Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.7 - Polynomial and Rational Inequalities - Exercise Set - Page 413: 53


$(-∞, -4] ∪ (-2, 1]$

Work Step by Step

Consider the Rational Inequality as follows: $\frac{(x+4)(x-1)}{x+2}≤0$ Here are the steps required for Solving Rational Inequalities: Step 1: One side must be zero and the other side can have only one fraction, so we simplify the fractions if there is more than one fraction. $\frac{(x+4)(x-1)}{x+2}≤0$ Step 2: Critical or Key Values are first evaluated. In order to this, set the numerator and denominator of the fraction equal to zero and solve. $x+4 = 0$ This implies $x =-4$ and $x-1=0$ This implies $ x =1$ Also, $x+2=0$ This implies $ x =-2$ These solutions are used as boundary points on a number line. Step 3: Locate the boundary points on a number line and divide the number line into intervals. The boundary points divide the number line into four intervals: $(-∞, -4), (-4, -2),(-2, 1), (1, ∞)$ Step 4: Now, one test value within each interval is chosen and $f$ is evaluated at that number. Intervals: $(-∞, -4), (-4, -2), (-2, 1), (1, ∞)$ Test value: $-5$ $-3$ $0$ $2$ Sign Change: Negative Positive Negative Positive $f (x)< 0?$: T F T F Step 5: Write the solution set, selecting the interval or intervals that satisfy the given inequality. We are interested in solving $f (x)≤ 0$. Based on our work done in Step 4, we see that $f (x)≤ 0$ for all x in $(-∞, -4]$ or $(-2, 1]$ . However, the inequality involves (less than or equal to), we must also include the solution of $f (x)= 0$ , namely -4 and 1 in the solution set. Conclusion: Thus, the interval notation of the given inequality is $(-∞, -4] ∪ (-2, 1]$ and the graph of the solution set on a number line is shown as follows:
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