Precalculus Examples

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

Step 1

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

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

Step 2

Solve for $y$ .

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Step 2.1

Add $x$ to both sides of the inequality.

$y^{2} \geq x$

Step 2.2

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

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

Step 2.3

Simplify the left side.

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Step 2.3.1

Pull terms out from under the radical.

$| y | \geq \sqrt{x}$

Step 2.4

Write $| y | \geq \sqrt{x}$ 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.

$y \geq 0$

Step 2.4.2

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

$y \geq \sqrt{x}$

Step 2.4.3

Find the domain of $y \geq \sqrt{x}$ and find the intersection with $y \geq 0$ .

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Step 2.4.3.1

Find the domain of $y \geq \sqrt{x}$ .

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Step 2.4.3.1.1

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

$x \geq 0$

Step 2.4.3.1.2

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

$[0, \infty)$

Step 2.4.3.2

Find the intersection of $y \geq 0$ and $[0, \infty)$ .

$y \geq 0$

Step 2.4.4

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

$y < 0$

Step 2.4.5

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

$- y \geq \sqrt{x}$

Step 2.4.6

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

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Step 2.4.6.1

Find the domain of $- y \geq \sqrt{x}$ .

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Step 2.4.6.1.1

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

$x \geq 0$

Step 2.4.6.1.2

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

$[0, \infty)$

Step 2.4.6.2

Find the intersection of $y < 0$ and $[0, \infty)$ .

$No solution$

Step 2.4.7

Write as a piecewise.

${\begin{cases} y \geq \sqrt{x} & y \geq 0 \end{cases}$

Step 2.5

Find the intersection of $y \geq \sqrt{x}$ and $y \geq 0$ .

$y \geq \sqrt{x}$ and $y \geq 0$

Step 2.6

Find the union of the solutions.

$y \geq N o (Max i m u m)$

Step 3

The domain is all real numbers.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${y | y \in ℝ}$

Step 4