Explanation Result used: 1. If | f ( x ) | > n , then f ( x ) < − n or f ( x ) > n . 2. If | f ( x ) | < n , then − n < f ( x ) < n . Calculation: The given inequality is | x 2 + x − 4 | < 2 . By result 2, the inequality becomes − 2 < x 2 + x − 4 < 2 . Consider the inequality x 2 + x − 4 > − 2 . Rewrite the above inequality with zero on one side. x 2 + x − 4 + 2 > − 2 + 2 x 2 + x − 2 > 0 ( x − 1 ) ( x + 2 ) > 0 Setting each factors equal to zero the critical values are found as follows. ( x − 1 ) ( x + 2 ) = 0 implies x = 1 and x = − 2 The critical values are x = 1 and x = − 2 . Therefore, the sign of the given function for each of the intervals x < − 2 , − 2 < x < 1 and x > 1 . The sign of the function for corresponding intervals are tabulated as follows. Interval Test value ( x − 1 ) ( x + 2 ) Sign x < − 2 ( − ∞ , − 2 ) –3 ( − ) ( − ) + − 2 < x < 1 ( − 2 , 1 ) − 1.5 ( − ) ( + ) – x > 1 ( 1 , ∞ ) 2 ( + ) ( + ) + From the above table, the values for which the product is greater than zero. Then the solution for the inequality x 2 + x − 4 > − 2 is x < − 2 or x > 1 . Consider the inequality x 2 + x − 4 < 2

Mylab Math With Pearson Etext -- 18 Week Standalone Access Card -- For Basic Technical Mathematics With Calculus

11th Edition

ISBN: 9780135902912

Author: Allyn J. Washington

Publisher: PEARSON

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Chapter 17.4, Problem 31E

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Mylab Math With Pearson Etext -- 18 Week Standalone Access Card -- For Basic Technical Mathematics With Calculus

Ch. 17.1 - For −6 < 3, determine the inequality if 1. 8 is...Ch. 17.1 - Prob. 2PE Ch. 17.1 - For the inequality −6 < 3, state the inequality...Ch. 17.1 - Prob. 4PE Ch. 17.1 - Prob. 5PE Ch. 17.1 - In Exercises 1–4, make the given changes in the...Ch. 17.1 - Prob. 2E Ch. 17.1 - Prob. 3E Ch. 17.1 - Prob. 4E Ch. 17.1 - In Exercises 5–12, for the inequality 4 < 9, state...

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