PP Section 2.4

# 6 3 1 inequality the solve x x 2 1 5 3 3x 6 negative

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6 3 1 inequality the Solve < + - x x

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-2 1 -5 0 3 3x + 6      negative           positive             positive x - 1        negative           negative            positive quotient      positive           negative          positive Answer: (-2, 1) 0 6 3 1 < + - x x
Solving Inequalities 4 3 2 1 inequality the Solve + - < + + x x x x Looking at the inequality we see that –2 and –4 are  NOT in the domain of the inequality. 0 4 3 2 1 < + - - + + x x x x

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We need a single fraction, so using a common  denominator:  0 4 3 2 1 < + - - + + x x x x ) 4 )( 2 ( ) 2 )( 3 ( ) 4 )( 1 ( 4 3 2 1 + + + - - + + = + - - + + x x x x x x x x x x ) 4 )( 2 ( 10 6 + + + = x x x
To construct a chart of signs we see that the  numerator is 0 when  x = -5/3 , and the denominator  when  x = -2, -4. -5/3 -4 -2 -6 -4.5 -3 0 6x+10       negative        negative      negative       positive     x+2      negative         negative      negative       positive     x+4      negative         negative      positive        positive quotient       negative        negative     positive      positive 0 ) 4 )( 2 ( 10 6 have ve So < + + + x x x What is the answer?
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