Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises: 67

Answer

$$f\left( x \right) = {x^2} - 3x + 4$$

Work Step by Step

$$\eqalign{ & f'\left( x \right) = 2x - 3 \cr & f\left( x \right) = \int {f'\left( x \right)} dx \cr & then \cr & f\left( x \right) = \int {\left( {2x - 3} \right)} dx \cr & find{\text{ the general solution}} \cr & f\left( x \right) = {x^2} - 3x + C \cr & {\text{using the initial condition }}F\left( 0 \right) = 4 \cr & 4 = {\left( 0 \right)^2} - 3\left( 0 \right) + C \cr & C = 4 \cr & {\text{the solution to the initial value problem is}} \cr & f\left( x \right) = {x^2} - 3x + 4 \cr} $$
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