Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.8 - Inequalities - 1.8 Exercises - Page 89: 69

Answer

SOLUTION $\frac{(2+x)(3-x)}{x(x-1)} \geq0$ INTERVAL [-2,0) (1,3] GRAPH in the real number line put four numbers. Put an open circle in 0 and 1. Put a close circle in -2 and 3. Connect the -2 and 0 and connect the 1 and three.

Work Step by Step

$\frac{6}{x-1}-\frac{6}{x} \geq1$ $\frac{6}{x-1}-\frac{6}{x}-1 \geq0$ $\frac{6(x)-6(x-1)-1(x^{2}-1)}{x(x-1)} \geq0$ $\frac{6x-6x+6-x^{2}+x}{x(x-1)} \geq0$ $\frac{x-x^{2}+6}{x(x-1)} \geq0$ $\frac{(2+x)(3-x)}{x(x-1)} \geq0$ You need to see what are the solutions of the inequality. Key numbers: NUMERATOR x=-2 x=3 DENOMINATOR x=0 x=1 INTERVALS $(- \infty,-2] [-2,0) (0,1) (1,3] [3,\infty)$ test the values to know if there is a solution. $(- \infty,-2]$ $y(-4)=\frac{(2-4)(3+4)}{-4(-4-1)} \geq0= y(-4)=-0.7$ No solution. $[-2,0)$ $y(-1)=\frac{(2-1)(3+1)}{-1(-1-1)} \geq0= y(-1)=2$ Solution. $(0,1)$ $y(0.5)=\frac{(2+0.5)(3-0.5)}{0.5(0.5-1)} \geq0= y(0.5)=-25$ No solution $(1,3]$ $y(2)=\frac{(2+2)(3-2)}{2(2-1)} \geq0= y(2)=2$ Solution. $[3,\infty)$ $y(6)=\frac{(2+6)(3-6)}{6(6-1)} \geq0= y(6)=-0.8$ No solution. CORRECT [-2,0) (1,3] GRAPH in the real number line put four numbers. Put an open circle in 0 and 1. Put a close circle in -2 and 3. Connect the -2 and 0 and connect the 1 and three.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.