Precalculus Examples

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

Step 1

Set the radicand in $\sqrt{x^{2} + y^{2} - 4}$ greater than or equal to $0$ to find where the expression is defined.

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

Step 2

Solve for $x$ .

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Step 2.1

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

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Step 2.1.1

Subtract $y^{2}$ from both sides of the inequality.

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

Step 2.1.2

Add $4$ to both sides of the inequality.

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

Step 2.2

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

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

Step 2.3

Simplify the equation.

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Step 2.3.1

Simplify the left side.

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Step 2.3.1.1

Pull terms out from under the radical.

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

Step 2.3.2

Simplify the right side.

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Step 2.3.2.1

Simplify $\sqrt{- y^{2} + 4}$ .

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Step 2.3.2.1.1

Simplify the expression.

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Step 2.3.2.1.1.1

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

$| x | \geq \sqrt{- y^{2} + 2^{2}}$

Step 2.3.2.1.1.2

Reorder $- y^{2}$ and $2^{2}$ .

$| x | \geq \sqrt{2^{2} - y^{2}}$

Step 2.3.2.1.2

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 $b = y$ .

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

Step 2.4

Write $| x | \geq \sqrt{(2 + y) (2 - y)}$ as a piecewise.

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Step 2.4.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

Step 2.4.2

In the piece where $x$ is non-negative, remove the absolute value.

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

Step 2.4.3

Find the domain of $x \geq \sqrt{(2 + y) (2 - y)}$ and find the intersection with $x \geq 0$ .

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Step 2.4.3.1

Find the domain of $x \geq \sqrt{(2 + y) (2 - y)}$ .

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Step 2.4.3.1.1

Set the radicand in $\sqrt{(2 + y) (2 - y)}$ greater than or equal to $0$ to find where the expression is defined.

$(2 + y) (2 - y) \geq 0$

Step 2.4.3.1.2

Solve for $x$ .

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Step 2.4.3.1.2.1

Simplify $(2 + y) (2 - y)$ .

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Step 2.4.3.1.2.1.1

Expand $(2 + y) (2 - y)$ using the FOIL Method.

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Step 2.4.3.1.2.1.1.1

Apply the distributive property.

$2 (2 - y) + y (2 - y) \geq 0$

Step 2.4.3.1.2.1.1.2

Apply the distributive property.

$2 \cdot 2 + 2 (- y) + y (2 - y) \geq 0$

Step 2.4.3.1.2.1.1.3

Apply the distributive property.

$2 \cdot 2 + 2 (- y) + y \cdot 2 + y (- y) \geq 0$

Step 2.4.3.1.2.1.2

Simplify and combine like terms.

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Step 2.4.3.1.2.1.2.1

Simplify each term.

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Step 2.4.3.1.2.1.2.1.1

Multiply $2$ by $2$ .

$4 + 2 (- y) + y \cdot 2 + y (- y) \geq 0$

Step 2.4.3.1.2.1.2.1.2

Multiply $- 1$ by $2$ .

$4 - 2 y + y \cdot 2 + y (- y) \geq 0$

Step 2.4.3.1.2.1.2.1.3

Move $2$ to the left of $y$ .

$4 - 2 y + 2 \cdot y + y (- y) \geq 0$

Step 2.4.3.1.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$4 - 2 y + 2 y - y \cdot y \geq 0$

Step 2.4.3.1.2.1.2.1.5

Multiply $y$ by $y$ by adding the exponents.

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Step 2.4.3.1.2.1.2.1.5.1

Move $y$ .

$4 - 2 y + 2 y - (y \cdot y) \geq 0$

Step 2.4.3.1.2.1.2.1.5.2

Multiply $y$ by $y$ .

$4 - 2 y + 2 y - y^{2} \geq 0$

Step 2.4.3.1.2.1.2.2

Add $- 2 y$ and $2 y$ .

$4 + 0 - y^{2} \geq 0$

Step 2.4.3.1.2.1.2.3

Add $4$ and $0$ .

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

Step 2.4.3.1.2.2

Subtract $4$ from both sides of the inequality.

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

Step 2.4.3.1.2.3

Divide each term in $- y^{2} \geq - 4$ by $- 1$ and simplify.

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Step 2.4.3.1.2.3.1

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

$\frac{- y^{2}}{- 1} \leq \frac{- 4}{- 1}$

Step 2.4.3.1.2.3.2

Simplify the left side.

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Step 2.4.3.1.2.3.2.1

Dividing two negative values results in a positive value.

$\frac{y^{2}}{1} \leq \frac{- 4}{- 1}$

Step 2.4.3.1.2.3.2.2

Divide $y^{2}$ by $1$ .

$y^{2} \leq \frac{- 4}{- 1}$

Step 2.4.3.1.2.3.3

Simplify the right side.

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Step 2.4.3.1.2.3.3.1

Divide $- 4$ by $- 1$ .

$y^{2} \leq 4$

Step 2.4.3.1.2.4

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

$\sqrt{y^{2}} \leq \sqrt{4}$

Step 2.4.3.1.2.5

Simplify the equation.

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Step 2.4.3.1.2.5.1

Simplify the left side.

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Step 2.4.3.1.2.5.1.1

Pull terms out from under the radical.

$| y | \leq \sqrt{4}$

Step 2.4.3.1.2.5.2

Simplify the right side.

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Step 2.4.3.1.2.5.2.1

Simplify $\sqrt{4}$ .

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Step 2.4.3.1.2.5.2.1.1

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

$| y | \leq \sqrt{2^{2}}$

Step 2.4.3.1.2.5.2.1.2

Pull terms out from under the radical.

$| y | \leq | 2 |$

Step 2.4.3.1.2.5.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$| y | \leq 2$

Step 2.4.3.1.2.6

Write $| y | \leq 2$ as a piecewise.

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Step 2.4.3.1.2.6.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$y \geq 0$

Step 2.4.3.1.2.6.2

In the piece where $y$ is non-negative, remove the absolute value.

$y \leq 2$

Step 2.4.3.1.2.6.3

To find the interval for the second piece, find where the inside of the absolute value is negative.

$y < 0$

Step 2.4.3.1.2.6.4

In the piece where $y$ is negative, remove the absolute value and multiply by $- 1$ .

$- y \leq 2$

Step 2.4.3.1.2.6.5

Write as a piecewise.

${\begin{cases} y \leq 2 & y \geq 0 \\ - y \leq 2 & y < 0 \end{cases}$

Step 2.4.3.1.2.7

Find the intersection of $y \leq 2$ and $y \geq 0$ .

$0 \leq y \leq 2$

Step 2.4.3.1.2.8

Solve $- y \leq 2$ when $y < 0$ .

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Step 2.4.3.1.2.8.1

Divide each term in $- y \leq 2$ by $- 1$ and simplify.

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Step 2.4.3.1.2.8.1.1

Divide each term in $- y \leq 2$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} \geq \frac{2}{- 1}$

Step 2.4.3.1.2.8.1.2

Simplify the left side.

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Step 2.4.3.1.2.8.1.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} \geq \frac{2}{- 1}$

Step 2.4.3.1.2.8.1.2.2

Divide $y$ by $1$ .

$y \geq \frac{2}{- 1}$

Step 2.4.3.1.2.8.1.3

Simplify the right side.

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Step 2.4.3.1.2.8.1.3.1

Divide $2$ by $- 1$ .

