Answer
$f(x)=x^2+x+4$
Work Step by Step
$f''(x)=2\hspace{8mm}f'(2)=5\hspace{8mm}f(2)=10$
$\int f''(x)dx=\int 2dx=2x+C=f'(x)$
$f'(2)=4+C=5$
$C=1$
$f'(x)=2x+1$
$\int f'(x)dx=\int (2x+1)dx=x^2+x+C=f(x)$
$f(2)=2^2+2+C=10$
$10-2-2^2=4$
$f(x)=x^2+x+4$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 4 - Integration - 4.1 Exercises - Page 251: 39
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