Find the Domain and Range f(x) = square root of 1/9-x^2

$f (x) = \sqrt{\frac{1}{9} - x^{2}}$

Step 1

Set the radicand in $\sqrt{\frac{1}{9} - x^{2}}$ greater than or equal to $0$ to find where the expression is defined.

$\frac{1}{9} - x^{2} \geq 0$

Step 2

Solve for $x$ .

Subtract $\frac{1}{9}$ from both sides of the inequality.

$- x^{2} \geq - \frac{1}{9}$

Divide each term in $- x^{2} \geq - \frac{1}{9}$ by $- 1$ and simplify.

Divide each term in $- x^{2} \geq - \frac{1}{9}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- \frac{1}{9}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- \frac{1}{9}}{- 1}$

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

$x^{2} \leq \frac{- \frac{1}{9}}{- 1}$

Simplify the right side.

Dividing two negative values results in a positive value.

$x^{2} \leq \frac{\frac{1}{9}}{1}$

Divide $\frac{1}{9}$ by $1$ .

$x^{2} \leq \frac{1}{9}$

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

$\sqrt{x^{2}} \leq \sqrt{\frac{1}{9}}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \leq \sqrt{\frac{1}{9}}$

Simplify the right side.

Simplify $\sqrt{\frac{1}{9}}$ .

Rewrite $\sqrt{\frac{1}{9}}$ as $\frac{\sqrt{1}}{\sqrt{9}}$ .

$| x | \leq \frac{\sqrt{1}}{\sqrt{9}}$

Any root of $1$ is $1$ .

$| x | \leq \frac{1}{\sqrt{9}}$

Simplify the denominator.

Rewrite $9$ as $3^{2}$ .

$| x | \leq \frac{1}{\sqrt{3^{2}}}$

Pull terms out from under the radical.

$| x | \leq \frac{1}{| 3 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$| x | \leq \frac{1}{3}$

Write $| x | \leq \frac{1}{3}$ 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.

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

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 $- 1$ .

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

Write as a piecewise.

${\begin{cases} x \leq \frac{1}{3} & x \geq 0 \\ - x \leq \frac{1}{3} & x < 0 \end{cases}$

Find the intersection of $x \leq \frac{1}{3}$ and $x \geq 0$ .

$0 \leq x \leq \frac{1}{3}$

Solve $- x \leq \frac{1}{3}$ when $x < 0$ .

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

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

$\frac{- x}{- 1} \geq \frac{\frac{1}{3}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{\frac{1}{3}}{- 1}$

Divide $x$ by $1$ .

$x \geq \frac{\frac{1}{3}}{- 1}$

Simplify the right side.

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

$x \geq - 1 \cdot \frac{1}{3}$

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

$x \geq - \frac{1}{3}$

Find the intersection of $x \geq - \frac{1}{3}$ and $x < 0$ .

$- \frac{1}{3} \leq x < 0$

Find the union of the solutions.

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

Step 3

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

Interval Notation:

$[- \frac{1}{3}, \frac{1}{3}]$

Set-Builder Notation:

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

Step 4

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

Interval Notation:

$[0, \frac{1}{3}]$

Set-Builder Notation:

${y | 0 \leq y \leq \frac{1}{3}}$

Step 5

Determine the domain and range.

Domain: $[- \frac{1}{3}, \frac{1}{3}], {x | - \frac{1}{3} \leq x \leq \frac{1}{3}}$

Range: $[0, \frac{1}{3}], {y | 0 \leq y \leq \frac{1}{3}}$

Step 6

Find the Domain and Range f(x) = square root of 1/9-x^2

Related Questions

Support LCMGCF.com by disabling your adblocker.