(a) Sketch the domain of the function f ( x , y ) = x 2 − y 2 x + 3

Question

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Answer 1

Question

Chapter 14, Problem 1CRE

To determine

(a)

Sketch the domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Expert Solution

Answer to Problem 1CRE

Solution:Domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ is $D = {(x, y) : x \geq \pm y; x \neq - 3}$

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 $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Formula:

a. Domain of Square root function, $\sqrt{x}$ is given as $x \geq 0$

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

Calculation:

Given function is $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ :

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.

$x + 3 \neq 0$ And $x^{2} - y^{2} \geq 0$

$x \neq - 3$ And $x^{2} \geq y^{2}$

$x \geq \pm y$

Thus, domain of the function is $D = {(x, y) : x \geq \pm y; x \neq - 3}$

Graph is as follows:

Loose-Leaf Version for Calculus: Early Transcendentals Single Variable, Chapter 14, Problem 1CRE

Conclusion: Domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ is $D = {(x, y) : x \geq \pm y; x \neq - 3}$

To determine

(b)

The value of $f (3, 1)$ and $f (- 5, - 3)$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Expert Solution

Answer to Problem 1CRE

Solution: The value of $f (3, 1) = \frac{\sqrt{2}}{3}$ 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 $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Calculation:

Given function is $f (x, y, z) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

$f (3, 1) = \frac{\sqrt{{(3)}^{2} - {(1)}^{2}}}{(3) + 3} = \frac{\sqrt{9 - 1}}{3 + 3} = \frac{\sqrt{8}}{6} = \frac{2 \sqrt{2}}{6} = \frac{\sqrt{2}}{3}$

$f (- 5, - 3) = \frac{\sqrt{{(- 5)}^{2} - {(- 3)}^{2}}}{(- 5) + 3} = \frac{\sqrt{25 - 9}}{- 5 + 3} = \frac{\sqrt{16}}{- 2} = \frac{4}{- 2} = - 2$

Conclusion: The value of $f (3, 1) = \frac{\sqrt{2}}{3}$ and $f (- 5, - 3) = - 2$

To determine

(c)

A point $(x, y)$ such that $f (x, y) = 1$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Expert Solution

Answer to Problem 1CRE

Solution: The point $(x, y)$ such that $f (x, y) = 1$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ Are $(- 3, 3 \sqrt{3})$ and $(- 3, - 3 \sqrt{3})$

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

$f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Calculation:

Given function is $f (x, y, z) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Also, $f (x, y) = 1$

$f (x, y, z) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3} = 1$

$\frac{\sqrt{x^{2} - y^{2}}}{x + 3} = 1$

$\sqrt{x^{2} - y^{2}} = x + 3$

Squaring both the sides, we get

$x^{2} - y^{2} = {(x + 3)}^{2}$

$x^{2} - y^{2} = x^{2} + 9 + 6 x$

$y^{2} = - (9 + 6 x)$

Let $x = - 6$ ,

Then $y^{2} = - (9 + 6 x) = - (9 - 36) = 27$

Thus, $y = \pm 3 \sqrt{3}$

Points are $(- 3, 3 \sqrt{3})$ and $(- 3, - 3 \sqrt{3})$

Conclusion: The point $(x, y)$ such that $f (x, y) = 1$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ are $(- 3, 3 \sqrt{3})$ and $(- 3, - 3 \sqrt{3})$

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Answer 2

Question

Chapter 14, Problem 1CRE

To determine

(a)

Sketch the domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Expert Solution

Answer to Problem 1CRE

Solution:Domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ is $D = {(x, y) : x \geq \pm y; x \neq - 3}$

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 $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Formula:

a. Domain of Square root function, $\sqrt{x}$ is given as $x \geq 0$

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

Calculation:

Given function is $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ :

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.

