Functions Modeling Change: A Preparation for Calculus, 5th Edition

Published by Wiley
ISBN 10: 1118583191
ISBN 13: 978-1-11858-319-7

Chapter 11 - Polynomial and Rational Functions - 11.1 Power Functions and Proportionality - Exercises and Problems for Section 11.1 - Exercises and Problems - Page 439: 3


Yes, $g(x)=-\frac{1}{6}x^9$

Work Step by Step

Use the fact that $(-x^3)^3=(-1 \cdot x^3)^3$, $(ab)^n=a^n(b^n)$, and $(a^m)^n=a^{mn}$ to get $$g(x)=\frac{(-1)^3(x^3)^3}{6}=-\frac{x^9}{6}=-\frac{1}{6}x^9$$ Therefore, this is a power function with $k=-\frac{1}{6}$ and $p=9$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.