Math / Algebra

Question19. The domain of $f(x, y)=\frac{\sqrt{y-x^{2}}}{y+x}$ consists of all points $(x, y) \in R^{2}$ such that (a) $y \geq x^{2}, y=-x$ (b) $y \leq x^{2}, y \neq-x$ (c) $y=x^{2}, y \neq x$ (d) $y \geq x^{2}, y \neq-x$ (e) Else

Studdy Solution

STEP 1

What is this asking? We need to find all the allowed $(x, y)$ pairs that we can plug into the function $f(x, y) = \frac{\sqrt{y - x^2}}{y + x}$ without breaking any math rules! Watch out! Square roots of negative numbers are a no-go, and dividing by zero is a big math no-no!

STEP 2

1. Numerator analysis
2. Denominator analysis
3. Combine the conditions

STEP 3

We've got a square root in the numerator: $\sqrt{y - x^2}$ .
Remember, we can't take the square root of a negative number!
So, whatever is under the square root has to be zero or positive.

STEP 4

This gives us our first condition: $y - x^2 \ge 0$ .
Adding $x^2$ to both sides, we get $y \ge x^2$ .
This means $y$ must be greater than or equal to $x$ squared.

STEP 5

Now, let's look at the denominator: $y + x$ .
We know we can't divide by zero, ever!

STEP 6

So, our second condition is that $y + x \ne 0$ .
Subtracting $x$ from both sides gives us $y \ne -x$ .
This means $y$ cannot be equal to negative $x$ .

STEP 7

We need both conditions to be true at the same time.
So, we need $y \ge x^2$ and $y \ne -x$ .

STEP 8

Let's put it all together!
The allowed $(x, y)$ values are all the points where $y$ is greater than or equal to $x$ squared, and $y$ is not equal to negative $x$ .

STEP 9

The answer is (d): $y \ge x^2, y \ne -x$ .

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Question19. The domain of $f(x, y)=\frac{\sqrt{y-x^{2}}}{y+x}$ consists of all points $(x, y) \in R^{2}$ such that (a) $y \geq x^{2}, y=-x$ (b) $y \leq x^{2}, y \neq-x$ (c) $y=x^{2}, y \neq x$ (d) $y \geq x^{2}, y \neq-x$ (e) Else

Studdy Solution

STEP 1

What is this asking? We need to find all the allowed $(x, y)$ pairs that we can plug into the function $f(x, y) = \frac{\sqrt{y - x^2}}{y + x}$ without breaking any math rules! Watch out! Square roots of negative numbers are a no-go, and dividing by zero is a big math no-no!

STEP 2

1. Numerator analysis
2. Denominator analysis
3. Combine the conditions

STEP 3

We've got a square root in the numerator: $\sqrt{y - x^2}$ .
Remember, we can't take the square root of a negative number!
So, whatever is under the square root has to be zero or positive.

STEP 4

This gives us our first condition: $y - x^2 \ge 0$ .
Adding $x^2$ to both sides, we get $y \ge x^2$ .
This means $y$ must be greater than or equal to $x$ squared.

STEP 5

Now, let's look at the denominator: $y + x$ .
We know we can't divide by zero, ever!

STEP 6

So, our second condition is that $y + x \ne 0$ .
Subtracting $x$ from both sides gives us $y \ne -x$ .
This means $y$ cannot be equal to negative $x$ .

STEP 7

We need both conditions to be true at the same time.
So, we need $y \ge x^2$ and $y \ne -x$ .

STEP 8

Let's put it all together!
The allowed $(x, y)$ values are all the points where $y$ is greater than or equal to $x$ squared, and $y$ is not equal to negative $x$ .

STEP 9

The answer is (d): $y \ge x^2, y \ne -x$ .

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Studdy Solution

STEP 1

STEP 2

1. Numerator analysis2. Denominator analysis3. Combine the conditions

STEP 3

We've got a square root in the numerator: y−x2\sqrt{y - x^2}y−x2​.Remember, we can't take the square root of a negative number!So, whatever is under the square root has to be zero or positive.

STEP 4

This gives us our **first condition**: y−x2≥0y - x^2 \ge 0y−x2≥0.Adding x2x^2x2 to both sides, we get y≥x2y \ge x^2y≥x2.This means yyy must be greater than or equal to xxx squared.

STEP 5

Now, let's look at the denominator: y+xy + xy+x.We know we can't divide by **zero**, ever!

STEP 6

So, our **second condition** is that y+x≠0y + x \ne 0y+x=0.Subtracting xxx from both sides gives us y≠−xy \ne -xy=−x.This means yyy cannot be equal to negative xxx.

STEP 7

We need *both* conditions to be true at the same time.So, we need y≥x2y \ge x^2y≥x2 *and* y≠−xy \ne -xy=−x.

STEP 8

Let's put it all together!The allowed (x,y)(x, y)(x,y) values are all the points where yyy is greater than or equal to xxx squared, and yyy is not equal to negative xxx.

STEP 9

The answer is (d): y≥x2,y≠−xy \ge x^2, y \ne -xy≥x2,y=−x.

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Numerator analysis
2. Denominator analysis
3. Combine the conditions

We've got a square root in the numerator: $\sqrt{y - x^2}$ .
Remember, we can't take the square root of a negative number!
So, whatever is under the square root has to be zero or positive.

This gives us our first condition: $y - x^2 \ge 0$ .
Adding $x^2$ to both sides, we get $y \ge x^2$ .
This means $y$ must be greater than or equal to $x$ squared.

Now, let's look at the denominator: $y + x$ .
We know we can't divide by zero, ever!

So, our second condition is that $y + x \ne 0$ .
Subtracting $x$ from both sides gives us $y \ne -x$ .
This means $y$ cannot be equal to negative $x$ .

We need both conditions to be true at the same time.
So, we need $y \ge x^2$ and $y \ne -x$ .

Let's put it all together!
The allowed $(x, y)$ values are all the points where $y$ is greater than or equal to $x$ squared, and $y$ is not equal to negative $x$ .

The answer is (d): $y \ge x^2, y \ne -x$ .