CALCULUS W/SAPLING ACCESS >IC<
CALCULUS W/SAPLING ACCESS >IC<
4th Edition
ISBN: 9781319323394
Author: Rogawski
Publisher: MAC HIGHER
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Question
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Chapter 15, Problem 1CRE
To determine

(a)

Sketch the domain of the function f(x,y) = x2− y2x + 3 

Expert Solution
Check Mark

Answer to Problem 1CRE

Solution:Domain of the function f(x,y) = x2− y2x + 3  is D = {(x, y) : x ≥ ±y  ; x ≠ - 3 }

Explanation of Solution

Domain: The domain of the function is defined as the set of complete possible values which will make the function work and gives output as real values.

Given: A function as f(x,y) = x2− y2x + 3 

Formula:

a. Domain of Square root function, x is given as x  ≥ 0 

b. The expression in the denominator can never be zero.

Calculation:

Given function is f(x,y) = x2− y2x + 3 

Domain of the functionf(x,y) = x2− y2x + 3 :

To find the domain, set the expression the expression inside the square root greater than equal to zero and the expression in the denominator not equal to zero.

x + 3 ≠ 0  And x2− y2≥ 0 

x ≠ - 3  And x2 ≥ y2 

x ≥ ±y 

Thus, domain of the function is D = {(x, y) : x ≥ ±y  ; x ≠ - 3 }

Graph is as follows:

CALCULUS W/SAPLING ACCESS >IC<, Chapter 15, Problem 1CRE

Conclusion: Domain of the function f(x,y) = x2− y2x + 3  is D = {(x, y) : x ≥ ±y  ; x ≠ - 3 }

To determine

(b)

The value of f(3,1) and f(−5,−3) for the function f(x,y) = x2− y2x + 3 

Expert Solution
Check Mark

Answer to Problem 1CRE

Solution: The value of f(3,1) = 23 and f(−5,−3) = −2

Explanation of Solution

Domain: The domain of the function is defined as the set of complete possible values which will make the function work and gives output as real values.

Given: A function as f(x,y) = x2− y2x + 3 

Calculation:

Given function is f(x,y,z) = x2− y2x + 3 

f(3,1) = (3)2− (1)2(3) + 3 = 9 − 13 + 3 = 86 = 226 = 23

f(−5,−3) = (−5)2− (−3)2(−5) + 3 = 25 − 9- 5 + 3 = 16−2 = 4−2 = −2

Conclusion: The value of f(3,1) = 23 and f(−5,−3) = −2

To determine

(c)

A point (x, y) such that f(x,y)=1 for the function f(x,y) = x2− y2x + 3 

Expert Solution
Check Mark

Answer to Problem 1CRE

Solution: The point (x, y) such that f(x,y)=1 for the function f(x,y) = x2− y2x + 3  Are (−3,33) and (−3,−33)

Explanation of Solution

Domain: The domain of the function is defined as the set of complete possible values which will make the function work and gives output as real values.

Given: A function as

f(x,y) = x2− y2x + 3 

Calculation:

Given function is f(x,y,z) = x2− y2x + 3 

Also, f(x,y)=1

f(x,y,z) = x2− y2x + 3 = 1

x2− y2x + 3 = 1

x2− y2 = x + 3

Squaring both the sides, we get

x2− y2 = (x + 3)2

x2− y2 = x2 + 9 + 6x

 y2 = -( 9 + 6x)

Let x = −6,

Then  y2 = -( 9 + 6x)=−(9−36)=27

Thus, y = ±33

Points are (−3,33) and (−3,−33)

Conclusion: The point (x, y) such that f(x,y)=1 for the function f(x,y) = x2− y2x + 3  are (−3,33) and (−3,−33)

