Precalculus Examples

$f (x) = \sqrt{36 - x^{2}}$ , $g (x) = \sqrt{1 + x}$ , $(f + g) (x)$

Step 1

Replace the function designators in $f + g$ with the actual functions.

$\sqrt{36 - x^{2}} + \sqrt{1 + x}$

Step 2

Evaluate $f + g$ at $x$ .

$\sqrt{36 - {(x)}^{2}} + \sqrt{1 + x}$

Step 3

Remove parentheses.

$\sqrt{36 - {(x)}^{2}} + \sqrt{1 + x}$

Step 4

Set the radicand in $\sqrt{36 - {(x)}^{2}}$ greater than or equal to $0$ to find where the expression is defined.

$36 - {(x)}^{2} \geq 0$

Step 5

Solve for $x$ .

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Step 5.1

Subtract $36$ from both sides of the inequality.

$- x^{2} \geq - 36$

Step 5.2

Divide each term in $- x^{2} \geq - 36$ by $- 1$ and simplify.

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Step 5.2.1

Divide each term in $- x^{2} \geq - 36$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 36}{- 1}$

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 36}{- 1}$

Step 5.2.2.2

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

$x^{2} \leq \frac{- 36}{- 1}$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

Divide $- 36$ by $- 1$ .

$x^{2} \leq 36$

Step 5.3

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

$\sqrt{x^{2}} \leq \sqrt{36}$

Step 5.4

Simplify the equation.

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Step 5.4.1

Simplify the left side.

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Step 5.4.1.1

Pull terms out from under the radical.

$| x | \leq \sqrt{36}$

Step 5.4.2

Simplify the right side.

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Step 5.4.2.1

Simplify $\sqrt{36}$ .

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Step 5.4.2.1.1

Rewrite $36$ as $6^{2}$ .

$| x | \leq \sqrt{6^{2}}$

Step 5.4.2.1.2

Pull terms out from under the radical.

$| x | \leq | 6 |$

Step 5.4.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $6$ is $6$ .

$| x | \leq 6$

Step 5.5

Write $| x | \leq 6$ as a piecewise.

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Step 5.5.1

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

$x \geq 0$

Step 5.5.2

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

$x \leq 6$

Step 5.5.3

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

$x < 0$

Step 5.5.4

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

$- x \leq 6$

Step 5.5.5

Write as a piecewise.

${\begin{cases} x \leq 6 & x \geq 0 \\ - x \leq 6 & x < 0 \end{cases}$

Step 5.6

Find the intersection of $x \leq 6$ and $x \geq 0$ .

$0 \leq x \leq 6$

Step 5.7

Solve $- x \leq 6$ when $x < 0$ .

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Step 5.7.1

Divide each term in $- x \leq 6$ by $- 1$ and simplify.

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Step 5.7.1.1

Divide each term in $- x \leq 6$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{6}{- 1}$

Step 5.7.1.2

Simplify the left side.

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Step 5.7.1.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{6}{- 1}$

Step 5.7.1.2.2

Divide $x$ by $1$ .

$x \geq \frac{6}{- 1}$

Step 5.7.1.3

Simplify the right side.

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Step 5.7.1.3.1

Divide $6$ by $- 1$ .

$x \geq - 6$

Step 5.7.2

Find the intersection of $x \geq - 6$ and $x < 0$ .

$- 6 \leq x < 0$

Step 5.8

Find the union of the solutions.

$- 6 \leq x \leq 6$

Step 6

Set the radicand in $\sqrt{1 + x}$ greater than or equal to $0$ to find where the expression is defined.

$1 + x \geq 0$

Step 7

Subtract $1$ from both sides of the inequality.

$x \geq - 1$

Step 8

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

Interval Notation:

$[- 1, 6]$

Set-Builder Notation:

${x | - 1 \leq x \leq 6}$

Step 9