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Pre- Calculus

Pre- Calculus. Chapter 1: Functions. 1.1 Introduction (p. 2-3). Copy definitions: (copy symbols if any apply). Natural Numbers Integers Rational Numbers Irrational Numbers Real Numbers. Being closed under operation(s).

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Pre- Calculus

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  1. Pre- Calculus Chapter 1: Functions

  2. 1.1 Introduction (p. 2-3) Copy definitions: (copy symbols if any apply) • Natural Numbers • Integers • Rational Numbers • Irrational Numbers • Real Numbers Being closed under operation(s) If two numbers of a particular set are under an operation, and the result of that operation is a number still in that same set, then it is closed under that operation.

  3. Ex:2 and 3 are natural numbers. Test all operations to see which operations natural numbers are closed under. 2 + 3 = 5 (natural number) 3 + 2 = 5 (natural number) Closed under addition Note: You must check the numbers in any order. Natural numbers are closed under: Addition and multiplication. Integers are closed under: Addition, subtraction and multiplication. Rational numbers are closed under: Addition, subtraction, multiplication, and division except for 0.

  4. 1.2 The Real Line (p. 3-12) All real numbers can be shown on acoordinate line. -3 0 3 Inequalities: The relative position of two points on a number line is used to define an inequality. a b We say: a is less than b or b is greater than a. a < b b > a

  5. Inequality properties: 1. Precisely one of a < b, b < a, or a = b holds. 2. If a > b, then a + c > b + c. ex: let a = 4, b = 1, and c = 10. Check the property. 3. If a > b and c > 0, then ac > bc. Use the same values in the example above. 4. If a > b and c < 0, then ac < bc. This time, let c = -10 Property 4 implies that when both sides of an inequality are multiplied by a negative number, the direction of the symbol must be reversed.

  6. Ex: 1 Find all real numbers satisfying 2x – 1 < 4x + 3. • Isolate x. • Don’t forget property 4 if you subtract 4x from both sides. Solution:-2 < x x -2

  7. Ex: 2 Find all real numbers x satisfying Solution:You can solve this as one inequality or break it up into two inequalities. (Now solve both) -2 < x Now, write the solution as one inequality. x 1 -2

  8. Intervals: Representing sets of numbers (table p. 7) open interval closed interval half open interval half open interval unbounded above unbounded above unbounded below unbounded below

  9. 3.) Find all the values of x satisfying the inequality Solution: • Set one side equal to zero: • Factor: (x – 3)(x – 1) > 0 • Use asign graphto determine where on the • coordinate line this holds true. (x – 3) - - - - - - - - - - - - - - - - - - - - - - - + + + + + + 0 (x – 1) - - - - - - - - - - - - - - - - - - - - + + + + + + + + + 0 (x – 3)(x – 1) 0 0 + + + + + + + + + + + + + + + + + - - - - o o 1 3 0 Since our inequality is “greater than” zero we want the positives. The values of x that are satisfied lie in that area.

  10. Ex: 4 Find all values of x that satisfy - - - - - - - - - + + + + + + + + + + + + + + + + + + + + + 0 (5 – x) (x – 1) (x – 3) (5 – x)(x – 1)(x – 3) - - - - - - - - - - - - - - - - - - - - + + + + + + + + + + + + 0 • - - - - - - - - - - - - - - - - - - - - - - - - + + + + + + + + + + + 0 + + + + + + + + - - - + + - - - - - - - - - - 0 0 0 0 1 3 5 Since the inequality is “less than or equal to zero” we want the negatives. Our answer written in interval notation:

  11. Ex:Find all values of x for which x+1 x-1 x 2-x fcn - - - - - - - - - - + + + + + + + + + + + + + 0 - - - - - - - - - - - - - - - + + + + + + + + + + + 0 • - - - - - - - - - - - - - + + + + + + + + + + + + 0 + + + + + + + + + + + + - - - - - - - - - - - - - 0 - - - - - - - - - - - + - + - - - - - - - - - - - - - 0 0 0 0 o o x -1 1 2 0 We want the positives and the endpoints unless it would make the denominator “0”. Solution:

  12. Absolute Values: The absolute value of a real number x, denoted |x|, describes the distance on the coordinate line from the number x to the number 0. Ex: 6 Determine all values of x for which Recall:When you take the absolute value of a number or an expression, the result is positive. However, we do not know whether the original number or expression was positive or negative. We must consider both possibilities. Notice the absolute value signs are now gone.

  13. Solve for x x – 2 = 3(2x + 1) -3(2x+1) = x – 2 x- 2 = 6x + 3 -6x – 3 = x - 2 -2 = 5x + 3 -3 = 7x - 2 -1 = 7x -5 = 5x -1/7 = x or -1 = x

  14. Distance formula: The distance between two numbers on a coordinate line can be found by We use absolute value because distance can never be negative. Midpoint formula: We can find the midpoint between two numbers on a coordinate line by Note: It does not matter which number you let be x1 or x2

  15. Ex: 7 Find all values of x that satisfy |2x – 1| < 3 This is just like absolute values involving equations. You must make two inequalities to account for both possibilities. -3 < 2x – 1 < 3 DO NOT CHANGE THE DIRECTION OF THE INEQUALITY SYMBOL!!! -3 < 2x – 1 and 2x – 1 < 3 -2 < 2x 2x < 4 -1 < x and x < 2 O O -1 0 1 2

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