Precalculus Examples

$y = \frac{\sqrt{9 - x^{2}}}{x^{2} + 7 x + 10}$

Step 1

Set the radicand in $\sqrt{9 - x^{2}}$ greater than or equal to $0$ to find where the expression is defined.

$9 - x^{2} \geq 0$

Step 2

Solve for $x$ .

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Step 2.1

Subtract $9$ from both sides of the inequality.

$- x^{2} \geq - 9$

Step 2.2

Divide each term in $- x^{2} \geq - 9$ by $- 1$ and simplify.

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Step 2.2.1

Divide each term in $- x^{2} \geq - 9$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 9}{- 1}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 9}{- 1}$

Step 2.2.2.2

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

$x^{2} \leq \frac{- 9}{- 1}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Divide $- 9$ by $- 1$ .

$x^{2} \leq 9$

Step 2.3

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

$\sqrt{x^{2}} \leq \sqrt{9}$

Step 2.4

Simplify the equation.

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Step 2.4.1

Simplify the left side.

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Step 2.4.1.1

Pull terms out from under the radical.

$| x | \leq \sqrt{9}$

Step 2.4.2

Simplify the right side.

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Step 2.4.2.1

Simplify $\sqrt{9}$ .

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Step 2.4.2.1.1

Rewrite $9$ as $3^{2}$ .

$| x | \leq \sqrt{3^{2}}$

Step 2.4.2.1.2

Pull terms out from under the radical.

$| x | \leq | 3 |$

Step 2.4.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$| x | \leq 3$

Step 2.5

Write $| x | \leq 3$ as a piecewise.

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Step 2.5.1

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

$x \geq 0$

Step 2.5.2

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

$x \leq 3$

Step 2.5.3

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

$x < 0$

Step 2.5.4

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

$- x \leq 3$

Step 2.5.5

Write as a piecewise.

${\begin{cases} x \leq 3 & x \geq 0 \\ - x \leq 3 & x < 0 \end{cases}$

Step 2.6

Find the intersection of $x \leq 3$ and $x \geq 0$ .

$0 \leq x \leq 3$

Step 2.7

Solve $- x \leq 3$ when $x < 0$ .

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Step 2.7.1

Divide each term in $- x \leq 3$ by $- 1$ and simplify.

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Step 2.7.1.1

Divide each term in $- x \leq 3$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{3}{- 1}$

Step 2.7.1.2

Simplify the left side.

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Step 2.7.1.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{3}{- 1}$

Step 2.7.1.2.2

Divide $x$ by $1$ .

$x \geq \frac{3}{- 1}$

Step 2.7.1.3

Simplify the right side.

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Step 2.7.1.3.1

Divide $3$ by $- 1$ .

$x \geq - 3$

Step 2.7.2

Find the intersection of $x \geq - 3$ and $x < 0$ .

$- 3 \leq x < 0$

Step 2.8

Find the union of the solutions.

$- 3 \leq x \leq 3$

Step 3

Set the denominator in $\frac{\sqrt{9 - x^{2}}}{x^{2} + 7 x + 10}$ equal to $0$ to find where the expression is undefined.

$x^{2} + 7 x + 10 = 0$

Step 4

Solve for $x$ .

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Step 4.1

Factor $x^{2} + 7 x + 10$ using the AC method.

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Step 4.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $10$ and whose sum is $7$ .

$2, 5$

Step 4.1.2

Write the factored form using these integers.

$(x + 2) (x + 5) = 0$

Step 4.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x + 2 = 0$

$x + 5 = 0$

Step 4.3

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 4.3.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 4.3.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 4.4

Set $x + 5$ equal to $0$ and solve for $x$ .

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Step 4.4.1

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Step 4.4.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 4.5

The final solution is all the values that make $(x + 2) (x + 5) = 0$ true.

$x = - 2, - 5$

Step 5

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

Interval Notation:

$[- 3, - 2) \cup (- 2, 3]$

Set-Builder Notation:

${x | - 3 \leq x \leq 3, x \neq - 2}$

Step 6