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Chapter 15 Solutions

CALCULUS W/SAPLING ACCESS >IC<

Ch. 15.1 - Prob. 1PQCh. 15.1 - Prob. 2PQCh. 15.1 - Prob. 3PQCh. 15.1 - Prob. 4PQCh. 15.1 - Prob. 5PQCh. 15.1 - Prob. 1ECh. 15.1 - Prob. 2ECh. 15.1 - Prob. 3ECh. 15.1 - Prob. 4ECh. 15.1 - Prob. 5E
Ch. 15.1 - Prob. 6ECh. 15.1 - Prob. 7ECh. 15.1 - Prob. 8ECh. 15.1 - Prob. 9ECh. 15.1 - Prob. 10ECh. 15.1 - Prob. 11ECh. 15.1 - Prob. 12ECh. 15.1 - Prob. 13ECh. 15.1 - Prob. 14ECh. 15.1 - Prob. 15ECh. 15.1 - Prob. 16ECh. 15.1 - Prob. 17ECh. 15.1 - Prob. 18ECh. 15.1 - Prob. 19ECh. 15.1 - Prob. 20ECh. 15.1 - Prob. 21ECh. 15.1 - Prob. 22ECh. 15.1 - Prob. 23ECh. 15.1 - Prob. 24ECh. 15.1 - Prob. 25ECh. 15.1 - Prob. 26ECh. 15.1 - Prob. 27ECh. 15.1 - Prob. 28ECh. 15.1 - Prob. 29ECh. 15.1 - Prob. 30ECh. 15.1 - Prob. 31ECh. 15.1 - Prob. 32ECh. 15.1 - Prob. 33ECh. 15.1 - Prob. 34ECh. 15.1 - Prob. 35ECh. 15.1 - Prob. 36ECh. 15.1 - Prob. 37ECh. 15.1 - Prob. 38ECh. 15.1 - Prob. 39ECh. 15.1 - Prob. 40ECh. 15.1 - Prob. 41ECh. 15.1 - Prob. 42ECh. 15.1 - Prob. 43ECh. 15.1 - Prob. 44ECh. 15.1 - Prob. 45ECh. 15.1 - Prob. 46ECh. 15.1 - Prob. 47ECh. 15.1 - Prob. 48ECh. 15.1 - Prob. 49ECh. 15.1 - Prob. 50ECh. 15.1 - Prob. 51ECh. 15.1 - Prob. 52ECh. 15.1 - Prob. 53ECh. 15.1 - Prob. 54ECh. 15.1 - Prob. 55ECh. 15.1 - Prob. 56ECh. 15.1 - Prob. 57ECh. 15.1 - Prob. 58ECh. 15.1 - Prob. 59ECh. 15.1 - Prob. 60ECh. 15.1 - Prob. 61ECh. 15.2 - Prob. 1PQCh. 15.2 - Prob. 2PQCh. 15.2 - Prob. 3PQCh. 15.2 - Prob. 4PQCh. 15.2 - Prob. 1ECh. 15.2 - Prob. 2ECh. 15.2 - Prob. 3ECh. 15.2 - Prob. 4ECh. 15.2 - Prob. 5ECh. 15.2 - Prob. 6ECh. 15.2 - Prob. 7ECh. 15.2 - Prob. 8ECh. 15.2 - Prob. 9ECh. 15.2 - Prob. 10ECh. 15.2 - Prob. 11ECh. 15.2 - Prob. 12ECh. 15.2 - Prob. 13ECh. 15.2 - Prob. 14ECh. 15.2 - Prob. 15ECh. 15.2 - Prob. 16ECh. 15.2 - Prob. 17ECh. 15.2 - Prob. 18ECh. 15.2 - Prob. 19ECh. 15.2 - Prob. 20ECh. 15.2 - Prob. 21ECh. 15.2 - Prob. 22ECh. 15.2 - Prob. 23ECh. 15.2 - Prob. 24ECh. 15.2 - Prob. 25ECh. 15.2 - Prob. 26ECh. 15.2 - Prob. 27ECh. 15.2 - Prob. 28ECh. 15.2 - Prob. 29ECh. 15.2 - Prob. 30ECh. 15.2 - Prob. 31ECh. 15.2 - Prob. 32ECh. 15.2 - Prob. 33ECh. 15.2 - Prob. 34ECh. 15.2 - Prob. 35ECh. 15.2 - Prob. 36ECh. 15.2 - Prob. 37ECh. 15.2 - Prob. 38ECh. 15.2 - Prob. 39ECh. 15.2 - Prob. 40ECh. 15.2 - Prob. 41ECh. 15.2 - 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