College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 3 - Section 3.1 - Functions - 3.1 Assess Your Understanding: 67

Answer

(a) $(f+g)(x)=5x+1$ (b) $(f-g(x)=x+7$ (c) $(f . g)(x)=6x^2-x-12$ (d) $(\frac{f}{g})(x)=\frac{3x+4}{2x-3}$ The domain is {$x/x\ne\frac{3}{2}$} (e) $(f+g)(3)=16$ (f) $(f-g(4)=11$ (g) $(f . g)(2)=21$ (h)$(\frac{f}{g})(1)=\frac{f(1)}{g(1)}=-7$

Work Step by Step

(a) $(f+g)(x)=$$f(x)+g(x)=3x+4+2x-3=5x+1$ (b) $(f-g)(x)=f(x)-g(x)=(3x+4)-(2x-3)=3x+4-2x+3=x+7$ (c) $(f . g)(x)=f(x).g(x)=(3x+4)(2x-3)=6x^2-9x+8x-12=6x^2-x-12$ (d) $(\frac{f}{g})(x)=\frac{f(x)}{g(x)}=\frac{3x+4}{2x-3}$ The domain is {$x/x\ne\frac{3}{2}$} (e) $(f+g)(3)=$$f(3)+g(3)=5(3)+1$=16 (f) $(f-g(4)=f(4)-g(4)=4+7=11$ (g) $(f . g)(2)=f(2).g(2)=6(2)^2-2-1=21$ (h) $(\frac{f}{g})(1)=\frac{f(1)}{g(1)}=\frac{3(1)+4}{2(1)-3}=\frac{7}{-1}=-7$
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.