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

Question

Want to see more full solutions like this?

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:

CALCULUS: EARLY TRANS 4TH ED W/ ACCESS, 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 $(- 3, - 3 \sqrt{3})$

Want to see more full solutions like this?

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

Students have asked these similar questions

Points z1 and z2 are shown on the graph.z1 is at (4 real,6 imaginary), z2 is at (-5 real, 2 imaginary)Part A: Identify the points in standard form and find the distance between them.Part B: Give the complex conjugate of z2 and explain how to find it geometrically.Part C: Find z2 − z1 geometrically and explain your steps.

A polar curve is represented by the equation r1 = 7 + 4cos θ.Part A: What type of limaçon is this curve? Justify your answer using the constants in the equation.Part B: Is the curve symmetrical to the polar axis or the line θ = pi/2 Justify your answer algebraically.Part C: What are the two main differences between the graphs of r1 = 7 + 4cos θ and r2 = 4 + 4cos θ?

A curve, described by x2 + y2 + 8x = 0, has a point A at (−4, 4) on the curve.Part A: What are the polar coordinates of A? Give an exact answer.Part B: What is the polar form of the equation? What type of polar curve is this?Part C: What is the directed distance when Ø = 5pi/6 Give an exact answer.

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:

CALCULUS: EARLY TRANS 4TH ED W/ ACCESS, 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 $(- 3, - 3 \sqrt{3})$

Want to see more full solutions like this?

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

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:

CALCULUS: EARLY TRANS 4TH ED W/ ACCESS, 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 $(- 3, - 3 \sqrt{3})$

Want to see more full solutions like this?

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

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:

CALCULUS: EARLY TRANS 4TH ED W/ ACCESS, 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 $(- 3, - 3 \sqrt{3})$

Want to see more full solutions like this?

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:

CALCULUS: EARLY TRANS 4TH ED W/ ACCESS, 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:

CALCULUS: EARLY TRANS 4TH ED W/ ACCESS, 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:

CALCULUS: EARLY TRANS 4TH ED W/ ACCESS, 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 $(- 3, - 3 \sqrt{3})$