$y \geq - 2$

Step 2.4.3.1.2.8.2

Find the intersection of $y \geq - 2$ and $y < 0$ .

$- 2 \leq y < 0$

Step 2.4.3.1.2.9

Find the union of the solutions.

$- 2 \leq y \leq 2$

Step 2.4.3.1.3

The domain is all values of $x$ that make the expression defined.

$[- 2, 2]$

Step 2.4.3.2

Find the intersection of $x \geq 0$ and $[- 2, 2]$ .

$0 \leq x \leq 2$

Step 2.4.4

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

Step 2.4.5

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

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

Step 2.4.6

Find the domain of $- x \geq \sqrt{(2 + y) (2 - y)}$ and find the intersection with $x < 0$ .

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Step 2.4.6.1

Find the domain of $- x \geq \sqrt{(2 + y) (2 - y)}$ .

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Step 2.4.6.1.1

Set the radicand in $\sqrt{(2 + y) (2 - y)}$ greater than or equal to $0$ to find where the expression is defined.

$(2 + y) (2 - y) \geq 0$

Step 2.4.6.1.2

Solve for $x$ .

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Step 2.4.6.1.2.1

Simplify $(2 + y) (2 - y)$ .

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Step 2.4.6.1.2.1.1

Expand $(2 + y) (2 - y)$ using the FOIL Method.

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Step 2.4.6.1.2.1.1.1

Apply the distributive property.

$2 (2 - y) + y (2 - y) \geq 0$

Step 2.4.6.1.2.1.1.2

Apply the distributive property.

$2 \cdot 2 + 2 (- y) + y (2 - y) \geq 0$

Step 2.4.6.1.2.1.1.3

Apply the distributive property.

$2 \cdot 2 + 2 (- y) + y \cdot 2 + y (- y) \geq 0$

Step 2.4.6.1.2.1.2

Simplify and combine like terms.

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Step 2.4.6.1.2.1.2.1

Simplify each term.

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Step 2.4.6.1.2.1.2.1.1

Multiply $2$ by $2$ .

$4 + 2 (- y) + y \cdot 2 + y (- y) \geq 0$

Step 2.4.6.1.2.1.2.1.2

Multiply $- 1$ by $2$ .

$4 - 2 y + y \cdot 2 + y (- y) \geq 0$

Step 2.4.6.1.2.1.2.1.3

Move $2$ to the left of $y$ .

$4 - 2 y + 2 \cdot y + y (- y) \geq 0$

Step 2.4.6.1.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$4 - 2 y + 2 y - y \cdot y \geq 0$

Step 2.4.6.1.2.1.2.1.5

Multiply $y$ by $y$ by adding the exponents.

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Step 2.4.6.1.2.1.2.1.5.1

Move $y$ .

$4 - 2 y + 2 y - (y \cdot y) \geq 0$

Step 2.4.6.1.2.1.2.1.5.2

Multiply $y$ by $y$ .

$4 - 2 y + 2 y - y^{2} \geq 0$

Step 2.4.6.1.2.1.2.2

Add $- 2 y$ and $2 y$ .

$4 + 0 - y^{2} \geq 0$

Step 2.4.6.1.2.1.2.3

Add $4$ and $0$ .

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

Step 2.4.6.1.2.2

Subtract $4$ from both sides of the inequality.

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

Step 2.4.6.1.2.3

Divide each term in $- y^{2} \geq - 4$ by $- 1$ and simplify.

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Step 2.4.6.1.2.3.1

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

$\frac{- y^{2}}{- 1} \leq \frac{- 4}{- 1}$

Step 2.4.6.1.2.3.2

Simplify the left side.

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Step 2.4.6.1.2.3.2.1

Dividing two negative values results in a positive value.

$\frac{y^{2}}{1} \leq \frac{- 4}{- 1}$

Step 2.4.6.1.2.3.2.2

Divide $y^{2}$ by $1$ .

$y^{2} \leq \frac{- 4}{- 1}$

Step 2.4.6.1.2.3.3

Simplify the right side.

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Step 2.4.6.1.2.3.3.1

Divide $- 4$ by $- 1$ .

$y^{2} \leq 4$

Step 2.4.6.1.2.4

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

$\sqrt{y^{2}} \leq \sqrt{4}$

Step 2.4.6.1.2.5

Simplify the equation.

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Step 2.4.6.1.2.5.1

Simplify the left side.

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Step 2.4.6.1.2.5.1.1

Pull terms out from under the radical.

$| y | \leq \sqrt{4}$

Step 2.4.6.1.2.5.2

Simplify the right side.

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Step 2.4.6.1.2.5.2.1

Simplify $\sqrt{4}$ .

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Step 2.4.6.1.2.5.2.1.1

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

$| y | \leq \sqrt{2^{2}}$

Step 2.4.6.1.2.5.2.1.2

Pull terms out from under the radical.

$| y | \leq | 2 |$

Step 2.4.6.1.2.5.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$| y | \leq 2$

Step 2.4.6.1.2.6

Write $| y | \leq 2$ as a piecewise.

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Step 2.4.6.1.2.6.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$y \geq 0$

Step 2.4.6.1.2.6.2

In the piece where $y$ is non-negative, remove the absolute value.

$y \leq 2$

Step 2.4.6.1.2.6.3

To find the interval for the second piece, find where the inside of the absolute value is negative.

$y < 0$

Step 2.4.6.1.2.6.4

In the piece where $y$ is negative, remove the absolute value and multiply by $- 1$ .

$- y \leq 2$

Step 2.4.6.1.2.6.5

Write as a piecewise.

${\begin{cases} y \leq 2 & y \geq 0 \\ - y \leq 2 & y < 0 \end{cases}$

Step 2.4.6.1.2.7

Find the intersection of $y \leq 2$ and $y \geq 0$ .

$0 \leq y \leq 2$

Step 2.4.6.1.2.8

Solve $- y \leq 2$ when $y < 0$ .

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Step 2.4.6.1.2.8.1

Divide each term in $- y \leq 2$ by $- 1$ and simplify.

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Step 2.4.6.1.2.8.1.1

Divide each term in $- y \leq 2$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- y}{- 1} \geq \frac{2}{- 1}$

Step 2.4.6.1.2.8.1.2

Simplify the left side.

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Step 2.4.6.1.2.8.1.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} \geq \frac{2}{- 1}$

Step 2.4.6.1.2.8.1.2.2

Divide $y$ by $1$ .

$y \geq \frac{2}{- 1}$

Step 2.4.6.1.2.8.1.3

Simplify the right side.

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Step 2.4.6.1.2.8.1.3.1

Divide $2$ by $- 1$ .

$y \geq - 2$

Step 2.4.6.1.2.8.2

Find the intersection of $y \geq - 2$ and $y < 0$ .

$- 2 \leq y < 0$

Step 2.4.6.1.2.9

Find the union of the solutions.

$- 2 \leq y \leq 2$

Step 2.4.6.1.3

The domain is all values of $x$ that make the expression defined.

$[- 2, 2]$

Step 2.4.6.2

Find the intersection of $x < 0$ and $[- 2, 2]$ .

$- 2 \leq x < 0$

Step 2.4.7

Write as a piecewise.

${\begin{cases} x \geq \sqrt{(2 + y) (2 - y)} & 0 \leq x \leq 2 \\ - x \geq \sqrt{(2 + y) (2 - y)} & - 2 \leq x < 0 \end{cases}$

Step 2.5

Solve $x \geq \sqrt{(2 + y) (2 - y)}$ when $0 \leq x \leq 2$ .

