Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 2 - Section 2.1 - Functions - Exercises: 20

Answer

$$f(-2)=-12$$ $$f(-1)=-3$$ $$f(0)=0$$ $$f(\frac{1}{2})=\frac{9}{8}$$

Work Step by Step

$$f(-2)=(-2)^3+2(-2)$$ $$f(-2)=-8-4$$ $$f(-2)=-12$$ $$f(-1)=(-1)^3+2(-1)$$ $$f(-1)=-1-2$$ $$f(-1)=-3$$ $$f(0)=(0)^3+2(0)$$ $$f(0)=0+0$$ $$f(0)=0$$ $$f(\frac{1}{2})=\frac{9}{8}$$ $$f(\frac{1}{2})=(\frac{1}{2})^3+2(\frac{1}{2})$$ $$f(\frac{1}{2})=\frac{1}{2^3}+\frac{2}{2}$$ $$f(\frac{1}{2})=\frac{1}{8}+1$$ $$f(\frac{1}{2})=\frac{1}{8}+1\times\frac{8}{8}$$ $$f(\frac{1}{2})=\frac{1}{8}+\frac{8}{8}$$ $$f(\frac{1}{2})=\frac{9}{8}$$
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