Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 3 - Applications of Differentiation - 3.1 Maximum and Minimum Values - 3.1 Execises - Page 211: 2


Absolute maximum: 15 at x=1 or (1,15) Absolute minimum: 7 at x=5 or (5,7)

Work Step by Step

f(x)=12+4x-2x^{2} [0,5] f'(x)=4-4x 0=4-4x 4x=4 x=1 Test with 0: f'(x)=4-4*0=4 (positive so increasing) Test with 2: f'(x)=4-4*2=-4 (negative so decreasing) So x=1 is a maximum: f(1)=12+4*1-1^2=15 Test the extremities: f(0)=12+4*0-0^2=12 f(5)=12+4*5-5^2=7 So x=5 is a minimum.
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