$x + 3 \neq 0$ And $x^{2} - y^{2} \geq 0$

$x \neq - 3$ And $x^{2} \geq y^{2}$

$x \geq \pm y$

Thus, domain of the function is $D = {(x, y) : x \geq \pm y; x \neq - 3}$

Graph is as follows:

Loose-Leaf Version for Calculus: Early Transcendentals Single Variable, Chapter 14, Problem 1CRE

Conclusion: Domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ is $D = {(x, y) : x \geq \pm y; x \neq - 3}$

To determine

(b)

The value of $f (3, 1)$ and $f (- 5, - 3)$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Expert Solution

Answer to Problem 1CRE

Solution: The value of $f (3, 1) = \frac{\sqrt{2}}{3}$ 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 $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Calculation:

Given function is $f (x, y, z) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

$f (3, 1) = \frac{\sqrt{{(3)}^{2} - {(1)}^{2}}}{(3) + 3} = \frac{\sqrt{9 - 1}}{3 + 3} = \frac{\sqrt{8}}{6} = \frac{2 \sqrt{2}}{6} = \frac{\sqrt{2}}{3}$

$f (- 5, - 3) = \frac{\sqrt{{(- 5)}^{2} - {(- 3)}^{2}}}{(- 5) + 3} = \frac{\sqrt{25 - 9}}{- 5 + 3} = \frac{\sqrt{16}}{- 2} = \frac{4}{- 2} = - 2$

Conclusion: The value of $f (3, 1) = \frac{\sqrt{2}}{3}$ and $f (- 5, - 3) = - 2$

To determine

(c)

A point $(x, y)$ such that $f (x, y) = 1$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Expert Solution

Answer to Problem 1CRE

Solution: The point $(x, y)$ such that $f (x, y) = 1$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ Are $(- 3, 3 \sqrt{3})$ and $(- 3, - 3 \sqrt{3})$

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

$f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Calculation:

Given function is $f (x, y, z) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Also, $f (x, y) = 1$

$f (x, y, z) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3} = 1$

$\frac{\sqrt{x^{2} - y^{2}}}{x + 3} = 1$

$\sqrt{x^{2} - y^{2}} = x + 3$

Squaring both the sides, we get

$x^{2} - y^{2} = {(x + 3)}^{2}$

$x^{2} - y^{2} = x^{2} + 9 + 6 x$

$y^{2} = - (9 + 6 x)$

Let $x = - 6$ ,

Then $y^{2} = - (9 + 6 x) = - (9 - 36) = 27$

Thus, $y = \pm 3 \sqrt{3}$

Points are $(- 3, 3 \sqrt{3})$ and $(- 3, - 3 \sqrt{3})$

Conclusion: The point $(x, y)$ such that $f (x, y) = 1$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ are $(- 3, 3 \sqrt{3})$ and $(- 3, - 3 \sqrt{3})$

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Answer 3

Question

Chapter 14, Problem 1CRE

To determine

(a)

Sketch the domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Expert Solution

Answer to Problem 1CRE

Solution:Domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ is $D = {(x, y) : x \geq \pm y; x \neq - 3}$

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 $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Formula:

a. Domain of Square root function, $\sqrt{x}$ is given as $x \geq 0$

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

Calculation:

Given function is $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ :

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.

$x + 3 \neq 0$ And $x^{2} - y^{2} \geq 0$

$x \neq - 3$ And $x^{2} \geq y^{2}$

$x \geq \pm y$

Thus, domain of the function is $D = {(x, y) : x \geq \pm y; x \neq - 3}$

Graph is as follows:

Loose-Leaf Version for Calculus: Early Transcendentals Single Variable, Chapter 14, Problem 1CRE

Conclusion: Domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ is $D = {(x, y) : x \geq \pm y; x \neq - 3}$

To determine

(b)

The value of $f (3, 1)$ and $f (- 5, - 3)$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Expert Solution

Answer to Problem 1CRE

Solution: The value of $f (3, 1) = \frac{\sqrt{2}}{3}$ 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 $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Calculation:

