Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 7 - Analytic Trigonometry - 7.4 Trigonometric Identities - 7.4 Assess Your Understanding - Page 477: 111

Answer

maximum $f(20)= 1250$

Work Step by Step

Step 1. Given $f(x)=-3x^2+120x+50$, $a_2=-3\lt0$, so there is a maximum. Step 2. The maximum happens at $x=-\frac{120}{2(-3)}=20$ and we have $f(20)=-3(20)^2+120(20)+50=1250$
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