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Step 2.5.1

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

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Step 2.5.1.1

Rewrite so $y$ is on the left side of the inequality.

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

Step 2.5.1.2

To remove the radical on the left side of the inequality, square both sides of the inequality.

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

Step 2.5.1.3

Simplify each side of the inequality.

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Step 2.5.1.3.1

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

${({((2 + y) (2 - y))}^{\frac{1}{2}})}^{2} \leq x^{2}$

Step 2.5.1.3.2

Simplify the left side.

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Step 2.5.1.3.2.1

Simplify ${({((2 + y) (2 - y))}^{\frac{1}{2}})}^{2}$ .

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Step 2.5.1.3.2.1.1

Multiply the exponents in ${({((2 + y) (2 - y))}^{\frac{1}{2}})}^{2}$ .

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Step 2.5.1.3.2.1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 2.5.1.3.2.1.1.2

Cancel the common factor of $2$ .

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Step 2.5.1.3.2.1.1.2.1

Cancel the common factor.

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

Step 2.5.1.3.2.1.1.2.2

Rewrite the expression.

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

Step 2.5.1.3.2.1.2

Expand $(2 + y) (2 - y)$ using the FOIL Method.

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Step 2.5.1.3.2.1.2.1

Apply the distributive property.

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

Step 2.5.1.3.2.1.2.2

Apply the distributive property.

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

Step 2.5.1.3.2.1.2.3

Apply the distributive property.

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

Step 2.5.1.3.2.1.3

Simplify and combine like terms.

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Step 2.5.1.3.2.1.3.1

Simplify each term.

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Step 2.5.1.3.2.1.3.1.1

Multiply $2$ by $2$ .

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

Step 2.5.1.3.2.1.3.1.2

Multiply $- 1$ by $2$ .

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

Step 2.5.1.3.2.1.3.1.3

Move $2$ to the left of $y$ .

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

Step 2.5.1.3.2.1.3.1.4

Rewrite using the commutative property of multiplication.

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

Step 2.5.1.3.2.1.3.1.5

Multiply $y$ by $y$ by adding the exponents.

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Step 2.5.1.3.2.1.3.1.5.1

Move $y$ .

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

Step 2.5.1.3.2.1.3.1.5.2

Multiply $y$ by $y$ .

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

Step 2.5.1.3.2.1.3.2

Add $- 2 y$ and $2 y$ .

${(4 + 0 - y^{2})}^{1} \leq x^{2}$

Step 2.5.1.3.2.1.3.3

Add $4$ and $0$ .

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

Step 2.5.1.3.2.1.4

Simplify.

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

Step 2.5.1.4

Solve for $y$ .

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Step 2.5.1.4.1

Subtract $4$ from both sides of the inequality.

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

Step 2.5.1.4.2

Divide each term in $- y^{2} \leq x^{2} - 4$ by $- 1$ and simplify.

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Step 2.5.1.4.2.1

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

$\frac{- y^{2}}{- 1} \geq \frac{x^{2}}{- 1} + \frac{- 4}{- 1}$

Step 2.5.1.4.2.2

Simplify the left side.

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Step 2.5.1.4.2.2.1

Dividing two negative values results in a positive value.

$\frac{y^{2}}{1} \geq \frac{x^{2}}{- 1} + \frac{- 4}{- 1}$

Step 2.5.1.4.2.2.2

Divide $y^{2}$ by $1$ .

$y^{2} \geq \frac{x^{2}}{- 1} + \frac{- 4}{- 1}$

Step 2.5.1.4.2.3

Simplify the right side.

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Step 2.5.1.4.2.3.1

Simplify each term.

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Step 2.5.1.4.2.3.1.1

Move the negative one from the denominator of $\frac{x^{2}}{- 1}$ .

$y^{2} \geq - 1 \cdot x^{2} + \frac{- 4}{- 1}$

Step 2.5.1.4.2.3.1.2

Rewrite $- 1 \cdot x^{2}$ as $- x^{2}$ .

$y^{2} \geq - x^{2} + \frac{- 4}{- 1}$

Step 2.5.1.4.2.3.1.3

Divide $- 4$ by $- 1$ .

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

Step 2.5.1.4.3

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

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

Step 2.5.1.4.4

Simplify the equation.

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Step 2.5.1.4.4.1

Simplify the left side.

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Step 2.5.1.4.4.1.1

Pull terms out from under the radical.

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

Step 2.5.1.4.4.2

Simplify the right side.

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Step 2.5.1.4.4.2.1

Simplify $\sqrt{- x^{2} + 4}$ .

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Step 2.5.1.4.4.2.1.1

Simplify the expression.

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Step 2.5.1.4.4.2.1.1.1

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

$| y | \geq \sqrt{- x^{2} + 2^{2}}$

Step 2.5.1.4.4.2.1.1.2

Reorder $- x^{2}$ and $2^{2}$ .

$| y | \geq \sqrt{2^{2} - x^{2}}$

Step 2.5.1.4.4.2.1.2

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 $b = x$ .

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

Step 2.5.1.4.5

Write $| y | \geq \sqrt{(2 + x) (2 - x)}$ as a piecewise.

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Step 2.5.1.4.5.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$y \geq 0$

Step 2.5.1.4.5.2

In the piece where $y$ is non-negative, remove the absolute value.

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

Step 2.5.1.4.5.3

Find the domain of $y \geq \sqrt{(2 + x) (2 - x)}$ and find the intersection with $y \geq 0$ .

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Step 2.5.1.4.5.3.1

Find the domain of $y \geq \sqrt{(2 + x) (2 - x)}$ .

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Step 2.5.1.4.5.3.1.1

Set the radicand in $\sqrt{(2 + x) (2 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$(2 + x) (2 - x) \geq 0$

Step 2.5.1.4.5.3.1.2

Solve for $y$ .

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Step 2.5.1.4.5.3.1.2.1

Simplify $(2 + x) (2 - x)$ .

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Step 2.5.1.4.5.3.1.2.1.1

Expand $(2 + x) (2 - x)$ using the FOIL Method.

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Step 2.5.1.4.5.3.1.2.1.1.1

Apply the distributive property.

$2 (2 - x) + x (2 - x) \geq 0$

Step 2.5.1.4.5.3.1.2.1.1.2

Apply the distributive property.

$2 \cdot 2 + 2 (- x) + x (2 - x) \geq 0$

Step 2.5.1.4.5.3.1.2.1.1.3

Apply the distributive property.

$2 \cdot 2 + 2 (- x) + x \cdot 2 + x (- x) \geq 0$

Step 2.5.1.4.5.3.1.2.1.2

Simplify and combine like terms.

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Step 2.5.1.4.5.3.1.2.1.2.1

Simplify each term.

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Step 2.5.1.4.5.3.1.2.1.2.1.1

Multiply $2$ by $2$ .

$4 + 2 (- x) + x \cdot 2 + x (- x) \geq 0$

Step 2.5.1.4.5.3.1.2.1.2.1.2

Multiply $- 1$ by $2$ .

$4 - 2 x + x \cdot 2 + x (- x) \geq 0$

Step 2.5.1.4.5.3.1.2.1.2.1.3

Move $2$ to the left of $x$ .

