Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises: 49

Answer

$f(1) = 8$

Work Step by Step

We know that: $f'(x)=3-4x$ Thus: $f(x) = 3x - 2x^2 + C$ Use the information given of $(2,5)$: $5 = 3(2) - 2(2)^2 + C$ $5 = 6 - 8 + C$ $5 = -2 + C$ $7 = C$ $f(x) = 3x - 2x^2 + 7$ Substitute $x = 1$ $f(1) = 3(1) - 2(1)^2 + 7$ $f(1) = 3 - 2 + 7$ $f(1) = 8$
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