## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 2 - Section 2.7 - Polynomial and Rational Inequalities - Exercise Set - Page 413: 73

#### Answer

$(-∞, -2) ∪ [-1, 2)$

#### Work Step by Step

Consider the Polynomial Inequality as follows: $\frac{1}{4(x+2)}$$≤$$-$$\frac{3}{4(x-2)} \frac{1}{4(x+2)}+\frac{3}{4(x-2)}≤0 \frac{4(x-2)+12(x+2)}{16(x+2)(x-2)}$$\leq$$0 \frac{4x-8+12x+24}{16(x^{2}-4)}$$\leq$$0 \frac{16(x+1)}{16(x^{2}-4)}$$\leq$$0$ $\frac{x+1}{x^{2}-4}≤0$ The graph of $f(x)$ is less than 0 from $-∞$ to $-2$ and from $-1$ to $2$ and goes to the infinities in $2$ and $-2$, they never touch that point. At this point, these infinities are not to be considered in the domain. Conclusion: Thus, the interval notation of the inequality is$(-∞, -2) ∪ [-1, 2)$

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