Section 1.7 Inequalities

Transcription

1 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 x < 9x+4 and sketch the solution set. x < 9x+4 x 9x < 9x+4 9x 6x < 4 6x 6 > 4 6 x > The solution set consists of all numbers that are greater than. In other words the solution of the inequality is the interval ( ),. EXAMPLE: Solve the inequality 5 6(x 4)+7 and sketch the solution set. 1

2 EXAMPLE: Solve the inequality 5 6(x 4)+7 and sketch the solution set. 5 6(x 4) (x 4) 1 6x x 1 6 6x 6 x The solution set consists of all numbers that are less than or equal to. In other words the solution of the inequality is the interval (,]. EXAMPLE: Solve the inequality 1 (x 6) 4 (x+) x and sketch the solution set. 4 1 (x 6) 4 (x+) 4 x ( 1 1 (x 6) 4 ) (x+) 1 ( 4 ) x 1 1 (x 6) 1 4 (x+) 1 ( 4 ) x 1 6(x 6) 16(x+) 9x 4 6x 6 16x 9x 4 6x 16x+9x 4+6+ x 44 ( 1)( x) ( 1)44 x 44 The solution set consists of all numbers that are less than or equal to 44. In other words the solution of the inequality is the interval (, 44]. -44

3 EXAMPLE: Solve the compound inequality 4 x < 1 and sketch the solution set. Therefore, the solution set is [, 5). 4 x < 1 4+ x + < 1+ 6 x < 15 6 x < 15 x < 5 EXAMPLE: Solve the compound inequality 1 < 5 x 4 1 < 5 x 4 4 ( 1) < 4 5 x < 5 x and sketch the solution set. 4 5 < 5 x < x 9 9 > x 9 > x 9 9 x < [ Therefore, the solution set is 9 ),. -9/

4 Nonlinear Inequalities EXAMPLE: Solve the inequality x 5x 6 and sketch the solution set. Solution: The corresponding equation x 5x+6 = (x )(x ) = 0 has the solutions and. As shown in the Figure below, the numbers and divide the real line into three intervals: (,), (,), and (, ). 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 and x at the value selected. For instance, if we use the test value x = 1 from the interval (,) shown in the Figure above, then substitution in the factors x and x gives x = 1 = 1, x = 1 = Both factors are negative on this interval, therefore x 5x+6 = (x )(x ) is positive on (,). Similarly, using the test values x = 1 and x = 4 from the intervals (,) and (, ), respectively, we get: + + Thus, the solution of the inequality x 5x 6 is {x x } = [,]. 4

5 EXAMPLE: Solve the inequality x(x 1)(x+) > 0 and sketch the solution set. Solution: The corresponding equation x(x 1)(x+) = 0 has the solutions 0,1, and. As shown in the Figure below, the numbers 0,1, and divide the real line into four intervals: (, ), (,0), (0,1), and (1, ) 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, x 1, and x+ at the value selected. For instance, if we use the test value x = from the interval (, ) shown in the Figure above, then substitution in the factors x, x 1, and x+ gives x =, x 1 = 1 = 4, x+ = + = 1 All three factors are negative on this interval, therefore x(x 1)(x+) is negative on (, ). Similarly, using the test values x = 1, x = 1/ and x = from the intervals (,0), (0,1) and (1, ), respectively, we get: Thus, the solution of the inequality x(x 1)(x+) > 0 is (,0) (1, ). EXAMPLE: Solve the inequality x(x 1) (x ) < 0. Solution: The corresponding equation x(x 1) (x ) = 0 has the solutions 0,1, and. As shown in the Figure below, the numbers 0,1, and divide the real line into four intervals: (,0), (0,1), (1,), and (, ). 0 1 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 and x at the value selected. For instance, if we use the test value x = 1 from the interval (,0) shown in the Figure above, then substitution in the factors x and x gives x = 1, x = 1 = 4 Bothfactorsarenegativeonthisinterval,(x 1) isalwaysnonnegative,thereforex(x 1) (x ) is positive on (,0). Similarly, using the test values x = 1/, x =, and x = 4 from the intervals (0, 1), (1, ), and (, ), respectively, we get: Thus, the solution of the inequality x(x 1) (x ) < 0 is (0,1) (1,). REMARK: The solution of the inequality x(x 1) (x ) 0 is [0,]. 5

