Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 12 - Section 12.3 - Higher Order Derivatives: Acceleration and Concavity - Exercises - Page 902: 45

Answer

$f'(x)=8x-1$ $f''(x)=8$ $f'''(x)=0$ $f^{(4)}(x)=\ldots=f^{(n)}(x)=0$

Work Step by Step

$f'''(x)=\displaystyle \frac{d}{dx}\left[f''(x)\right]$ $f^{(4)}(x)=\displaystyle \frac{d}{dx}\left[f'''(x)\right]$ $\displaystyle \quad f^{(5)}(x)=\frac{d}{dx}\left[f^{(4)}(x)\right]$ and so on, assuming all these derivatives exist. --- $f(x)=4x^{2}-x+1$ $f'(x)=8x-1$ $f''(x)=8$ $f'''(x)=0$ $f^{(4)}(x)=\ldots=f^{(n)}(x)=0$
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