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 rational inequality is x + 1 x − 2 − x − 3 x − 1 < 0 . By the procedure mentioned above, find the solution set of the inequality. Change the inequlity sign by equal sign to form the related equation as x + 1 x − 2 − x − 3 x − 1 = 0 . The values of x for which the related function is not defined or is 0 are called the critical values. The rational equation is undefined at the points x = 2 and x = 1 . Find the value of x from the equation x + 1 x − 2 − x − 3 x − 1 = 0 by multiplying by ( x − 2 ) ( x − 1 ) on both sides. x + 1 x − 2 − x − 3 x − 1 = 0 ( x − 2 ) ( x − 1 ) ⋅ ( x + 1 x − 2 − x − 3 x − 1 ) = 0 ⋅ ( x − 2 ) ( x − 1 ) ( x + 1 ) ( x − 1 ) − ( x − 3 ) ( x − 2 ) = 0 x 2 − 1 − x 2 + 5 x − 6 = 0 On further simplification, 5 x − 7 = 0 x = 7 5 Thus, the critical values of the rational equation are 1 , 7 5 , and 2 . The critical values divide the x axis into the intervals as ( − ∞ , 1 ) , ( 1 , 7 5 ) , ( 7 5 , 2 ) and ( 2 , ∞ ) . Find the sign of the rational function on each interval by selecting test values as follows. Let the polynomial function be g ( x ) = x + 1 x − 2 − x − 3 x − 1 . Obtain the sign of polynomial g ( x ) = x + 1 x − 2 − x − 3 x − 1 at x = 0 on the interval ( − ∞ , 1 )

College Algebra: Graphs and Models (6th Edition)

6th Edition

ISBN: 9780134179032

Author: Marvin L. Bittinger, Judith A. Beecher, David J. Ellenbogen, Judith A. Penna

Publisher: PEARSON

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Chapter 4.6, Problem 62E

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To solve: The inequality x+1x−2−x−3x−1<0 by listing the critical values of the related function.

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Chapter 4.6, Problem 61E

Chapter 4.6, Problem 63E

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Chapter 4 Solutions

College Algebra: Graphs and Models (6th Edition)

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

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The inequality x + 1 x − 2 − x − 3 x − 1 < 0 by listing the critical values of the related function.

Solution Summary: The author explains how to solve the inequality by listing the critical values of the related function.

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To solve: The inequality x+1x−2−x−3x−1<0 by listing the critical values of the related function.

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The inequality x + 1 x − 2 − x − 3 x − 1 &lt; 0 by listing the critical values of the related function.

Solution Summary: The author explains how to solve the inequality by listing the critical values of the related function.

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To solve: The inequality x+1x−2−x−3x−1<0 by listing the critical values of the related function.

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Chapter 4 Solutions

The inequality x + 1 x − 2 − x − 3 x − 1 < 0 by listing the critical values of the related function.