Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Appendix A - Appendix A Exercises - Page 1158: 29

Answer

$$\left( { - \infty ,4} \right] \cup \left[ {5,6} \right)$$

Work Step by Step

$$\eqalign{ & \frac{{{x^2} - 9x + 20}}{{x - 6}} \leqslant 0 \cr & {\text{Factor the numerator}} \cr & \frac{{\left( {x - 5} \right)\left( {x - 4} \right)}}{{x - 6}} \leqslant 0 \cr & {\text{The critical values are}} \cr & x = 4,\,\,x = 5,\,\,\,x = 6 \cr & \cr & {\text{ These critical values partition the number line into}} \cr & {\text{four intervals:}}\,\,\,\left( { - \infty ,4} \right],\,\,\left[ {4,5} \right]{\text{,}}\,\,\left[ {5,6} \right)\,{\text{ and }}\left( {6,\infty } \right).{\text{ }} \cr & {\text{Testing the intervals:}} \cr & {\text{*For the interval }}\left( { - \infty ,4} \right]{\text{ use the test value }}x = 3 \cr & \frac{{\left( {3 - 5} \right)\left( {3 - 4} \right)}}{{3 - 6}} = - \frac{2}{3},\,\,\,negative \cr & {\text{*For the interval }}\left[ {4,5} \right]{\text{ use the test value }}x = 4.5 \cr & \frac{{\left( {4.5 - 5} \right)\left( {4.5 - 4} \right)}}{{4.5 - 6}} = \frac{1}{6},\,\,\,positive \cr & {\text{*For the interval }}\left[ {5,6} \right){\text{ use the test value }}x = 5.5 \cr & \frac{{\left( {5.5 - 5} \right)\left( {5.5 - 4} \right)}}{{5.5 - 6}} = - \frac{3}{2},\,\,\,negative \cr & {\text{*For the interval }}\left( {6,\infty } \right){\text{ use the test value }}x = 7 \cr & \frac{{\left( {7 - 5} \right)\left( {7 - 4} \right)}}{{7 - 6}} = 6,\,\,\,positive \cr & \cr & {\text{Therefore, }}\frac{{\left( {x - 5} \right)\left( {x - 4} \right)}}{{x - 6}} \leqslant 0{\text{ on the intervals}} \cr & \left( { - \infty ,4} \right]{\text{ and }}\left[ {5,6} \right) \cr & \cr & then \cr & {\text{The solution set of the inequailty }}\frac{{{x^2} - 9x + 20}}{{x - 6}} \leqslant 0{\text{ is}} \cr & \left( { - \infty ,4} \right] \cup \left[ {5,6} \right) \cr & \cr & {\text{Graph}} \cr & \cr} $$
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