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Section_1.7-Inequalities

Section_1.7-Inequalities - Section 1.7 Inequalities Linear...

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Section 1.7 Inequalities Linear Inequalities An inequality is linear if each term is constant or a multiple of the variable. EXAMPLE: Solve the inequality 3 x < 9 x + 4 and sketch the solution set. Solution: We have 3 x < 9 x + 4 3 x - 9 x < 9 x + 4 - 9 x - 6 x < 4 ( - 1 6 ) ( - 6 x ) > ( - 1 6 ) (4) x > - 2 3 The solution set consists of all numbers greater than - 2 3 . In other words the solution of the inequality is the interval ( - 2 3 , ) . EXAMPLE: Solve the inequalities 4 3 x - 2 < 13 and sketch the solution set. 1

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EXAMPLE: Solve the inequalities 4 3 x - 2 < 13 and sketch the solution set. Solution: We have 4 3 x - 2 < 13 4 + 2 3 x - 2 + 2 < 13 + 2 6 3 x < 15 ( 1 3 ) (6) ( 1 3 ) (3 x ) < ( 1 3 ) (15) 2 x < 5 Therefore, the solution set is [2 , 5) . EXAMPLE: Solve the inequalities - 4 < 5 - 3 x 17 and sketch the solution set. Solution: We have - 4 < 5 - 3 x 17 - 4 - 5 < 5 - 3 x - 5 17 - 5 - 9 < - 3 x 12 ( - 1 3 ) (12) ( - 1 3 ) ( - 3 x ) < ( - 1 3 ) ( - 9) - 4 x < 3 Therefore, the solution set is [ - 4 , 3) . 2
Nonlinear Inequalities EXAMPLE: Solve the inequality x 2 5 x - 6 and sketch the solution set. Solution: The corresponding equation x 2 - 5 x + 6 = ( x - 2)( x - 3) = 0 has the solutions 2 and 3 . As shown in the Figure below, the numbers 2 and 3 divide the real line into three intervals: ( -∞ , 2) , (2 , 3) , and (3 , ). - 2 3 On each of these intervals we determine the signs of the factors using test values . We choose a number inside each interval and check the sign of the factors x - 2 and x - 3 at the value selected. For instance, if we use the test value x = 1 from the interval ( -∞ , 2) shown in Figure above, then substitution in the factors x - 2 and x - 3 gives x - 2 = 1 - 2 = - 1 , x - 3 = 1 - 3 = - 2

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