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Summer Assignment Review
Graphing Compound Inequalities
And Absolute Value Inequalities
Compound Inequalities
AND compound inequalities should
create a range of values.
Don’t assume that the given
inequality creates this range.
βˆ’2 ≀ π‘₯ ≀ 3
βˆ’2 β‰₯ π‘₯ β‰₯ 3
π‘₯ β‰₯ βˆ’2 𝐴𝑁𝐷 π‘₯ ≀ 3
π‘₯ ≀ βˆ’2 𝐴𝑁𝐷 π‘₯ β‰₯ 3
π‘₯ falls between -2 and 3 on the
number line.
π‘₯ can not be both to the left of -2
and to the right of 3 on the number
line . There is no solution.
Compound Inequalities
OR compound inequalities should
create two ranges of values.
π‘₯ ≀ βˆ’2 𝑂𝑅
π‘₯β‰₯3
π‘₯ is either to the left of -2 or to the
right of 3 on the number line.
AND compound inequalities should
create a range of values.
π‘₯ β‰₯ βˆ’2 𝑂𝑅 π‘₯ ≀ 3
π‘₯ can be anywhere on the number
line. The solution is All Real
Numbers.
Practice
Graph each compound inequality.
1)
βˆ’3 < π‘₯ < 4
2)
π‘₯ β‰₯ βˆ’3 π‘œπ‘Ÿ π‘₯ < 3
3)
π‘₯ < βˆ’4 π‘œπ‘Ÿ π‘₯ β‰₯ 2
4)
π‘₯ ≀ 0 π‘Žπ‘›π‘‘ π‘₯ β‰₯ 3
5)
π‘₯ > βˆ’1 π‘Žπ‘›π‘‘ π‘₯ ≀ 5
All Real
Numbers
No Solutions
Absolute Value Inequalities
When we say that π‘₯ = 3, we are saying that the distance between x and 0 is 3.
When we say that π‘₯ < 3, we are saying that the distance between x and 0 is less than 3.
The inequalities < and ≀ lead to β€œand” relationships
βˆ’3 < π‘₯ < 3
When we say that π‘₯ > 3, we are saying that the distance between x and 0 is greater than 3.
The inequalities > and β‰₯ lead to β€œor” relationships
π‘₯ < βˆ’3 𝑂𝑅 π‘₯ > 3
Solving Absolute Value Inequalities
Case 2 - modified method: Drop the
absolute value bars, flip the inequality
symbol and change the sign of the term
on the right.
2π‘₯ βˆ’ 1 < βˆ’7
2π‘₯ < βˆ’6
π‘₯ < βˆ’3
Solve and Graph an Abs. Val. Ineq.
Example : Solve π‘₯ βˆ’ 5 β‰₯ 7 . Graph your solution.
Step 1: Is it and or or?
This is β‰₯ so it is β€œor”. Set up two inequalities.
π‘₯βˆ’5 β‰₯7
For the second, flip the inequality and
change the sign of the 7
π‘₯βˆ’5β‰₯7
π‘₯ βˆ’ 5 ≀ βˆ’7
Step 2: Solve both
π‘₯ ≀ βˆ’2
π‘₯ β‰₯ 12
Step 3: Graph
π‘₯ β‰₯ 12
-3 -2 -1
or
π‘₯ ≀ βˆ’2
· · · 11 12 13
Solve and Graph an Abs. Val. Ineq.
Example : Solve βˆ’4π‘₯ βˆ’ 5 + 3 < 9 . Graph your solution.
Step 1: Is it and or or?
We don’t know yet, get the abs. val. alone.
βˆ’4π‘₯ βˆ’ 5 + 3 < 9
βˆ’4π‘₯ βˆ’ 5 < 6
Step 2: Solve
Now :
It is < , so it is an β€œand” inequality. Drop
the abs. val. Bars and put -6 < on the left.
βˆ’6 < βˆ’4π‘₯ βˆ’ 5 < 6
+5
+5
+5
βˆ’1 < βˆ’4π‘₯ < 11
-4
-4
-4
0.25 > π‘₯ > βˆ’2.75
βˆ’2.75 < π‘₯ < 0.25
Step 3: Graph
Practice
π‘₯ > 5 π‘œπ‘Ÿ π‘₯ < βˆ’11
-12 -11 -10
··· 4
5 6
βˆ’5 < 𝑀 < 6
-6 -5 -4 · · · 5 6 7
βˆ’0.2 ≀ π‘š ≀ 2.6