/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 Describe the domain and range of... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 28

Describe the domain and range of the function. $$ g(x, y)=x \sqrt{y} $$

Short Answer

Expert verified
The domain of the function \(g(x, y) = x \sqrt{y}\) is \(D = \{(x,y) : x \in \mathbb{R}, y \geq 0\}\) and the range is all real numbers \(R = \{ g(x,y) : g(x,y) \in \mathbb{R} \}\).

Step by step solution

01

Determine the domain

The domain of the function is the set of all possible input values for which the function is defined. First, note that the variable x does not impact the definition of the function, because x can be any real number. The square root function sqrt(y) is undefined for negative values of y. Therefore, to satisfy this condition, y has to be greater than or equal to 0. So the domain will be \(D = \{(x,y) : x \in \mathbb{R}, y \geq 0\}\).
02

Determine the range

The range of a function is the set of all possible output values (y-values). The function \(g(x, y) = x \sqrt{y}\) can generate every possible real number. Here's why: For y>0, as x varies from negative to positive infinity, the function will also vary from negative to positive infinity. For y=0, the function is always 0 regardless of the value of x. Therefore, the range of the function is all real numbers \(R = \{ g(x,y) : g(x,y) \in \mathbb{R} \}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Differentiate implicitly to find the first partial derivatives of \(z\) \(x z+y z+x y=0\)

Use the gradient to find a unit normal vector to the graph of the equation at the given point. Sketch your results $$ 9 x^{2}+4 y^{2}=40,(2,-1) $$

Find \(\partial w / \partial s\) and \(\partial w / \partial t\) using the appropriate Chain Rule, and evaluate each partial derivative at the given values of \(s\) and \(t\) $$ \begin{array}{l} \text { Function } \\ \hline w=y^{3}-3 x^{2} y \\ x=e^{s}, \quad y=e^{t} \end{array} $$ $$ \frac{\text { Point }}{s=0, \quad t=1} $$

Find the gradient of the function and the maximum value of the directional derivative at the given point. $$ \frac{\text { Function }}{w=x y^{2} z^{2}} \frac{\text { Point }}{(2,1,1)} $$

In Exercises 19 and \(20,\) use the gradient to find the directional derivative of the function at \(P\) in the direction of \(Q\). $$ g(x, y)=x^{2}+y^{2}+1, \quad P(1,2), Q(3,6) $$
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.