## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 1 - Section 1.8 - Inequalities - 1.8 Exercises: 65

#### Answer

SOLUTION $\frac{(2+x)(2-x)}{x}\lt0$ INTERVAL (-2, 0) (2, $\infty)$ GRAPH In the real number line there should be the numbers -2, 0 and 2. In those three there should be an opne circle. The you connect -2 and 0. The you draw an arrow from 2 to $\infty$.

#### Work Step by Step

$\frac{4}{x}\lt(x)$ $\frac{4}{x}-x\lt0$ $\frac{4}{x}-\frac{x^{2}}{x}\lt0$ $\frac{4-x^{2}}{x}\lt0$ $\frac{(2+x)(2-x)}{x}\lt0$ In this inequality the answer can’t be equal to 0 therefore the answer go in parenthses. The solutions of the numerator are x=-2 and x=2 The solutions of the denominator is x=0 INTERVAL $(-\infty, -2) (-2,0) (0,2) (2,\infty)$ Test the intervals to know which one is the solution $(-\infty, -2)$ $y(-4)=\frac{(2-4)(2+4)}{-4}\lt0$ $y(-4)=3$ The result is greater than 0 so it is not a solution. (-2,0) $y(-1)=\frac{(2-1)(2+1)}{-1}\lt0$ $y(-1)=-3$ The result is less than 0 so the interval is a solution. (0,2) $y(1)=\frac{(2+1)(2-1)}{1}\lt0$ $y(1)=3$ It is not a solution $(2, \infty)$ $y(4)=\frac{(2+4)(2-4)}{4}\lt0$ $y(4)= -3$ It is a solution. SOLUTION (-2,0) $(2,\infty)$ GRAPH In the real number line there should be the numbers -2, 0 and 2. In those three there should be an opne circle. The you connect -2 and 0. The you draw an arrow from 2 to $\infty$.

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