Find the Domain and Range f(x) = square root of x^2-6-6

$f (x) = \sqrt{x^{2} - 6} - 6$

Step 1

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

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

Step 2

Solve for $x$ .

Add $6$ to both sides of the inequality.

$x^{2} \geq 6$

Take the square root of both sides of the inequality to eliminate the exponent on the left side.

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

Simplify the left side.

Pull terms out from under the radical.

$| x | \geq \sqrt{6}$

Write $| x | \geq \sqrt{6}$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

In the piece where $x$ is non-negative, remove the absolute value.

$x \geq \sqrt{6}$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \geq \sqrt{6}$

Write as a piecewise.

${\begin{cases} x \geq \sqrt{6} & x \geq 0 \\ - x \geq \sqrt{6} & x < 0 \end{cases}$

Find the intersection of $x \geq \sqrt{6}$ and $x \geq 0$ .

$x \geq \sqrt{6}$

Divide each term in $- x \geq \sqrt{6}$ by $- 1$ and simplify.

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

$\frac{- x}{- 1} \leq \frac{\sqrt{6}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{\sqrt{6}}{- 1}$

Divide $x$ by $1$ .

$x \leq \frac{\sqrt{6}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\sqrt{6}}{- 1}$ .

$x \leq - 1 \cdot \sqrt{6}$

Rewrite $- 1 \cdot \sqrt{6}$ as $- \sqrt{6}$ .

$x \leq - \sqrt{6}$

Find the union of the solutions.

$x \leq - \sqrt{6}$ or $x \geq \sqrt{6}$

Step 3

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

Interval Notation:

$(- \infty, - \sqrt{6}] \cup [\sqrt{6}, \infty)$

Set-Builder Notation:

${x | x \leq - \sqrt{6}, x \geq \sqrt{6}}$

Step 4

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

Interval Notation:

$[- 6, \infty)$

Set-Builder Notation:

${y | y \geq - 6}$

Step 5

Determine the domain and range.

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

Range: $[- 6, \infty), {y | y \geq - 6}$

Step 6

Find the Domain and Range f(x) = square root of x^2-6-6

Related Questions

Support LCMGCF.com by disabling your adblocker.