Find the Domain f(x,y) = square root of 16-x^2-4y^2

$f (x, y) = \sqrt{16 - x^{2} - 4 y^{2}}$

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{16 - x^{2} - 4 y^{2}}$ as ${(16 - x^{2} - 4 y^{2})}^{\frac{1}{2}}$ .

$f (x, y) = {(16 - x^{2} - 4 y^{2})}^{\frac{1}{2}}$

Subtract ${(16 - x^{2} - 4 y^{2})}^{\frac{1}{2}}$ from both sides of the equation.

$f (x, y) - {(16 - x^{2} - 4 y^{2})}^{\frac{1}{2}} = 0$

Set the base in ${(16 - x^{2} - 4 y^{2})}^{\frac{1}{2}}$ greater than or equal to $0$ to find where the expression is defined.

$16 - x^{2} - 4 y^{2} \geq 0$

Solve for $x$ .

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Move all terms not containing $x$ to the right side of the inequality.

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Subtract $16$ from both sides of the inequality.

$- x^{2} - 4 y^{2} \geq - 16$

Add $4 y^{2}$ to both sides of the inequality.

$- x^{2} \geq - 16 + 4 y^{2}$

Multiply each term in $- x^{2} \geq - 16 + 4 y^{2}$ by $- 1$

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Multiply each term in $- x^{2} \geq - 16 + 4 y^{2}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$(- x^{2}) \cdot - 1 \leq (- 16) \cdot - 1 + 4 y^{2} \cdot - 1$

Multiply $(- x^{2}) \cdot - 1$ .

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Multiply $- 1$ by $- 1$ .

$1 x^{2} \leq (- 16) \cdot - 1 + 4 y^{2} \cdot - 1$

Multiply $x^{2}$ by $1$ .

$x^{2} \leq (- 16) \cdot - 1 + 4 y^{2} \cdot - 1$

Simplify each term.

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Multiply $- 16$ by $- 1$ .

$x^{2} \leq 16 + 4 y^{2} \cdot - 1$

Multiply $- 1$ by $4$ .

$x^{2} \leq 16 - 4 y^{2}$

Take the square root of both sides of the inequality to eliminate the exponent on the left side.

$x \leq \pm \sqrt{16 - 4 y^{2}}$

The complete solution is the result of both the positive and negative portions of the solution.

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Simplify the right side of the equation.

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Factor $4$ out of $16 - 4 y^{2}$ .

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Factor $4$ out of $16$ .

$x \leq \pm \sqrt{4 (4) - 4 y^{2}}$

Factor $4$ out of $- 4 y^{2}$ .

$x \leq \pm \sqrt{4 (4) + 4 (- y^{2})}$

Factor $4$ out of $4 (4) + 4 (- y^{2})$ .

$x \leq \pm \sqrt{4 (4 - y^{2})}$

Rewrite $4$ as $2^{2}$ .

$x \leq \pm \sqrt{4 (2^{2} - y^{2})}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2$ and $b = y$ .

$x \leq \pm \sqrt{4 (2 + y) (2 - y)}$

Rewrite $4 (2 + y) (2 - y)$ as $2^{2} ((2 + y) (2 - y))$ .

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Rewrite $4$ as $2^{2}$ .

$x \leq \pm \sqrt{2^{2} (2 + y) (2 - y)}$

Add parentheses.

$x \leq \pm \sqrt{2^{2} ((2 + y) (2 - y))}$

Pull terms out from under the radical.

$x \leq \pm 2 \sqrt{(2 + y) (2 - y)}$

The complete solution is the result of both the positive and negative portions of the solution.

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First, use the positive value of the $\pm$ to find the first solution.

$x \leq 2 \sqrt{(2 + y) (2 - y)}$

Next, use the negative value of the $\pm$ to find the second solution. Since this is an inequality, flip the direction of the inequality sign on the $-$ portion of the solution.

$x \geq - 2 \sqrt{(2 + y) (2 - y)}$

The complete solution is the result of both the positive and negative portions of the solution.

$x \leq 2 \sqrt{(2 + y) (2 - y)}$ and $x \geq - 2 \sqrt{(2 + y) (2 - y)}$

Find the intersection.

$- 2 \sqrt{(2 + y) (2 - y)} \leq x \leq 2 \sqrt{(2 + y) (2 - y)}$

The domain is all real numbers.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Find the Domain f(x,y) = square root of 16-x^2-4y^2

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