Find the Domain and Range f(x) = square root of 3x-12+2

$f (x) = \sqrt{3 x - 12} + 2$

Step 1

Set the radicand in $\sqrt{3 x - 12}$ greater than or equal to $0$ to find where the expression is defined.

$3 x - 12 \geq 0$

Step 2

Solve for $x$ .

Add $12$ to both sides of the inequality.

$3 x \geq 12$

Divide each term in $3 x \geq 12$ by $3$ and simplify.

Divide each term in $3 x \geq 12$ by $3$ .

$\frac{3 x}{3} \geq \frac{12}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} \geq \frac{12}{3}$

Divide $x$ by $1$ .

$x \geq \frac{12}{3}$

Simplify the right side.

Divide $12$ by $3$ .

$x \geq 4$

Step 3

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

Interval Notation:

$[4, \infty)$

Set-Builder Notation:

${x | x \geq 4}$

Step 4

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

Interval Notation:

$[2, \infty)$

Set-Builder Notation:

${y | y \geq 2}$

Step 5

Determine the domain and range.

Domain: $[4, \infty), {x | x \geq 4}$

Range: $[2, \infty), {y | y \geq 2}$

Step 6