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 )

6th Edition

ISBN: 2810000027802

Author: BITTINGER

Publisher: PEARSON

See similar textbooks

Concept explainers

Equations and Inequations

Equations and inequalities describe the relationship between two mathematical expressions.

Linear Functions

A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.

Videos

Question

Chapter 4, Problem 87RE

To determine

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

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 4, Problem 86RE

Chapter 4, Problem 88RE

Chapter 4 Solutions

COLLEGE ALGEBRA (PRINT UPGRADE)

Ch. 4.1 - Determine the leading term, the leading...Ch. 4.1 - Prob. 2E Ch. 4.1 - Determine the leading term, the leading...Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Determine the leading term, the leading...Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E Ch. 4.1 - Determine the leading term, the leading...Ch. 4.1 - Prob. 10E

Knowledge Booster

$Algebra$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, algebra and related others by exploring similar questions and additional content below.