Find the Domain and Range f(x) = square root of 6-x

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Algebra

$f (x) = \sqrt{6 - x}$

Set the radicand in $\sqrt{6 - x}$ greater than or equal to $0$ to find where the expression is defined.

$6 - x \geq 0$

Solve for $x$ .

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Subtract $6$ from both sides of the inequality.

$- x \geq - 6$

Multiply each term in $- x \geq - 6$ by $- 1$

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

$(- x) \cdot - 1 \leq (- 6) \cdot - 1$

Multiply $(- x) \cdot - 1$ .

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Multiply $- 1$ by $- 1$ .

$1 x \leq (- 6) \cdot - 1$

Multiply $x$ by $1$ .

$x \leq (- 6) \cdot - 1$

Multiply $- 6$ by $- 1$ .

$x \leq 6$

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

Interval Notation:

$(- \infty, 6]$

Set-Builder Notation:

${x | x \leq 6}$

The range of an even indexed radical starts at its starting point $(6, 0)$ and extends to infinity.

Interval Notation:

$[0, \infty)$

Set-Builder Notation:

${y | y \geq 0}$

Determine the domain and range.

Domain: $(- \infty, 6], {x | x \leq 6}$

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

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