College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 3, Polynomial and Rational Functions - Section 3.7 - Polynomial and Rational Inequalities - 3.7 Exercises - Page 352: 34

Answer

$[-2, -1) \cup [9, \infty)$

Work Step by Step

1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$ where $f$ is a Rational function. $\displaystyle \frac{x}{2}\geq \frac{5}{x+1} + 4$, $\displaystyle\frac{x}{2} -\frac{5}{x+1} -4\geq 0$ , $\displaystyle\frac{x^2+x-10-8x-8}{2(x+1)}\geq 0$ , $\displaystyle \frac{(x-9)(x+2)}{2(x+1)}\geq 0$ , $f(x)=\displaystyle \frac{(x-9)(x+2)}{2(x+1)}\geq 0$, 2.The cut points are: $\displaystyle \frac{(x-9)(x+2)}{2(x+1)}=0$ $x=-2$ or $x=-1$ or $x=9$ 3. Make a table or diagram: use the test values to make a table or diagram of the sign of each factor in each interval. 4. Test each interval's sign of $f(x)$ with a test value, $\begin{array}{llll} Intervals: & a=test.v. & f(a),signs & f(a) \geq 0 ? \\ & & \displaystyle \displaystyle \frac{(a-9)(a+2)}{2(a+1)} & \\ (-\infty,-2) & -5 & \frac{(-)(-)}{(-)}=(-) & F\\ (-2, -1) & -\frac{3}{2} & \frac{(-)(+)}{(-)}=(+) & T\\ (-1, 9) & 0 & \frac{(-)(+)}{(+)}=(-) & F\\ (9,\infty) & 10 & \frac{(+)(+)}{(+)}=(+) & T\\ \end{array}$ 5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points unless they are zeros of the denominator. Solution set: $[-2, -1) \cup [9, \infty)$
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.