Find the Domain and Range g(x) = square root of 1/3(x-1)

$g (x) = \sqrt{\frac{1}{3} (x - 1)}$

Set the radicand in $\sqrt{\frac{1}{3} \cdot (x - 1)}$ greater than or equal to $0$ to find where the expression is defined.

$\frac{1}{3} \cdot (x - 1) \geq 0$

Solve for $x$ .

Simplify.

Apply the distributive property.

$\frac{1}{3} x + \frac{1}{3} \cdot - 1 \geq 0$

Combine $\frac{1}{3}$ and $x$ .

$\frac{x}{3} + \frac{1}{3} \cdot - 1 \geq 0$

Combine $\frac{1}{3}$ and $- 1$ .

$\frac{x}{3} + \frac{- 1}{3} \geq 0$

Move the negative in front of the fraction.

$\frac{x}{3} - \frac{1}{3} \geq 0$

Multiply each term in $\frac{x}{3} - \frac{1}{3} \geq 0$ by $3$ to eliminate the fractions.

Multiply each term in $\frac{x}{3} - \frac{1}{3} \geq 0$ by $3$ .

$\frac{x}{3} \cdot 3 - \frac{1}{3} \cdot 3 \geq 0 \cdot 3$

Simplify the left side.

Simplify each term.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{x}{3} \cdot 3 - \frac{1}{3} \cdot 3 \geq 0 \cdot 3$

Rewrite the expression.

$x - \frac{1}{3} \cdot 3 \geq 0 \cdot 3$

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{1}{3}$ into the numerator.

$x + \frac{- 1}{3} \cdot 3 \geq 0 \cdot 3$

Cancel the common factor.

$x + \frac{- 1}{3} \cdot 3 \geq 0 \cdot 3$

Rewrite the expression.

$x - 1 \geq 0 \cdot 3$

Simplify the right side.

Multiply $0$ by $3$ .

$x - 1 \geq 0$

Add $1$ to both sides of the inequality.

$x \geq 1$

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

Interval Notation:

$[1, \infty)$

Set-Builder Notation:

${x | x \geq 1}$

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

Interval Notation:

$[0, \infty)$

Set-Builder Notation:

${y | y \geq 0}$

Determine the domain and range.

Domain: $[1, \infty), {x | x \geq 1}$

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

Find the Domain and Range g(x) = square root of 1/3(x-1)

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