Answer
$-\frac{558}{81}$.
Work Step by Step
Since $ f(x)=\frac{12}{x}-x^3$, then we get
$$ f'(x)=-\frac{12}{x^2}-3x^2, \quad f''(x)=\frac{24}{x^3}-6x, \quad f'''(x)=-\frac{72}{x^4}-6.$$
Hence, $ f'''(-3)=-\frac{72}{81}-6=-\frac{558}{81}$.
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 3 - Differentiation - 3.5 Higher Derivatives - Exercises - Page 135: 23
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