Precalculus Examples

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

Step 1

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

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

Step 2

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

$y \geq x^{2}$

Step 3

Set the denominator in $\frac{\sqrt{y - x^{2}}}{1 - x^{2}}$ equal to $0$ to find where the expression is undefined.

$1 - x^{2} = 0$

Step 4

Solve for $y$ .

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Step 4.1

Subtract $1$ from both sides of the equation.

$- x^{2} = - 1$

Step 4.2

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

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Step 4.2.1

Divide each term in $- x^{2} = - 1$ by $- 1$ .

$\frac{- x^{2}}{- 1} = \frac{- 1}{- 1}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} = \frac{- 1}{- 1}$

Step 4.2.2.2

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

$x^{2} = \frac{- 1}{- 1}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Divide $- 1$ by $- 1$ .

$x^{2} = 1$

Step 4.3

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

$x = \pm \sqrt{1}$

Step 4.4

Any root of $1$ is $1$ .

$x = \pm 1$

Step 4.5

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

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Step 4.5.1

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

$x = 1$

Step 4.5.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 1$

Step 4.5.3

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

$x = 1, - 1$

Step 5

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

Interval Notation:

$(- \infty, - 1) \cup (- 1, 1) \cup (1, \infty)$

Set-Builder Notation:

${y | y \neq 1, - 1}$