$4 - 2 x + 2 \cdot x + x (- x) \geq 0$

Step 2.5.1.4.5.3.1.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$4 - 2 x + 2 x - x \cdot x \geq 0$

Step 2.5.1.4.5.3.1.2.1.2.1.5

Multiply $x$ by $x$ by adding the exponents.

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Step 2.5.1.4.5.3.1.2.1.2.1.5.1

Move $x$ .

$4 - 2 x + 2 x - (x \cdot x) \geq 0$

Step 2.5.1.4.5.3.1.2.1.2.1.5.2

Multiply $x$ by $x$ .

$4 - 2 x + 2 x - x^{2} \geq 0$

Step 2.5.1.4.5.3.1.2.1.2.2

Add $- 2 x$ and $2 x$ .

$4 + 0 - x^{2} \geq 0$

Step 2.5.1.4.5.3.1.2.1.2.3

Add $4$ and $0$ .

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

Step 2.5.1.4.5.3.1.2.2

Subtract $4$ from both sides of the inequality.

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

Step 2.5.1.4.5.3.1.2.3

Divide each term in $- x^{2} \geq - 4$ by $- 1$ and simplify.

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Step 2.5.1.4.5.3.1.2.3.1

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

$\frac{- x^{2}}{- 1} \leq \frac{- 4}{- 1}$

Step 2.5.1.4.5.3.1.2.3.2

Simplify the left side.

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Step 2.5.1.4.5.3.1.2.3.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 4}{- 1}$

Step 2.5.1.4.5.3.1.2.3.2.2

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

$x^{2} \leq \frac{- 4}{- 1}$

Step 2.5.1.4.5.3.1.2.3.3

Simplify the right side.

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Step 2.5.1.4.5.3.1.2.3.3.1

Divide $- 4$ by $- 1$ .

$x^{2} \leq 4$

Step 2.5.1.4.5.3.1.2.4

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

$\sqrt{x^{2}} \leq \sqrt{4}$

Step 2.5.1.4.5.3.1.2.5

Simplify the equation.

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Step 2.5.1.4.5.3.1.2.5.1

Simplify the left side.

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Step 2.5.1.4.5.3.1.2.5.1.1

Pull terms out from under the radical.

$| x | \leq \sqrt{4}$

Step 2.5.1.4.5.3.1.2.5.2

Simplify the right side.

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Step 2.5.1.4.5.3.1.2.5.2.1

Simplify $\sqrt{4}$ .

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Step 2.5.1.4.5.3.1.2.5.2.1.1

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

$| x | \leq \sqrt{2^{2}}$

Step 2.5.1.4.5.3.1.2.5.2.1.2

Pull terms out from under the radical.

$| x | \leq | 2 |$

Step 2.5.1.4.5.3.1.2.5.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$| x | \leq 2$

Step 2.5.1.4.5.3.1.2.6

Write $| x | \leq 2$ as a piecewise.

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Step 2.5.1.4.5.3.1.2.6.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

Step 2.5.1.4.5.3.1.2.6.2

In the piece where $x$ is non-negative, remove the absolute value.

$x \leq 2$

Step 2.5.1.4.5.3.1.2.6.3

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

Step 2.5.1.4.5.3.1.2.6.4

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \leq 2$

Step 2.5.1.4.5.3.1.2.6.5

Write as a piecewise.

${\begin{cases} x \leq 2 & x \geq 0 \\ - x \leq 2 & x < 0 \end{cases}$

Step 2.5.1.4.5.3.1.2.7

Find the intersection of $x \leq 2$ and $x \geq 0$ .

$0 \leq x \leq 2$

Step 2.5.1.4.5.3.1.2.8

Solve $- x \leq 2$ when $x < 0$ .

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Step 2.5.1.4.5.3.1.2.8.1

Divide each term in $- x \leq 2$ by $- 1$ and simplify.

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Step 2.5.1.4.5.3.1.2.8.1.1

Divide each term in $- x \leq 2$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{2}{- 1}$

Step 2.5.1.4.5.3.1.2.8.1.2

Simplify the left side.

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Step 2.5.1.4.5.3.1.2.8.1.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{2}{- 1}$

Step 2.5.1.4.5.3.1.2.8.1.2.2

Divide $x$ by $1$ .

$x \geq \frac{2}{- 1}$

Step 2.5.1.4.5.3.1.2.8.1.3

Simplify the right side.

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Step 2.5.1.4.5.3.1.2.8.1.3.1

Divide $2$ by $- 1$ .

$x \geq - 2$

Step 2.5.1.4.5.3.1.2.8.2

Find the intersection of $x \geq - 2$ and $x < 0$ .

$- 2 \leq x < 0$

Step 2.5.1.4.5.3.1.2.9

Find the union of the solutions.

$- 2 \leq x \leq 2$

Step 2.5.1.4.5.3.1.3

The domain is all values of $y$ that make the expression defined.

$[- 2, 2]$

Step 2.5.1.4.5.3.2

Find the intersection of $y \geq 0$ and $[- 2, 2]$ .

$0 \leq y \leq 2$

Step 2.5.1.4.5.4

To find the interval for the second piece, find where the inside of the absolute value is negative.

$y < 0$

Step 2.5.1.4.5.5

In the piece where $y$ is negative, remove the absolute value and multiply by $- 1$ .

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

Step 2.5.1.4.5.6

Find the domain of $- y \geq \sqrt{(2 + x) (2 - x)}$ and find the intersection with $y < 0$ .

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Step 2.5.1.4.5.6.1

Find the domain of $- y \geq \sqrt{(2 + x) (2 - x)}$ .

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Step 2.5.1.4.5.6.1.1

Set the radicand in $\sqrt{(2 + x) (2 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$(2 + x) (2 - x) \geq 0$

Step 2.5.1.4.5.6.1.2

Solve for $y$ .

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Step 2.5.1.4.5.6.1.2.1

Simplify $(2 + x) (2 - x)$ .

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Step 2.5.1.4.5.6.1.2.1.1

Expand $(2 + x) (2 - x)$ using the FOIL Method.

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Step 2.5.1.4.5.6.1.2.1.1.1

Apply the distributive property.

$2 (2 - x) + x (2 - x) \geq 0$

Step 2.5.1.4.5.6.1.2.1.1.2

Apply the distributive property.

$2 \cdot 2 + 2 (- x) + x (2 - x) \geq 0$

Step 2.5.1.4.5.6.1.2.1.1.3

Apply the distributive property.

$2 \cdot 2 + 2 (- x) + x \cdot 2 + x (- x) \geq 0$

Step 2.5.1.4.5.6.1.2.1.2

Simplify and combine like terms.

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Step 2.5.1.4.5.6.1.2.1.2.1

Simplify each term.

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Step 2.5.1.4.5.6.1.2.1.2.1.1

Multiply $2$ by $2$ .

$4 + 2 (- x) + x \cdot 2 + x (- x) \geq 0$

Step 2.5.1.4.5.6.1.2.1.2.1.2

Multiply $- 1$ by $2$ .

$4 - 2 x + x \cdot 2 + x (- x) \geq 0$

Step 2.5.1.4.5.6.1.2.1.2.1.3

Move $2$ to the left of $x$ .

$4 - 2 x + 2 \cdot x + x (- x) \geq 0$

Step 2.5.1.4.5.6.1.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$4 - 2 x + 2 x - x \cdot x \geq 0$

Step 2.5.1.4.5.6.1.2.1.2.1.5

Multiply $x$ by $x$ by adding the exponents.

