Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.7 - Polynomial and Rational Inequalities - Exercise Set - Page 414: 77

Answer

a. $f(35)\approx160\ ft$. $g(35) \approx185\ ft$. b. the upper blue curve is for the wet pavement; the lower red curve is for the dry pavement. c. very well. d. $x\gt76\ mph$; see explanations.
1582369601

Work Step by Step

a. With $x=35\ mph$, for the dry pavement, we have $f(35)=0.0875(35)^2-0.4(35)+66.6=159.7875\approx160\ ft$ Similarly, for the wet pavement, we have $g(35)=0.0875(35)^2+1.9(35)+11.6=185.2875\approx185\ ft$. b. Clearly, the stopping distance is larger for web the pavement than that for the dry pavement. Based on the graph, we can identify that the upper blue curve is for the wet pavement, while the lower red curve is for the dry pavement. c. The actual data shows that the stopping distances for dry pavement is 160 feet and for the wet pavement is 185 feet when the speed is 35 mph. Thus, the model in part-a fits the data very well. d. Let $f(x)\gt540$; we have $0.0875x^2-0.4x+66.6\gt540$, or $0.0875x^2-0.4x-473.4\gt0$ Solving the inequality graphically as shown in the figure, we have $x\gt76\ mph$. The corresponding point can be found on the graph by locating $x=76\ mph$ and then by finding the vertical coordinate on the red curve, which should give a value of about 540 feet.
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.