Alg2 Ch 1.6 Class Notes

Ch 1.6 Class Notes Inequalities Objective: Given an inequality, transform it to a simpler, equivalent inequality so that you can draw a graph of its solution set. Inequalities: What result when the “=” sign in an equation is replaced by one of the order signs, <, >, ≤, ≥. Since the solution set usually contains an infinite (ie: ϵ) number of solutions, it is customary to draw a graph rather than write the set. (For example: 3x – 5 < 6 would have an infinite number of solutions.) EXAMPLE 1: Graph the solution set of 3 5 16 x   . Solution: 3 16 5 2 7 3 1 xx x       *The open circle at the 7 indicates that the endpoint of the ray is not included. A closed circle would be used for inequalities with ≤ or ≥, where the endpoint is included. PROPERTY Multiplication Property of Order: If x < y, then xz < yz, if z is positive. If x > y, then xz > yz, if z is positive. If x < y, then xz > yz, if z is negative. If x > y, then xz < yz, if z is negative. If x < y, then xz = yz, if z is zero. If x > y, then xz = yz, if z is zero. EXAMPLE 2: Graph the solution set of 2 18 x   . Solution: 2 18 (If , then , if is neg 9 ative.) Reverse order sign because is negat i ve xx x y xz yz z z       Ch 1.6 Class Notes (Cont’d) EXAMPLE 3: Graph the solution set of 3 2 5 11 x    . Solution: This inequality has three members. 3 2 5 11 3 ( 5) 2 5 ( 5) 11 ( 5) Add -5 to all three members. 2 2 6 Multiply by 1/2 to eliminate numerical coefficient 3 2. 1 x x x x                     The graph is all numbers between 1 and 3, including the -1 but not including the 3. EXAMPLE 4: Graph the solution set of 29 x  . Solution: │x│means “the distance between the origin and x.” So inequality really says, “x is the number that is more than 29 units from the origin.” 29 or Therefore, the inequality 29 is equivalent to the inequalities, . ( 29 becomes 29 because "If , then , if is negat 29 ive.")x x x combined x x x y xz yz z           The graph is all numbers greater than 29 and less than -29, but not including these numbers. EXAMPLE 5: Graph the solution set of 29 x  . Solution: The solution set will contain all number that are closer to the origin than 29 units (ie:, all those points between -29 and 29). 29 is equivalent to 29 and 29, which is equivalent to 29 2 9 x x x x        . (If x < y, then xz > yz, if z is positive; if x < y, then xz > yz if z is negative.) (all numbers between 29 and -29, but not including 29 and -29) We can therefore transform inequalities to eliminate the absolute value sign. Ch 1.6 Class Notes (Cont’d) CONCLUSION: If c is a non-negative constant, then │expression│ > c is equivalent to: expression > c or expression < -c If c is a non-negative constant, then │expression│< c is equivalent to: expression < c and expression > -c (or to –c < expression < c).  To remember these transformations, recall that the absolute value of a number is its distance from the origin. EXAMPLE 6: Graph the solution set of 3 5 13 x   . 3 5 13 x  (│expression│ < c equivalent to -c < expression < c) 8 6 3 13 3 5 13 ( expression ) 8 3 18 Add 5 to all three members. Divide all three members by 3 to eliminate numerical coefficient 3. xx c c x                Graphs: a) Domain of x is {real numbers}: ** 8 22 3 3    (all numbers between -2 2/3 and 6, including – 2 2/3 and 6) b) Domain of x is {positive numbers}: (all numbers between 0 and 6, but not including 0) c) Domain of x is {integers}: (all integers, and only integers from -2 to 6)