Find the Domain and Range square root of x^2-6x

$\sqrt{x^{2} - 6 x}$

Step 1

Set the radicand in $\sqrt{x^{2} - 6 x}$ greater than or equal to $0$ to find where the expression is defined.

$x^{2} - 6 x \geq 0$

Step 2

Solve for $x$ .

Convert the inequality to an equation.

$x^{2} - 6 x = 0$

Factor $x$ out of $x^{2} - 6 x$ .

Factor $x$ out of $x^{2}$ .

$x \cdot x - 6 x = 0$

Factor $x$ out of $- 6 x$ .

$x \cdot x + x \cdot - 6 = 0$

Factor $x$ out of $x \cdot x + x \cdot - 6$ .

$x (x - 6) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x - 6 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

The final solution is all the values that make $x (x - 6) = 0$ true.

$x = 0, 6$

Use each root to create test intervals.

$x < 0$

$0 < x < 6$

$x > 6$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < 0$ to see if it makes the inequality true.

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Replace $x$ with $- 2$ in the original inequality.

${(- 2)}^{2} - 6 \cdot - 2 \geq 0$

The left side $16$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $0 < x < 6$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 6$ and see if this value makes the original inequality true.

$x = 3$

Replace $x$ with $3$ in the original inequality.

${(3)}^{2} - 6 \cdot 3 \geq 0$

The left side $- 9$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 6$ to see if it makes the inequality true.

Choose a value on the interval $x > 6$ and see if this value makes the original inequality true.

$x = 8$

Replace $x$ with $8$ in the original inequality.

${(8)}^{2} - 6 \cdot 8 \geq 0$

The left side $16$ is greater than the right side $0$ , which means that the given statement is always true.

True

Compare the intervals to determine which ones satisfy the original inequality.

$x < 0$ True

$0 < x < 6$ False

$x > 6$ True

$x < 0$ True

$0 < x < 6$ False

$x > 6$ True

The solution consists of all of the true intervals.

$x \leq 0$ or $x \geq 6$

Step 3

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

Interval Notation:

$(- \infty, 0] \cup [6, \infty)$

Set-Builder Notation:

${x | x \leq 0, x \geq 6}$

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: $(- \infty, 0] \cup [6, \infty), {x | x \leq 0, x \geq 6}$

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

Step 6

Find the Domain and Range square root of x^2-6x

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