Find the Domain and Range f(x) = square root of 12x+24

$f (x) = \sqrt{12 x + 24}$

Step 1

Set the radicand in $\sqrt{12 x + 24}$ greater than or equal to $0$ to find where the expression is defined.

$12 x + 24 \geq 0$

Step 2

Solve for $x$ .

Subtract $24$ from both sides of the inequality.

$12 x \geq - 24$

Divide each term in $12 x \geq - 24$ by $12$ and simplify.

Divide each term in $12 x \geq - 24$ by $12$ .

$\frac{12 x}{12} \geq \frac{- 24}{12}$

Simplify the left side.

Cancel the common factor of $12$ .

Cancel the common factor.

$\frac{12 x}{12} \geq \frac{- 24}{12}$

Divide $x$ by $1$ .

$x \geq \frac{- 24}{12}$

Simplify the right side.

Divide $- 24$ by $12$ .

$x \geq - 2$

Step 3

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

Interval Notation:

$[- 2, \infty)$

Set-Builder Notation:

${x | x \geq - 2}$

Step 4

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$[0, \infty)$

Set-Builder Notation:

${y | y \geq 0}$

Step 5

Determine the domain and range.

Domain: $[- 2, \infty), {x | x \geq - 2}$

Range: $[0, \infty), {y | y \geq 0}$

Step 6