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Step 2.5.1.4.5.6.1.2.1.2.1.5.1

Move $x$ .

$4 - 2 x + 2 x - (x \cdot x) \geq 0$

Step 2.5.1.4.5.6.1.2.1.2.1.5.2

Multiply $x$ by $x$ .

$4 - 2 x + 2 x - x^{2} \geq 0$

Step 2.5.1.4.5.6.1.2.1.2.2

Add $- 2 x$ and $2 x$ .

$4 + 0 - x^{2} \geq 0$

Step 2.5.1.4.5.6.1.2.1.2.3

Add $4$ and $0$ .

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

Step 2.5.1.4.5.6.1.2.2

Subtract $4$ from both sides of the inequality.

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

Step 2.5.1.4.5.6.1.2.3

Divide each term in $- x^{2} \geq - 4$ by $- 1$ and simplify.

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Step 2.5.1.4.5.6.1.2.3.1

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

$\frac{- x^{2}}{- 1} \leq \frac{- 4}{- 1}$

Step 2.5.1.4.5.6.1.2.3.2

Simplify the left side.

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Step 2.5.1.4.5.6.1.2.3.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 4}{- 1}$

Step 2.5.1.4.5.6.1.2.3.2.2

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

$x^{2} \leq \frac{- 4}{- 1}$

Step 2.5.1.4.5.6.1.2.3.3

Simplify the right side.

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Step 2.5.1.4.5.6.1.2.3.3.1

Divide $- 4$ by $- 1$ .

$x^{2} \leq 4$

Step 2.5.1.4.5.6.1.2.4

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

$\sqrt{x^{2}} \leq \sqrt{4}$

Step 2.5.1.4.5.6.1.2.5

Simplify the equation.

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Step 2.5.1.4.5.6.1.2.5.1

Simplify the left side.

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Step 2.5.1.4.5.6.1.2.5.1.1

Pull terms out from under the radical.

$| x | \leq \sqrt{4}$

Step 2.5.1.4.5.6.1.2.5.2

Simplify the right side.

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Step 2.5.1.4.5.6.1.2.5.2.1

Simplify $\sqrt{4}$ .

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Step 2.5.1.4.5.6.1.2.5.2.1.1

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

$| x | \leq \sqrt{2^{2}}$

Step 2.5.1.4.5.6.1.2.5.2.1.2

Pull terms out from under the radical.

$| x | \leq | 2 |$

Step 2.5.1.4.5.6.1.2.5.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$| x | \leq 2$

Step 2.5.1.4.5.6.1.2.6

Write $| x | \leq 2$ as a piecewise.

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Step 2.5.1.4.5.6.1.2.6.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

Step 2.5.1.4.5.6.1.2.6.2

In the piece where $x$ is non-negative, remove the absolute value.

$x \leq 2$

Step 2.5.1.4.5.6.1.2.6.3

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

Step 2.5.1.4.5.6.1.2.6.4

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \leq 2$

Step 2.5.1.4.5.6.1.2.6.5

Write as a piecewise.

${\begin{cases} x \leq 2 & x \geq 0 \\ - x \leq 2 & x < 0 \end{cases}$

Step 2.5.1.4.5.6.1.2.7

Find the intersection of $x \leq 2$ and $x \geq 0$ .

$0 \leq x \leq 2$

Step 2.5.1.4.5.6.1.2.8

Solve $- x \leq 2$ when $x < 0$ .

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Step 2.5.1.4.5.6.1.2.8.1

Divide each term in $- x \leq 2$ by $- 1$ and simplify.

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Step 2.5.1.4.5.6.1.2.8.1.1

Divide each term in $- x \leq 2$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{2}{- 1}$

Step 2.5.1.4.5.6.1.2.8.1.2

Simplify the left side.

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Step 2.5.1.4.5.6.1.2.8.1.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{2}{- 1}$

Step 2.5.1.4.5.6.1.2.8.1.2.2

Divide $x$ by $1$ .

$x \geq \frac{2}{- 1}$

Step 2.5.1.4.5.6.1.2.8.1.3

Simplify the right side.

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Step 2.5.1.4.5.6.1.2.8.1.3.1

Divide $2$ by $- 1$ .

$x \geq - 2$

Step 2.5.1.4.5.6.1.2.8.2

Find the intersection of $x \geq - 2$ and $x < 0$ .

$- 2 \leq x < 0$

Step 2.5.1.4.5.6.1.2.9

Find the union of the solutions.

$- 2 \leq x \leq 2$

Step 2.5.1.4.5.6.1.3

The domain is all values of $y$ that make the expression defined.

$[- 2, 2]$

Step 2.5.1.4.5.6.2

Find the intersection of $y < 0$ and $[- 2, 2]$ .

$- 2 \leq y < 0$

Step 2.5.1.4.5.7

Write as a piecewise.

${\begin{cases} y \geq \sqrt{(2 + x) (2 - x)} & 0 \leq y \leq 2 \\ - y \geq \sqrt{(2 + x) (2 - x)} & - 2 \leq y < 0 \end{cases}$

Step 2.5.1.4.6

Find the intersection of $y \geq \sqrt{(2 + x) (2 - x)}$ and $0 \leq y \leq 2$ .

No solution

Step 2.5.1.4.7

Solve $- y \geq \sqrt{(2 + x) (2 - x)}$ when $- 2 \leq y < 0$ .

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Step 2.5.1.4.7.1

Divide each term in $- y \geq \sqrt{(2 + x) (2 - x)}$ by $- 1$ and simplify.

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Step 2.5.1.4.7.1.1

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

$\frac{- y}{- 1} \leq \frac{\sqrt{(2 + x) (2 - x)}}{- 1}$

Step 2.5.1.4.7.1.2

Simplify the left side.

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Step 2.5.1.4.7.1.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} \leq \frac{\sqrt{(2 + x) (2 - x)}}{- 1}$

Step 2.5.1.4.7.1.2.2

Divide $y$ by $1$ .

$y \leq \frac{\sqrt{(2 + x) (2 - x)}}{- 1}$

Step 2.5.1.4.7.1.3

Simplify the right side.

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Step 2.5.1.4.7.1.3.1

Move the negative one from the denominator of $\frac{\sqrt{(2 + x) (2 - x)}}{- 1}$ .

$y \leq - 1 \cdot \sqrt{(2 + x) (2 - x)}$

Step 2.5.1.4.7.1.3.2

Rewrite $- 1 \cdot \sqrt{(2 + x) (2 - x)}$ as $- \sqrt{(2 + x) (2 - x)}$ .

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

Step 2.5.1.4.7.2

Find the intersection of $y \leq - \sqrt{(2 + x) (2 - x)}$ and $- 2 \leq y < 0$ .

$- 2 \leq y < N o (Min i m u m)$

Step 2.5.1.4.8

Find the union of the solutions.

$- 2 \leq y < i N o Min m^{2} u$

Step 2.5.2

Find the intersection of $- 2 \leq y < i N o Min m^{2} u$ and $0 \leq x \leq 2$ .

$0 \leq x \leq N o (Min i m u m)$

Step 2.6

Solve $- x \geq \sqrt{(2 + y) (2 - y)}$ when $- 2 \leq x < 0$ .

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Step 2.6.1

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

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Step 2.6.1.1

Rewrite so $y$ is on the left side of the inequality.

