Find the Domain and Range y=- square root of 64-x^2

$y = - \sqrt{64 - x^{2}}$

Step 1

Set the radicand in $\sqrt{64 - x^{2}}$ greater than or equal to $0$ to find where the expression is defined.

$64 - x^{2} \geq 0$

Step 2

Solve for $x$ .

Subtract $64$ from both sides of the inequality.

$- x^{2} \geq - 64$

Divide each term in $- x^{2} \geq - 64$ by $- 1$ and simplify.

Divide each term in $- x^{2} \geq - 64$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 64}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 64}{- 1}$

Divide $x^{2}$ by $1$ .

$x^{2} \leq \frac{- 64}{- 1}$

Simplify the right side.

Divide $- 64$ by $- 1$ .

$x^{2} \leq 64$

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

$\sqrt{x^{2}} \leq \sqrt{64}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \leq \sqrt{64}$

Simplify the right side.

Simplify $\sqrt{64}$ .

Rewrite $64$ as $8^{2}$ .

$| x | \leq \sqrt{8^{2}}$

Pull terms out from under the radical.

$| x | \leq | 8 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $8$ is $8$ .

$| x | \leq 8$

Write $| x | \leq 8$ 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.

$x \leq 8$

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 $- 1$ .

$- x \leq 8$

Write as a piecewise.

${\begin{cases} x \leq 8 & x \geq 0 \\ - x \leq 8 & x < 0 \end{cases}$

Find the intersection of $x \leq 8$ and $x \geq 0$ .

$0 \leq x \leq 8$

Solve $- x \leq 8$ when $x < 0$ .

Divide each term in $- x \leq 8$ by $- 1$ and simplify.

Divide each term in $- x \leq 8$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{8}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{8}{- 1}$

Divide $x$ by $1$ .

$x \geq \frac{8}{- 1}$

Simplify the right side.

Divide $8$ by $- 1$ .

$x \geq - 8$

Find the intersection of $x \geq - 8$ and $x < 0$ .

$- 8 \leq x < 0$

Find the union of the solutions.

$- 8 \leq x \leq 8$

Step 3

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

Interval Notation:

$[- 8, 8]$

Set-Builder Notation:

${x | - 8 \leq x \leq 8}$

Step 4

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

Interval Notation:

$[- 8, 0]$

Set-Builder Notation:

${y | - 8 \leq y \leq 0}$

Step 5

Determine the domain and range.

Domain: $[- 8, 8], {x | - 8 \leq x \leq 8}$

Range: $[- 8, 0], {y | - 8 \leq y \leq 0}$

Step 6

Find the Domain and Range y=- square root of 64-x^2

Related Questions

Support LCMGCF.com by disabling your adblocker.