Precalculus Examples

$f (x) = \sqrt{x + 2} + \sqrt{5 - x}$

Step 1

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

$x + 2 \geq 0$

Step 2

Subtract $2$ from both sides of the inequality.

$x \geq - 2$

Step 3

Set the radicand in $\sqrt{5 - x}$ greater than or equal to $0$ to find where the expression is defined.

$5 - x \geq 0$

Step 4

Solve for $x$ .

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Step 4.1

Subtract $5$ from both sides of the inequality.

$- x \geq - 5$

Step 4.2

Divide each term in $- x \geq - 5$ by $- 1$ and simplify.

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Step 4.2.1

Divide each term in $- x \geq - 5$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{- 5}{- 1}$

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.

$\frac{x}{1} \leq \frac{- 5}{- 1}$

Step 4.2.2.2

Divide $x$ by $1$ .

$x \leq \frac{- 5}{- 1}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Divide $- 5$ by $- 1$ .

$x \leq 5$

Step 5

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

Interval Notation:

$[- 2, 5]$

Set-Builder Notation:

${x | - 2 \leq x \leq 5}$

Step 6