Find the Domain and Range y^2=x^2-3

$y^{2} = x^{2} - 3$

Step 1

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

$y = \pm \sqrt{x^{2} - 3}$

Step 2

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$y = \sqrt{x^{2} - 3}$

Next, use the negative value of the $\pm$ to find the second solution.

$y = - \sqrt{x^{2} - 3}$

The complete solution is the result of both the positive and negative portions of the solution.

$y = \sqrt{x^{2} - 3}$

$y = - \sqrt{x^{2} - 3}$

$y = \sqrt{x^{2} - 3}$

$y = - \sqrt{x^{2} - 3}$

Step 3

Set the radicand in $\sqrt{x^{2} - 3}$ greater than or equal to $0$ to find where the expression is defined.

$x^{2} - 3 \geq 0$

Step 4

Solve for $x$ .

Add $3$ to both sides of the inequality.

$x^{2} \geq 3$

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

$\sqrt{x^{2}} \geq \sqrt{3}$

Simplify the left side.

Pull terms out from under the radical.

$| x | \geq \sqrt{3}$

Write $| x | \geq \sqrt{3}$ 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.

$x \geq \sqrt{3}$

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 $- 1$ .

$- x \geq \sqrt{3}$

Write as a piecewise.

${\begin{cases} x \geq \sqrt{3} & x \geq 0 \\ - x \geq \sqrt{3} & x < 0 \end{cases}$

Find the intersection of $x \geq \sqrt{3}$ and $x \geq 0$ .

$x \geq \sqrt{3}$

Divide each term in $- x \geq \sqrt{3}$ by $- 1$ and simplify.

Divide each term in $- x \geq \sqrt{3}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{\sqrt{3}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{\sqrt{3}}{- 1}$

Divide $x$ by $1$ .

$x \leq \frac{\sqrt{3}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\sqrt{3}}{- 1}$ .

$x \leq - 1 \cdot \sqrt{3}$

Rewrite $- 1 \cdot \sqrt{3}$ as $- \sqrt{3}$ .

$x \leq - \sqrt{3}$

Find the union of the solutions.

$x \leq - \sqrt{3}$ or $x \geq \sqrt{3}$

Step 5

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

Interval Notation:

$(- \infty, - \sqrt{3}] \cup [\sqrt{3}, \infty)$

Set-Builder Notation:

${x | x \leq - \sqrt{3}, x \geq \sqrt{3}}$

Step 6

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

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${y | y \in ℝ}$

Step 7

Determine the domain and range.

Domain: $(- \infty, - \sqrt{3}] \cup [\sqrt{3}, \infty), {x | x \leq - \sqrt{3}, x \geq \sqrt{3}}$

Range: $(- \infty, \infty), {y | y \in ℝ}$

Step 8

Find the Domain and Range y^2=x^2-3

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