Given function is $f (x, y, z) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

$f (3, 1) = \frac{\sqrt{{(3)}^{2} - {(1)}^{2}}}{(3) + 3} = \frac{\sqrt{9 - 1}}{3 + 3} = \frac{\sqrt{8}}{6} = \frac{2 \sqrt{2}}{6} = \frac{\sqrt{2}}{3}$

$f (- 5, - 3) = \frac{\sqrt{{(- 5)}^{2} - {(- 3)}^{2}}}{(- 5) + 3} = \frac{\sqrt{25 - 9}}{- 5 + 3} = \frac{\sqrt{16}}{- 2} = \frac{4}{- 2} = - 2$

Conclusion: The value of $f (3, 1) = \frac{\sqrt{2}}{3}$ and $f (- 5, - 3) = - 2$

To determine

(c)

A point $(x, y)$ such that $f (x, y) = 1$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Expert Solution

Answer to Problem 1CRE

Solution: The point $(x, y)$ such that $f (x, y) = 1$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ Are $(- 3, 3 \sqrt{3})$ and $(- 3, - 3 \sqrt{3})$

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

$f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Calculation:

Given function is $f (x, y, z) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Also, $f (x, y) = 1$

$f (x, y, z) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3} = 1$

$\frac{\sqrt{x^{2} - y^{2}}}{x + 3} = 1$

$\sqrt{x^{2} - y^{2}} = x + 3$

Squaring both the sides, we get

$x^{2} - y^{2} = {(x + 3)}^{2}$

$x^{2} - y^{2} = x^{2} + 9 + 6 x$

$y^{2} = - (9 + 6 x)$

Let $x = - 6$ ,

Then $y^{2} = - (9 + 6 x) = - (9 - 36) = 27$

Thus, $y = \pm 3 \sqrt{3}$

Points are $(- 3, 3 \sqrt{3})$ and $(- 3, - 3 \sqrt{3})$

Conclusion: The point $(x, y)$ such that $f (x, y) = 1$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ are $(- 3, 3 \sqrt{3})$ and $(- 3, - 3 \sqrt{3})$

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Answer 4

Question

Chapter 14, Problem 1CRE

To determine

(a)

Sketch the domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Expert Solution

Answer to Problem 1CRE

Solution:Domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ is $D = {(x, y) : x \geq \pm y; x \neq - 3}$

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 $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Formula:

a. Domain of Square root function, $\sqrt{x}$ is given as $x \geq 0$

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

Calculation:

Given function is $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ :

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.

$x + 3 \neq 0$ And $x^{2} - y^{2} \geq 0$

$x \neq - 3$ And $x^{2} \geq y^{2}$

$x \geq \pm y$

Thus, domain of the function is $D = {(x, y) : x \geq \pm y; x \neq - 3}$

Graph is as follows:

Loose-Leaf Version for Calculus: Early Transcendentals Single Variable, Chapter 14, Problem 1CRE

Conclusion: Domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ is $D = {(x, y) : x \geq \pm y; x \neq - 3}$

To determine

(b)

The value of $f (3, 1)$ and $f (- 5, - 3)$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Expert Solution

Answer to Problem 1CRE

Solution: The value of $f (3, 1) = \frac{\sqrt{2}}{3}$ 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 $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Calculation:

Given function is $f (x, y, z) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

$f (3, 1) = \frac{\sqrt{{(3)}^{2} - {(1)}^{2}}}{(3) + 3} = \frac{\sqrt{9 - 1}}{3 + 3} = \frac{\sqrt{8}}{6} = \frac{2 \sqrt{2}}{6} = \frac{\sqrt{2}}{3}$

$f (- 5, - 3) = \frac{\sqrt{{(- 5)}^{2} - {(- 3)}^{2}}}{(- 5) + 3} = \frac{\sqrt{25 - 9}}{- 5 + 3} = \frac{\sqrt{16}}{- 2} = \frac{4}{- 2} = - 2$