6 EXAMPLE: Solve the following inequalities: (a) (x+)() 8 (b) (x+)(x+4) 1 (c) (x+)(x+4) 1 Solution: (a) We have (x+)() 8 x +4x+ 8 x +4x+ 8 0 x +4x 5 0 (x+5)(x 1) 0 The corresponding equation (x + 5)(x 1) = 0 has the solutions 5 and 1. As shown in the Figurebelow, thenumbers 5and1dividethereallineintothreeintervals: (, 5), ( 5,1), and (1, ) 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 + 5 and x 1 at the value selected. For instance, if we use the test value x = 6 from the interval (, 5) shown in the Figure above, then substitution in the factors x+5 and x 1 gives x+5 = 6+5 = 1, x 1 = 6 1 = 7 Both factors are negative on this interval, therefore (x+5)(x 1) is positive on (,5). Similarly, using the test values x = 0 and x = from the intervals ( 5,1) and (1, ), respectively, we get: Thus, the solution of the inequality (x+)() 8 is {x x 5 or x 1} = (, 5] [1, ). (b, c) We have (x+)(x+4) 1 x +6x+8 1 x +6x x } +6x+9 {{} 0 x + x + (x+) 0 Note that (x + ) is either positive or zero. Therefore, the only solution of the inequality (x + )(x + 4) 1 is x =. In a similar way one can show that the solution of the inequality (x+)(x+4) 1 is all real numbers. 6

7 EXAMPLE: Solve the inequality 1+x 1 and sketch the solution set. 1+x 1 1+x x 0 1+x () 1+x 1+x 0 0 x 0 As shown in the Figure below, the numbers 0 (at which the numerator of which the denominator of and (1, ). x x is 0) and 1 (at is 0) divide the real line into three intervals: (,0), (0,1), 0 1 x On each of these intervals we determine the sign of using test values. We choose a x number inside each interval and check the sign of at the value selected. For instance, if we use the test value x = 1 from the interval (,0) shown in the Figure above, then substitution in x gives ( 1) 1 ( 1) = = 1 < 0 and x = from the intervals (0,1), and (1, ), respec- Similarly, using the test values x = 1 tively, we get: Thus, the solution of the inequality 1+x 1 is [0,1). EXAMPLE: Solve the inequality x 1 and sketch the solution set. 7

8 EXAMPLE: Solve the inequality x 1 and sketch the solution set. x 1 x 1 0 x 0 x () x x x 4 0 As shown in the Figure below, the numbers 4 (at which the numerator of x 4 is 0) and 1 (at whichthedenominatorof x 4 is0)dividethereallineintothreeintervals: (, 1), ( 1,4), and (4, ) On each of these intervals we determine the sign of x 4 using test values. We choose a number inside each interval and check the sign of x 4 at the value selected. For instance, if we use the test value x = from the interval (, 1) shown in the Figure above, then substitution in x 4 gives 4 +1 = 6 1 = 6 > 0 Similarly, using the test values x = 0 and x = 5 from the intervals ( 1,4), and (4, ), respectively, we get: Thus, the solution of the inequality x 1 is ( 1,4]. EXAMPLE: Solve the inequality 7 5x 8x+9 < and sketch the solution set. 8

9 EXAMPLE: Solve the inequality 7 5x 8x+9 < and sketch the solution set. 7 5x 8x+9 < 7 5x 8x+9 < 0 (7 5x) (8x+9) < 0 (7 5x) (8x+9) 1 15x 16x 18 < 0 < 0 < 0 As shown in the Figure below, the numbers /1 (at which the numerator of is 0) and 9/8 (at which the denominator of is 0) divide the real line into three intervals: (, 9/8), ( 9/8, /1), and (/1, ). -9/8 /1 On each of these intervals we determine the sign of using test values. We choose a number inside each interval and check the sign of at the value selected. For instance, if we use the test value x = from the interval (, 9/8) shown in the Figure above, then substitution in gives = 1( ) (8( )+9) = +6 ( 16+9) = 65 ( 7) < 0 Similarly, using the test values x = 0 and x = 1 from the intervals ( 9/8,/1), and (/1, ), respectively, we get: -9/8 + /1 Thus, the solution of the inequality 7 5x 8x+9 < ( is, 9 ) ( ) 8 1,. -9/8 /1 9

10 Absolute Value Inequalities EXAMPLE: Solve the inequality x 5 < and sketch the solution set. Solution: The inequality x 5 < is equivalent to < x 5 < +5 < x 5+5 < +5 < x < 7 The solution set is the open interval (,7). EXAMPLE: Solve the inequality x 5 <. Solution: The inequality x 5 < has no solutions, since is always nonnegative. So, the solution set is the empty set. EXAMPLE: Solve the inequality 7 x < 1. 10

11 EXAMPLE: Solve the inequality 7 x < 1. Solution: The inequality 7 x < 1 is equivalent to The solution set is the open interval 1 < 7 x < < 7 x 7 < < x < 6 8 > x > 6 < x < 8 (, 8 ). EXAMPLE: Solve the inequality x+ 4 and sketch the solution set. Solution: The inequality x+ 4 is equivalent to x+ 4 or x+ 4 x x 6 x x The solution set is { x x or x } [ ) = (, ], EXAMPLE: Solve the inequality 4 5x x x+9 +7 }{{} 9 The inequality 5x is equivalent to 5x The solution set is { x 9 4 x x ( 4 9 ) 4 (5x+9) x x x = 45 0 x 7 0 } [ = 9 4, ].