Find the Domain and Range f(x) = square root of 25-x^2

$f (x) = \sqrt{25 - x^{2}}$

Set the radicand in $\sqrt{25 - x^{2}}$ greater than or equal to $0$ to find where the expression is defined.

$25 - x^{2} \geq 0$

Solve for $x$ .

Subtract $25$ from both sides of the inequality.

$- x^{2} \geq - 25$

Divide each term in $- x^{2} \geq - 25$ by $- 1$ and simplify.

Divide each term in $- x^{2} \geq - 25$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 25}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 25}{- 1}$

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

$x^{2} \leq \frac{- 25}{- 1}$

Simplify the right side.

Divide $- 25$ by $- 1$ .

$x^{2} \leq 25$

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

$\sqrt{x^{2}} \leq \sqrt{25}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| x | \leq \sqrt{25}$

Simplify the right side.

Simplify $\sqrt{25}$ .

Rewrite $25$ as $5^{2}$ .

$| x | \leq \sqrt{5^{2}}$

Pull terms out from under the radical.

$| x | \leq | 5 |$

The absolute value is the distance between a number and zero. The distance between $0$ and $5$ is $5$ .

$| x | \leq 5$

Write $| x | \leq 5$ 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.

$x \leq 5$

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 $- 1$ .

$- x \leq 5$

Write as a piecewise.

${\begin{cases} x \leq 5 & x \geq 0 \\ - x \leq 5 & x < 0 \end{cases}$

Find the intersection of $x \leq 5$ and $x \geq 0$ .

$0 \leq x \leq 5$

Solve $- x \leq 5$ when $x < 0$ .

Divide each term in $- x \leq 5$ by $- 1$ and simplify.

Divide each term in $- x \leq 5$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{5}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{5}{- 1}$

Divide $x$ by $1$ .

$x \geq \frac{5}{- 1}$

Simplify the right side.

Divide $5$ by $- 1$ .

$x \geq - 5$

Find the intersection of $x \geq - 5$ and $x < 0$ .

$- 5 \leq x < 0$

Find the union of the solutions.

$- 5 \leq x \leq 5$

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

Interval Notation:

$[- 5, 5]$

Set-Builder Notation:

${x | - 5 \leq x \leq 5}$

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

Interval Notation:

$[0, 5]$

Set-Builder Notation:

${y | 0 \leq y \leq 5}$

Determine the domain and range.

Domain: $[- 5, 5], {x | - 5 \leq x \leq 5}$

Range: $[0, 5], {y | 0 \leq y \leq 5}$

Find the Domain and Range f(x) = square root of 25-x^2

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