Conclusion: The value of $f (3, 1) = \frac{\sqrt{2}}{3}$ and $f (- 5, - 3) = - 2$

To determine

(c)

A point $(x, y)$ such that $f (x, y) = 1$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Expert Solution

Answer to Problem 1CRE

Solution: The point $(x, y)$ such that $f (x, y) = 1$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ Are $(- 3, 3 \sqrt{3})$ and $(- 3, - 3 \sqrt{3})$

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

$f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Calculation:

Given function is $f (x, y, z) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Also, $f (x, y) = 1$

$f (x, y, z) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3} = 1$

$\frac{\sqrt{x^{2} - y^{2}}}{x + 3} = 1$

$\sqrt{x^{2} - y^{2}} = x + 3$

Squaring both the sides, we get

$x^{2} - y^{2} = {(x + 3)}^{2}$

$x^{2} - y^{2} = x^{2} + 9 + 6 x$

$y^{2} = - (9 + 6 x)$

Let $x = - 6$ ,

Then $y^{2} = - (9 + 6 x) = - (9 - 36) = 27$

Thus, $y = \pm 3 \sqrt{3}$

Points are $(- 3, 3 \sqrt{3})$ and $(- 3, - 3 \sqrt{3})$

Conclusion: The point $(x, y)$ such that $f (x, y) = 1$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ are $(- 3, 3 \sqrt{3})$ and $(- 3, - 3 \sqrt{3})$

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 14, Problem 1CRE

To determine

(a)

Sketch the domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Expert Solution

Answer to Problem 1CRE

Solution:Domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ is $D = {(x, y) : x \geq \pm y; x \neq - 3}$

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 $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Formula:

a. Domain of Square root function, $\sqrt{x}$ is given as $x \geq 0$

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

Calculation:

Given function is $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ :

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.

$x + 3 \neq 0$ And $x^{2} - y^{2} \geq 0$

$x \neq - 3$ And $x^{2} \geq y^{2}$

$x \geq \pm y$

Thus, domain of the function is $D = {(x, y) : x \geq \pm y; x \neq - 3}$

Graph is as follows:

Loose-Leaf Version for Calculus: Early Transcendentals Single Variable, Chapter 14, Problem 1CRE

Conclusion: Domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ is $D = {(x, y) : x \geq \pm y; x \neq - 3}$

To determine

(b)

The value of $f (3, 1)$ and $f (- 5, - 3)$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Expert Solution

Answer to Problem 1CRE

Solution: The value of $f (3, 1) = \frac{\sqrt{2}}{3}$ 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 $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Calculation:

Given function is $f (x, y, z) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

$f (3, 1) = \frac{\sqrt{{(3)}^{2} - {(1)}^{2}}}{(3) + 3} = \frac{\sqrt{9 - 1}}{3 + 3} = \frac{\sqrt{8}}{6} = \frac{2 \sqrt{2}}{6} = \frac{\sqrt{2}}{3}$

$f (- 5, - 3) = \frac{\sqrt{{(- 5)}^{2} - {(- 3)}^{2}}}{(- 5) + 3} = \frac{\sqrt{25 - 9}}{- 5 + 3} = \frac{\sqrt{16}}{- 2} = \frac{4}{- 2} = - 2$

Conclusion: The value of $f (3, 1) = \frac{\sqrt{2}}{3}$ and $f (- 5, - 3) = - 2$

To determine

(c)

A point $(x, y)$ such that $f (x, y) = 1$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Expert Solution

Answer to Problem 1CRE

Solution: The point $(x, y)$ such that $f (x, y) = 1$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ Are $(- 3, 3 \sqrt{3})$ and $(- 3, - 3 \sqrt{3})$

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

$f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Calculation:

Given function is $f (x, y, z) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Also, $f (x, y) = 1$

