation.
• But then we have to take the given equation get rid of the |abs| bars, multiply one side by a negative 1 and then solve that equation.
• Each absolute value equation will result in 2 other equations that we can solve. We cannot solve with the |abs| bars remaining in the equation they must be eliminated.
Solve
• |3a + 2| = 6• This is saying that 3a+2 is a distance of 6 units from 0.
Solve
• |5x-3|+12 = 19
we must isolate the |abs| bars first
Solve
• |4x + 7| = -8– Understanding the idea of absolute value this
would say that 4x+7 is a distance of -8 from 0. Can distance be negative? So can this be solved?
– This is somewhat of a trick question that often gets asked on tests/quizzes.
• When you have absolute value bars on both sides of the equation you simply solve them the way they are without the |abs| bars.
• Then create a second equation by removing the bars from both sides and multiplying one side by a negative 1 (make sure you distribute it)
• |a| = |b| this means that the values of a and b are the same distance from 0, thus they must be equal to one another or opposites of one another
Solve
• |6a + 4| = |4a +6|
Solve
2 1 5
3 2 6x x
11.4 Absolute Value Inequalities
• Every absolute value inequality needs to be changed into two separate inequalities that do not include the absolute value bars.
• IF |a|< 4 , this means that a is somewhere between -4 and 4 on the number line.
• If |a| > 5, this means that a is somewhere to the right of 5 or to the left of 5 on the number line.
Solve and Graph
• Here we are saying that 2x+1 is less than 7 units away from 0 so it could be to the left of zero or to the right of zero. So it must appear between -7 and 7 on the number line so we transform the inequality into…
| 2 1| 7x
7 2 1 7x
Solve |4a-3|<5
Solve |-8x+4|<12
Solve 4>|y-5|
• Things have flipped but if we re-write the inequality so that |abs| comes first we can better decide how to address this problem
An absolute value is greater than
• When asked to solve |x -2|> 6. • We are stating that x-2 is further than 6 units away
from 0, and that can occur to the right of zero and to the left of zero. So we address it as two inequalities again.
• Take the current inequality without the |abs| and solve.
• Repeat the process for the second inequality but multiply the constant side by a negative one and solve
• Combine your answers with the word “or”
Solve and Graph
| 2 9 | 11x
Solve and Graph1
3 53
t
Solve and Graph
| 3 9 | 2 14a
Solve and Graph
• After we isolate the |abs| we recognize that the quantity is > -9. Ask yourself how often will that quantity be greater than -9 units away from zero. And this is your answer.
| 2 2 | 2 11a
Summary of absolute value inequalities
• Blue box on page 770 is a good graphic to have memorized for a test or quiz if you struggle with the concepts. I think it is much more informative and useful to develop an understanding of the concepts and then you won’t forget something that you were suppose to memorize. Memorization fails us every once and a while.