Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 4 - Exponential and Logarithmic Functions - Section 4.1 Composite Functions - 4.1 Assess Your Understanding - Page 280: 61


$(f\circ g) (x)=11$ and $(g\circ f) (x)=2$

Work Step by Step

Let us consider $f(x)=2x^3-3x^2+4x-1$ and $g(x)=2$ Then we have: $(f\circ g) (x) =f[g(x)]=f [2]=(2)^3-(3)(2)^2+4(2)-1=16-12+8-1=11$ and $(g\circ f) (x) = g[f(x)]=g(2x^3-3x^2+4x-1)=2$ Therefore, $(f\circ g) (x)=11$ and $(g\circ f) (x)=2$
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.