$f (x, y, z) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3} = 1$

$\frac{\sqrt{x^{2} - y^{2}}}{x + 3} = 1$

$\sqrt{x^{2} - y^{2}} = x + 3$

Squaring both the sides, we get

$x^{2} - y^{2} = {(x + 3)}^{2}$

$x^{2} - y^{2} = x^{2} + 9 + 6 x$

$y^{2} = - (9 + 6 x)$

Let $x = - 6$ ,

Then $y^{2} = - (9 + 6 x) = - (9 - 36) = 27$

Thus, $y = \pm 3 \sqrt{3}$

Points are $(- 3, 3 \sqrt{3})$ and $(- 3, - 3 \sqrt{3})$

Conclusion: The point $(x, y)$ such that $f (x, y) = 1$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ are $(- 3, 3 \sqrt{3})$ and $(- 3, - 3 \sqrt{3})$

Answer 6

Chapter 14, Problem 1CRE

To determine

(a)

Sketch the domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Expert Solution

Answer to Problem 1CRE

Solution:Domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ is $D = {(x, y) : x \geq \pm y; x \neq - 3}$

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 $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Formula:

a. Domain of Square root function, $\sqrt{x}$ is given as $x \geq 0$

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

Calculation:

Given function is $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ :

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.

$x + 3 \neq 0$ And $x^{2} - y^{2} \geq 0$

$x \neq - 3$ And $x^{2} \geq y^{2}$

$x \geq \pm y$

Thus, domain of the function is $D = {(x, y) : x \geq \pm y; x \neq - 3}$

Graph is as follows:

Loose-Leaf Version for Calculus: Early Transcendentals Single Variable, Chapter 14, Problem 1CRE

Conclusion: Domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ is $D = {(x, y) : x \geq \pm y; x \neq - 3}$

Answer 7

Chapter 14, Problem 1CRE

To determine

(a)

Sketch the domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Answer 8

Expert Solution

Answer to Problem 1CRE

Solution:Domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ is $D = {(x, y) : x \geq \pm y; x \neq - 3}$

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

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 $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Formula:

a. Domain of Square root function, $\sqrt{x}$ is given as $x \geq 0$

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

Calculation:

Given function is $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ :

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.

$x + 3 \neq 0$ And $x^{2} - y^{2} \geq 0$

$x \neq - 3$ And $x^{2} \geq y^{2}$

$x \geq \pm y$

Thus, domain of the function is $D = {(x, y) : x \geq \pm y; x \neq - 3}$

Graph is as follows:

Loose-Leaf Version for Calculus: Early Transcendentals Single Variable, Chapter 14, Problem 1CRE

Conclusion: Domain of the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ is $D = {(x, y) : x \geq \pm y; x \neq - 3}$

Answer 13

To determine

(b)

The value of $f (3, 1)$ and $f (- 5, - 3)$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Expert Solution

Answer to Problem 1CRE

Solution: The value of $f (3, 1) = \frac{\sqrt{2}}{3}$ 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 $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Calculation:

Given function is $f (x, y, z) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

$f (3, 1) = \frac{\sqrt{{(3)}^{2} - {(1)}^{2}}}{(3) + 3} = \frac{\sqrt{9 - 1}}{3 + 3} = \frac{\sqrt{8}}{6} = \frac{2 \sqrt{2}}{6} = \frac{\sqrt{2}}{3}$

$f (- 5, - 3) = \frac{\sqrt{{(- 5)}^{2} - {(- 3)}^{2}}}{(- 5) + 3} = \frac{\sqrt{25 - 9}}{- 5 + 3} = \frac{\sqrt{16}}{- 2} = \frac{4}{- 2} = - 2$

Conclusion: The value of $f (3, 1) = \frac{\sqrt{2}}{3}$ and $f (- 5, - 3) = - 2$

Answer 14

To determine

(b)

The value of $f (3, 1)$ and $f (- 5, - 3)$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Answer 15

