Explanation Procedure used: (1) . Find an equivalent inequality with 0 on one side. (2). Change the inequality symbol to an equals sign and solve the related equation. (3). Find values of the variables for which the related rational function is not defined. (4). The found in steps (2) and (3) are called the critical values. Use the critical values to divide the x axis into intervals. Then determine the function’s sign in each interval using an x value from the interval or using the graph of the equation. (5). Select the intervals for which the inequality is satisfied and write intervals notation or set builder notation for the solution set. If the inequality symbol ≤ or ≥ , then the solutions to step (2) should be include in the solution set. The x values found in step (3) are never include in the solution set. Calculation: The given polynomial inequality is | 1 − 1 x 2 | < 3 . By the procedure from 1 to 5, find the solution set of the inequality as follows. Simplify the above rational inequality as follows. | 1 − 1 x 2 | < 3 − 3 < 1 − 1 x 2 < 3 − 3 < x 2 − 1 x 2 < 3 − 3 < ( x − 1 ) ( x + 1 ) x 2 < 3 Case 1: First consider the inequality − 3 < ( x − 1 ) ( x + 1 ) x 2 . An equivalent inequality with 0 on one side is 0 < ( x − 1 ) ( x + 1 ) x 2 + 3 . Change an inequality sign by equal sign to form related equation as ( x − 1 ) ( x + 1 ) x 2 + 3 = 0 . The rational equation is undefined at x = 0 . Find the value of x from the equation ( x − 1 ) ( x + 1 ) x 2 + 3 = 0 . ( x − 1 ) ( x + 1 ) x 2 + 3 = 0 x 2 ⋅ ( ( x − 1 ) ( x + 1 ) x 2 + 3 ) = 0 ⋅ x 2 ( x − 1 ) ( x + 1 ) + 3 x 2 = 0 x 2 − 1 + 3 x 2 = 0 On further simplification, 4 x 2 = 1 x 2 = 1 4 x = ± 1 4 x = ± 1 2 Thus, the critical values of the rational equation are − 1 2 , 0 and 1 2 . The critical values divide the x axis into the intervals as ( − ∞ , − 1 2 ) , ( − 1 2 , 0 ) , ( 0 , 1 2 ) and ( 1 2 , ∞ ) . Find the sign of the rational function on each interval by selecting test value as follows. Let the polynomial function is denoted by g ( x ) = ( x − 1 ) ( x + 1 ) x 2 + 3 . Obtain the sign of polynomial g ( x ) at x = − 1 on the interval ( − ∞ , − 1 2 ) . g ( − 1 ) = ( − 1 − 1 ) ( − 1 + 1 ) ( − 1 ) 2 + 3 = 0 + 3 = 3 > 0 Obtain the sign of polynomial g ( x ) at x = − 1 4 on the interval ( − 1 2 , 0 ) . g ( − 1 4 ) = ( − 1 4 − 1 ) ( − 1 4 + 1 ) ( − 1 4 ) 2 + 3 = − 15 + 3 = − 12 < 0 Obtain the sign of polynomial g ( x ) at x = 1 4 on the interval ( 0 , 1 2 )

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ISBN: 2810000027802

Author: BITTINGER

Publisher: PEARSON

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Chapter 4, Problem 87RE

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To solve: The rational inequality |1−1x2|<3.

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Chapter 4, Problem 86RE

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