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

Step 2.6.1.2

To remove the radical on the left side of the inequality, square both sides of the inequality.

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

Step 2.6.1.3

Simplify each side of the inequality.

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Step 2.6.1.3.1

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

${({((2 + y) (2 - y))}^{\frac{1}{2}})}^{2} \leq {(- x)}^{2}$

Step 2.6.1.3.2

Simplify the left side.

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Step 2.6.1.3.2.1

Simplify ${({((2 + y) (2 - y))}^{\frac{1}{2}})}^{2}$ .

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Step 2.6.1.3.2.1.1

Multiply the exponents in ${({((2 + y) (2 - y))}^{\frac{1}{2}})}^{2}$ .

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Step 2.6.1.3.2.1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 2.6.1.3.2.1.1.2

Cancel the common factor of $2$ .

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Step 2.6.1.3.2.1.1.2.1

Cancel the common factor.

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

Step 2.6.1.3.2.1.1.2.2

Rewrite the expression.

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

Step 2.6.1.3.2.1.2

Expand $(2 + y) (2 - y)$ using the FOIL Method.

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Step 2.6.1.3.2.1.2.1

Apply the distributive property.

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

Step 2.6.1.3.2.1.2.2

Apply the distributive property.

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

Step 2.6.1.3.2.1.2.3

Apply the distributive property.

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

Step 2.6.1.3.2.1.3

Simplify and combine like terms.

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Step 2.6.1.3.2.1.3.1

Simplify each term.

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Step 2.6.1.3.2.1.3.1.1

Multiply $2$ by $2$ .

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

Step 2.6.1.3.2.1.3.1.2

Multiply $- 1$ by $2$ .

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

Step 2.6.1.3.2.1.3.1.3

Move $2$ to the left of $y$ .

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

Step 2.6.1.3.2.1.3.1.4

Rewrite using the commutative property of multiplication.

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

Step 2.6.1.3.2.1.3.1.5

Multiply $y$ by $y$ by adding the exponents.

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Step 2.6.1.3.2.1.3.1.5.1

Move $y$ .

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

Step 2.6.1.3.2.1.3.1.5.2

Multiply $y$ by $y$ .

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

Step 2.6.1.3.2.1.3.2

Add $- 2 y$ and $2 y$ .

${(4 + 0 - y^{2})}^{1} \leq {(- x)}^{2}$

Step 2.6.1.3.2.1.3.3

Add $4$ and $0$ .

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

Step 2.6.1.3.2.1.4

Simplify.

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

Step 2.6.1.3.3

Simplify the right side.

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Step 2.6.1.3.3.1

Simplify ${(- x)}^{2}$ .

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Step 2.6.1.3.3.1.1

Apply the product rule to $- x$ .

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

Step 2.6.1.3.3.1.2

Raise $- 1$ to the power of $2$ .

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

Step 2.6.1.3.3.1.3

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

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

Step 2.6.1.4

Solve for $y$ .

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Step 2.6.1.4.1

Subtract $4$ from both sides of the inequality.

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

Step 2.6.1.4.2

Divide each term in $- y^{2} \leq x^{2} - 4$ by $- 1$ and simplify.

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Step 2.6.1.4.2.1

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

$\frac{- y^{2}}{- 1} \geq \frac{x^{2}}{- 1} + \frac{- 4}{- 1}$

Step 2.6.1.4.2.2

Simplify the left side.

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Step 2.6.1.4.2.2.1

Dividing two negative values results in a positive value.

$\frac{y^{2}}{1} \geq \frac{x^{2}}{- 1} + \frac{- 4}{- 1}$

Step 2.6.1.4.2.2.2

Divide $y^{2}$ by $1$ .

$y^{2} \geq \frac{x^{2}}{- 1} + \frac{- 4}{- 1}$

Step 2.6.1.4.2.3

Simplify the right side.

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Step 2.6.1.4.2.3.1

Simplify each term.

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Step 2.6.1.4.2.3.1.1

Move the negative one from the denominator of $\frac{x^{2}}{- 1}$ .

$y^{2} \geq - 1 \cdot x^{2} + \frac{- 4}{- 1}$

Step 2.6.1.4.2.3.1.2

Rewrite $- 1 \cdot x^{2}$ as $- x^{2}$ .

$y^{2} \geq - x^{2} + \frac{- 4}{- 1}$

Step 2.6.1.4.2.3.1.3

Divide $- 4$ by $- 1$ .

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

Step 2.6.1.4.3

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

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

Step 2.6.1.4.4

Simplify the equation.

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Step 2.6.1.4.4.1

Simplify the left side.

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Step 2.6.1.4.4.1.1

Pull terms out from under the radical.

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

Step 2.6.1.4.4.2

Simplify the right side.

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Step 2.6.1.4.4.2.1

Simplify $\sqrt{- x^{2} + 4}$ .

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Step 2.6.1.4.4.2.1.1

Simplify the expression.

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Step 2.6.1.4.4.2.1.1.1

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

$| y | \geq \sqrt{- x^{2} + 2^{2}}$

Step 2.6.1.4.4.2.1.1.2

Reorder $- x^{2}$ and $2^{2}$ .

$| y | \geq \sqrt{2^{2} - x^{2}}$

Step 2.6.1.4.4.2.1.2

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 $b = x$ .

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

Step 2.6.1.4.5

Write $| y | \geq \sqrt{(2 + x) (2 - x)}$ as a piecewise.

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Step 2.6.1.4.5.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$y \geq 0$

Step 2.6.1.4.5.2

In the piece where $y$ is non-negative, remove the absolute value.

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

Step 2.6.1.4.5.3

Find the domain of $y \geq \sqrt{(2 + x) (2 - x)}$ and find the intersection with $y \geq 0$ .

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Step 2.6.1.4.5.3.1

Find the domain of $y \geq \sqrt{(2 + x) (2 - x)}$ .

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Step 2.6.1.4.5.3.1.1

Set the radicand in $\sqrt{(2 + x) (2 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$(2 + x) (2 - x) \geq 0$

Step 2.6.1.4.5.3.1.2

Solve for $y$ .

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Step 2.6.1.4.5.3.1.2.1

Simplify $(2 + x) (2 - x)$ .

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Step 2.6.1.4.5.3.1.2.1.1

Expand $(2 + x) (2 - x)$ using the FOIL Method.

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Step 2.6.1.4.5.3.1.2.1.1.1

Apply the distributive property.

$2 (2 - x) + x (2 - x) \geq 0$

Step 2.6.1.4.5.3.1.2.1.1.2

Apply the distributive property.

$2 \cdot 2 + 2 (- x) + x (2 - x) \geq 0$

Step 2.6.1.4.5.3.1.2.1.1.3

Apply the distributive property.

$2 \cdot 2 + 2 (- x) + x \cdot 2 + x (- x) \geq 0$

Step 2.6.1.4.5.3.1.2.1.2

Simplify and combine like terms.

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Step 2.6.1.4.5.3.1.2.1.2.1

Simplify each term.

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Step 2.6.1.4.5.3.1.2.1.2.1.1

Multiply $2$ by $2$ .

$4 + 2 (- x) + x \cdot 2 + x (- x) \geq 0$

Step 2.6.1.4.5.3.1.2.1.2.1.2

Multiply $- 1$ by $2$ .

$4 - 2 x + x \cdot 2 + x (- x) \geq 0$

Step 2.6.1.4.5.3.1.2.1.2.1.3

Move $2$ to the left of $x$ .