Expert Solution

Answer to Problem 1CRE

Solution: The value of $f (3, 1) = \frac{\sqrt{2}}{3}$ and $f (- 5, - 3) = - 2$

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

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 $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Calculation:

Given function is $f (x, y, z) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

$f (3, 1) = \frac{\sqrt{{(3)}^{2} - {(1)}^{2}}}{(3) + 3} = \frac{\sqrt{9 - 1}}{3 + 3} = \frac{\sqrt{8}}{6} = \frac{2 \sqrt{2}}{6} = \frac{\sqrt{2}}{3}$

$f (- 5, - 3) = \frac{\sqrt{{(- 5)}^{2} - {(- 3)}^{2}}}{(- 5) + 3} = \frac{\sqrt{25 - 9}}{- 5 + 3} = \frac{\sqrt{16}}{- 2} = \frac{4}{- 2} = - 2$

Conclusion: The value of $f (3, 1) = \frac{\sqrt{2}}{3}$ and $f (- 5, - 3) = - 2$

Answer 20

To determine

(c)

A point $(x, y)$ such that $f (x, y) = 1$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Expert Solution

Answer to Problem 1CRE

Solution: The point $(x, y)$ such that $f (x, y) = 1$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ Are $(- 3, 3 \sqrt{3})$ and $(- 3, - 3 \sqrt{3})$

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

$f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Calculation:

Given function is $f (x, y, z) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Also, $f (x, y) = 1$

$f (x, y, z) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3} = 1$

$\frac{\sqrt{x^{2} - y^{2}}}{x + 3} = 1$

$\sqrt{x^{2} - y^{2}} = x + 3$

Squaring both the sides, we get

$x^{2} - y^{2} = {(x + 3)}^{2}$

$x^{2} - y^{2} = x^{2} + 9 + 6 x$

$y^{2} = - (9 + 6 x)$

Let $x = - 6$ ,

Then $y^{2} = - (9 + 6 x) = - (9 - 36) = 27$

Thus, $y = \pm 3 \sqrt{3}$

Points are $(- 3, 3 \sqrt{3})$ and $(- 3, - 3 \sqrt{3})$

Conclusion: The point $(x, y)$ such that $f (x, y) = 1$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ are $(- 3, 3 \sqrt{3})$ and $(- 3, - 3 \sqrt{3})$

Answer 21

To determine

(c)

A point $(x, y)$ such that $f (x, y) = 1$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Answer 22

Expert Solution

Answer to Problem 1CRE

Solution: The point $(x, y)$ such that $f (x, y) = 1$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ Are $(- 3, 3 \sqrt{3})$ and $(- 3, - 3 \sqrt{3})$

Answer 23

Expert Solution

Answer 24

Expert Solution

Answer 25

Expert Solution

Answer 26

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

$f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Calculation:

Given function is $f (x, y, z) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$

Also, $f (x, y) = 1$

$f (x, y, z) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3} = 1$

$\frac{\sqrt{x^{2} - y^{2}}}{x + 3} = 1$

$\sqrt{x^{2} - y^{2}} = x + 3$

Squaring both the sides, we get

$x^{2} - y^{2} = {(x + 3)}^{2}$

$x^{2} - y^{2} = x^{2} + 9 + 6 x$

$y^{2} = - (9 + 6 x)$

Let $x = - 6$ ,

Then $y^{2} = - (9 + 6 x) = - (9 - 36) = 27$

Thus, $y = \pm 3 \sqrt{3}$

Points are $(- 3, 3 \sqrt{3})$ and $(- 3, - 3 \sqrt{3})$

Conclusion: The point $(x, y)$ such that $f (x, y) = 1$ for the function $f (x, y) = \frac{\sqrt{x^{2} - y^{2}}}{x + 3}$ are $(- 3, 3 \sqrt{3})$ and $(- 3, - 3 \sqrt{3})$