Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.1 Functions of Several Variables - 14.1 Exercises: 9

Answer

(a) $g(2,-1)=1$ (b) Domain: $\mathbb{R^2}.$ (c) Range: $[-1,1]$.

Work Step by Step

(a) By direct calculation $$g(2,-1)=\cos(2+2\cdot(-1))=\cos0=1.$$ (b) The argument of cosine can be any real number so $x+2y$ can take any real value. Also, any real $x$ and $y$ can be taken into consideration because using the formula $x+2y$ will just produce another real number which is a 'legal' argument for cosine. So the domain is $$\mathcal{D}=\mathbb{R}×\mathbb{R}=\mathbb{R}^2.$$ (c) Cosine can only take values between $-1$ and $1$. Since the argument of the domain will cover all real numbers, the whole range of the cosine will be covered as well so we get for the range $$\mathcal{R}=[-1,1].$$
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.