$4 - 2 x + 2 \cdot x + x (- x) \geq 0$

Step 2.6.1.4.5.3.1.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$4 - 2 x + 2 x - x \cdot x \geq 0$

Step 2.6.1.4.5.3.1.2.1.2.1.5

Multiply $x$ by $x$ by adding the exponents.

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Step 2.6.1.4.5.3.1.2.1.2.1.5.1

Move $x$ .

$4 - 2 x + 2 x - (x \cdot x) \geq 0$

Step 2.6.1.4.5.3.1.2.1.2.1.5.2

Multiply $x$ by $x$ .

$4 - 2 x + 2 x - x^{2} \geq 0$

Step 2.6.1.4.5.3.1.2.1.2.2

Add $- 2 x$ and $2 x$ .

$4 + 0 - x^{2} \geq 0$

Step 2.6.1.4.5.3.1.2.1.2.3

Add $4$ and $0$ .

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

Step 2.6.1.4.5.3.1.2.2

Subtract $4$ from both sides of the inequality.

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

Step 2.6.1.4.5.3.1.2.3

Divide each term in $- x^{2} \geq - 4$ by $- 1$ and simplify.

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Step 2.6.1.4.5.3.1.2.3.1

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

$\frac{- x^{2}}{- 1} \leq \frac{- 4}{- 1}$

Step 2.6.1.4.5.3.1.2.3.2

Simplify the left side.

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Step 2.6.1.4.5.3.1.2.3.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 4}{- 1}$

Step 2.6.1.4.5.3.1.2.3.2.2

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

$x^{2} \leq \frac{- 4}{- 1}$

Step 2.6.1.4.5.3.1.2.3.3

Simplify the right side.

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Step 2.6.1.4.5.3.1.2.3.3.1

Divide $- 4$ by $- 1$ .

$x^{2} \leq 4$

Step 2.6.1.4.5.3.1.2.4

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

$\sqrt{x^{2}} \leq \sqrt{4}$

Step 2.6.1.4.5.3.1.2.5

Simplify the equation.

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Step 2.6.1.4.5.3.1.2.5.1

Simplify the left side.

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Step 2.6.1.4.5.3.1.2.5.1.1

Pull terms out from under the radical.

$| x | \leq \sqrt{4}$

Step 2.6.1.4.5.3.1.2.5.2

Simplify the right side.

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Step 2.6.1.4.5.3.1.2.5.2.1

Simplify $\sqrt{4}$ .

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Step 2.6.1.4.5.3.1.2.5.2.1.1

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

$| x | \leq \sqrt{2^{2}}$

Step 2.6.1.4.5.3.1.2.5.2.1.2

Pull terms out from under the radical.

$| x | \leq | 2 |$

Step 2.6.1.4.5.3.1.2.5.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$| x | \leq 2$

Step 2.6.1.4.5.3.1.2.6

Write $| x | \leq 2$ as a piecewise.

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Step 2.6.1.4.5.3.1.2.6.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

Step 2.6.1.4.5.3.1.2.6.2

In the piece where $x$ is non-negative, remove the absolute value.

$x \leq 2$

Step 2.6.1.4.5.3.1.2.6.3

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

Step 2.6.1.4.5.3.1.2.6.4

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \leq 2$

Step 2.6.1.4.5.3.1.2.6.5

Write as a piecewise.

${\begin{cases} x \leq 2 & x \geq 0 \\ - x \leq 2 & x < 0 \end{cases}$

Step 2.6.1.4.5.3.1.2.7

Find the intersection of $x \leq 2$ and $x \geq 0$ .

$0 \leq x \leq 2$

Step 2.6.1.4.5.3.1.2.8

Solve $- x \leq 2$ when $x < 0$ .

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Step 2.6.1.4.5.3.1.2.8.1

Divide each term in $- x \leq 2$ by $- 1$ and simplify.

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Step 2.6.1.4.5.3.1.2.8.1.1

Divide each term in $- x \leq 2$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{2}{- 1}$

Step 2.6.1.4.5.3.1.2.8.1.2

Simplify the left side.

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Step 2.6.1.4.5.3.1.2.8.1.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{2}{- 1}$

Step 2.6.1.4.5.3.1.2.8.1.2.2

Divide $x$ by $1$ .

$x \geq \frac{2}{- 1}$

Step 2.6.1.4.5.3.1.2.8.1.3

Simplify the right side.

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Step 2.6.1.4.5.3.1.2.8.1.3.1

Divide $2$ by $- 1$ .

$x \geq - 2$

Step 2.6.1.4.5.3.1.2.8.2

Find the intersection of $x \geq - 2$ and $x < 0$ .

$- 2 \leq x < 0$

Step 2.6.1.4.5.3.1.2.9

Find the union of the solutions.

$- 2 \leq x \leq 2$

Step 2.6.1.4.5.3.1.3

The domain is all values of $y$ that make the expression defined.

$[- 2, 2]$

Step 2.6.1.4.5.3.2

Find the intersection of $y \geq 0$ and $[- 2, 2]$ .

$0 \leq y \leq 2$

Step 2.6.1.4.5.4

To find the interval for the second piece, find where the inside of the absolute value is negative.

$y < 0$

Step 2.6.1.4.5.5

In the piece where $y$ is negative, remove the absolute value and multiply by $- 1$ .

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

Step 2.6.1.4.5.6

Find the domain of $- y \geq \sqrt{(2 + x) (2 - x)}$ and find the intersection with $y < 0$ .

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Step 2.6.1.4.5.6.1

Find the domain of $- y \geq \sqrt{(2 + x) (2 - x)}$ .

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Step 2.6.1.4.5.6.1.1

Set the radicand in $\sqrt{(2 + x) (2 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$(2 + x) (2 - x) \geq 0$

Step 2.6.1.4.5.6.1.2

Solve for $y$ .

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Step 2.6.1.4.5.6.1.2.1

Simplify $(2 + x) (2 - x)$ .

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Step 2.6.1.4.5.6.1.2.1.1

Expand $(2 + x) (2 - x)$ using the FOIL Method.

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Step 2.6.1.4.5.6.1.2.1.1.1

Apply the distributive property.

$2 (2 - x) + x (2 - x) \geq 0$

Step 2.6.1.4.5.6.1.2.1.1.2

Apply the distributive property.

$2 \cdot 2 + 2 (- x) + x (2 - x) \geq 0$

Step 2.6.1.4.5.6.1.2.1.1.3

Apply the distributive property.

$2 \cdot 2 + 2 (- x) + x \cdot 2 + x (- x) \geq 0$

Step 2.6.1.4.5.6.1.2.1.2

Simplify and combine like terms.

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Step 2.6.1.4.5.6.1.2.1.2.1

Simplify each term.

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Step 2.6.1.4.5.6.1.2.1.2.1.1

Multiply $2$ by $2$ .

$4 + 2 (- x) + x \cdot 2 + x (- x) \geq 0$

Step 2.6.1.4.5.6.1.2.1.2.1.2

Multiply $- 1$ by $2$ .

$4 - 2 x + x \cdot 2 + x (- x) \geq 0$

Step 2.6.1.4.5.6.1.2.1.2.1.3

Move $2$ to the left of $x$ .

$4 - 2 x + 2 \cdot x + x (- x) \geq 0$

Step 2.6.1.4.5.6.1.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$4 - 2 x + 2 x - x \cdot x \geq 0$

Step 2.6.1.4.5.6.1.2.1.2.1.5

Multiply $x$ by $x$ by adding the exponents.

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Step 2.6.1.4.5.6.1.2.1.2.1.5.1

Move $x$ .

$4 - 2 x + 2 x - (x \cdot x) \geq 0$

Step 2.6.1.4.5.6.1.2.1.2.1.5.2

Multiply $x$ by $x$ .

$4 - 2 x + 2 x - x^{2} \geq 0$

Step 2.6.1.4.5.6.1.2.1.2.2

Add $- 2 x$ and $2 x$ .

$4 + 0 - x^{2} \geq 0$

Step 2.6.1.4.5.6.1.2.1.2.3

Add $4$ and $0$ .

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

Step 2.6.1.4.5.6.1.2.2

Subtract $4$ from both sides of the inequality.

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

Step 2.6.1.4.5.6.1.2.3

Divide each term in $- x^{2} \geq - 4$ by $- 1$ and simplify.

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Step 2.6.1.4.5.6.1.2.3.1

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

$\frac{- x^{2}}{- 1} \leq \frac{- 4}{- 1}$

Step 2.6.1.4.5.6.1.2.3.2

Simplify the left side.

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Step 2.6.1.4.5.6.1.2.3.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 4}{- 1}$

Step 2.6.1.4.5.6.1.2.3.2.2

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

$x^{2} \leq \frac{- 4}{- 1}$

Step 2.6.1.4.5.6.1.2.3.3

Simplify the right side.

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Step 2.6.1.4.5.6.1.2.3.3.1

Divide $- 4$ by $- 1$ .

$x^{2} \leq 4$

Step 2.6.1.4.5.6.1.2.4

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

$\sqrt{x^{2}} \leq \sqrt{4}$

Step 2.6.1.4.5.6.1.2.5

Simplify the equation.

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Step 2.6.1.4.5.6.1.2.5.1

Simplify the left side.

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Step 2.6.1.4.5.6.1.2.5.1.1

Pull terms out from under the radical.

$| x | \leq \sqrt{4}$

Step 2.6.1.4.5.6.1.2.5.2

Simplify the right side.

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Step 2.6.1.4.5.6.1.2.5.2.1

Simplify $\sqrt{4}$ .

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Step 2.6.1.4.5.6.1.2.5.2.1.1

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

$| x | \leq \sqrt{2^{2}}$

Step 2.6.1.4.5.6.1.2.5.2.1.2

Pull terms out from under the radical.

$| x | \leq | 2 |$

Step 2.6.1.4.5.6.1.2.5.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$| x | \leq 2$

Step 2.6.1.4.5.6.1.2.6

Write $| x | \leq 2$ as a piecewise.

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Step 2.6.1.4.5.6.1.2.6.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

Step 2.6.1.4.5.6.1.2.6.2

In the piece where $x$ is non-negative, remove the absolute value.

$x \leq 2$

Step 2.6.1.4.5.6.1.2.6.3

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

Step 2.6.1.4.5.6.1.2.6.4

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \leq 2$

Step 2.6.1.4.5.6.1.2.6.5

Write as a piecewise.

${\begin{cases} x \leq 2 & x \geq 0 \\ - x \leq 2 & x < 0 \end{cases}$

Step 2.6.1.4.5.6.1.2.7

Find the intersection of $x \leq 2$ and $x \geq 0$ .

$0 \leq x \leq 2$

Step 2.6.1.4.5.6.1.2.8

Solve $- x \leq 2$ when $x < 0$ .

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Step 2.6.1.4.5.6.1.2.8.1

Divide each term in $- x \leq 2$ by $- 1$ and simplify.

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Step 2.6.1.4.5.6.1.2.8.1.1

Divide each term in $- x \leq 2$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{2}{- 1}$

Step 2.6.1.4.5.6.1.2.8.1.2

Simplify the left side.

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Step 2.6.1.4.5.6.1.2.8.1.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{2}{- 1}$

Step 2.6.1.4.5.6.1.2.8.1.2.2

Divide $x$ by $1$ .

$x \geq \frac{2}{- 1}$

Step 2.6.1.4.5.6.1.2.8.1.3

Simplify the right side.

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Step 2.6.1.4.5.6.1.2.8.1.3.1

Divide $2$ by $- 1$ .

$x \geq - 2$

Step 2.6.1.4.5.6.1.2.8.2

Find the intersection of $x \geq - 2$ and $x < 0$ .

$- 2 \leq x < 0$

Step 2.6.1.4.5.6.1.2.9

Find the union of the solutions.

$- 2 \leq x \leq 2$

Step 2.6.1.4.5.6.1.3

The domain is all values of $y$ that make the expression defined.

$[- 2, 2]$

Step 2.6.1.4.5.6.2

Find the intersection of $y < 0$ and $[- 2, 2]$ .

$- 2 \leq y < 0$

Step 2.6.1.4.5.7

Write as a piecewise.

${\begin{cases} y \geq \sqrt{(2 + x) (2 - x)} & 0 \leq y \leq 2 \\ - y \geq \sqrt{(2 + x) (2 - x)} & - 2 \leq y < 0 \end{cases}$

Step 2.6.1.4.6

Find the intersection of $y \geq \sqrt{(2 + x) (2 - x)}$ and $0 \leq y \leq 2$ .

No solution

Step 2.6.1.4.7

Solve $- y \geq \sqrt{(2 + x) (2 - x)}$ when $- 2 \leq y < 0$ .

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Step 2.6.1.4.7.1

Divide each term in $- y \geq \sqrt{(2 + x) (2 - x)}$ by $- 1$ and simplify.

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Step 2.6.1.4.7.1.1

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

$\frac{- y}{- 1} \leq \frac{\sqrt{(2 + x) (2 - x)}}{- 1}$

Step 2.6.1.4.7.1.2

Simplify the left side.

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Step 2.6.1.4.7.1.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} \leq \frac{\sqrt{(2 + x) (2 - x)}}{- 1}$

Step 2.6.1.4.7.1.2.2

Divide $y$ by $1$ .

$y \leq \frac{\sqrt{(2 + x) (2 - x)}}{- 1}$

Step 2.6.1.4.7.1.3

Simplify the right side.

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Step 2.6.1.4.7.1.3.1

Move the negative one from the denominator of $\frac{\sqrt{(2 + x) (2 - x)}}{- 1}$ .

$y \leq - 1 \cdot \sqrt{(2 + x) (2 - x)}$

Step 2.6.1.4.7.1.3.2

Rewrite $- 1 \cdot \sqrt{(2 + x) (2 - x)}$ as $- \sqrt{(2 + x) (2 - x)}$ .

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

Step 2.6.1.4.7.2

Find the intersection of $y \leq - \sqrt{(2 + x) (2 - x)}$ and $- 2 \leq y < 0$ .

$- 2 \leq y < N o (Min i m u m)$

Step 2.6.1.4.8

Find the union of the solutions.

$- 2 \leq y < i N o Min m^{2} u$

Step 2.6.2

Find the intersection of $- 2 \leq y < i N o Min m^{2} u$ and $- 2 \leq x < 0$ .

$- 2 \leq x < N o (Min i m u m)$

Step 2.7

Find the union of the solutions.

$- 2 \leq x < N o (Max i m u m)$

Step 3

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression defined.

No solution