Find the Inverse y = square root of 9-x^2

$y = \sqrt{9 - x^{2}}$

Step 1

Interchange the variables.

$x = \sqrt{9 - y^{2}}$

Step 2

Solve for $y$ .

Rewrite the equation as $\sqrt{9 - y^{2}} = x$ .

$\sqrt{9 - y^{2}} = x$

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{9 - y^{2}}}^{2} = x^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{9 - y^{2}}$ as ${(9 - y^{2})}^{\frac{1}{2}}$ .

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

Simplify the left side.

Simplify ${({(9 - y^{2})}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(9 - y^{2})}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(9 - y^{2})}^{\frac{1}{2} \cdot 2} = x^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(9 - y^{2})}^{\frac{1}{2} \cdot 2} = x^{2}$

Rewrite the expression.

${(9 - y^{2})}^{1} = x^{2}$

${(9 - y^{2})}^{1} = x^{2}$

${(9 - y^{2})}^{1} = x^{2}$

Simplify.

$9 - y^{2} = x^{2}$

$9 - y^{2} = x^{2}$

$9 - y^{2} = x^{2}$

$9 - y^{2} = x^{2}$

Solve for $y$ .

Subtract $9$ from both sides of the equation.

$- y^{2} = x^{2} - 9$

Divide each term in $- y^{2} = x^{2} - 9$ by $- 1$ and simplify.

Divide each term in $- y^{2} = x^{2} - 9$ by $- 1$ .

$\frac{- y^{2}}{- 1} = \frac{x^{2}}{- 1} + \frac{- 9}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y^{2}}{1} = \frac{x^{2}}{- 1} + \frac{- 9}{- 1}$

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

$y^{2} = \frac{x^{2}}{- 1} + \frac{- 9}{- 1}$

$y^{2} = \frac{x^{2}}{- 1} + \frac{- 9}{- 1}$

Simplify the right side.

Simplify each term.

Move the negative one from the denominator of $\frac{x^{2}}{- 1}$ .

$y^{2} = - 1 \cdot x^{2} + \frac{- 9}{- 1}$

Rewrite $- 1 \cdot x^{2}$ as $- x^{2}$ .

$y^{2} = - x^{2} + \frac{- 9}{- 1}$

Divide $- 9$ by $- 1$ .

$y^{2} = - x^{2} + 9$

$y^{2} = - x^{2} + 9$

$y^{2} = - x^{2} + 9$

$y^{2} = - x^{2} + 9$

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

$y = \pm \sqrt{- x^{2} + 9}$

Simplify $\pm \sqrt{- x^{2} + 9}$ .

Simplify the expression.

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

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

Reorder $- x^{2}$ and $3^{2}$ .

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

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3$ and $b = x$ .

$y = \pm \sqrt{(3 + x) (3 - x)}$

$y = \pm \sqrt{(3 + x) (3 - x)}$

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.

$y = \sqrt{(3 + x) (3 - x)}$

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

$y = - \sqrt{(3 + x) (3 - x)}$

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

$y = \sqrt{(3 + x) (3 - x)}$

$y = - \sqrt{(3 + x) (3 - x)}$

$y = \sqrt{(3 + x) (3 - x)}$

$y = - \sqrt{(3 + x) (3 - x)}$

$y = \sqrt{(3 + x) (3 - x)}$

$y = - \sqrt{(3 + x) (3 - x)}$

$y = \sqrt{(3 + x) (3 - x)}$

$y = - \sqrt{(3 + x) (3 - x)}$

Step 3

Replace $y$ with $f^{- 1} (x)$ to show the final answer.

$f^{- 1} (x) = \sqrt{(3 + x) (3 - x)}, - \sqrt{(3 + x) (3 - x)}$

Step 4

Verify if $f^{- 1} (x) = \sqrt{(3 + x) (3 - x)}, - \sqrt{(3 + x) (3 - x)}$ is the inverse of $f (x) = \sqrt{9 - x^{2}}$ .

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = \sqrt{9 - x^{2}}$ and $f^{- 1} (x) = \sqrt{(3 + x) (3 - x)}, - \sqrt{(3 + x) (3 - x)}$ and compare them.

Find the range of $f (x) = \sqrt{9 - x^{2}}$ .

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

Interval Notation:

$[0, 3]$

$[0, 3]$

Find the domain of $\sqrt{(3 + x) (3 - x)}$ .

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

$(3 + x) (3 - x) \geq 0$

Solve for $x$ .

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$3 + x = 0$

$3 - x = 0$

Set $3 + x$ equal to $0$ and solve for $x$ .

Set $3 + x$ equal to $0$ .

$3 + x = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

$x = - 3$

Set $3 - x$ equal to $0$ and solve for $x$ .

Set $3 - x$ equal to $0$ .

$3 - x = 0$

Solve $3 - x = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$- x = - 3$

Divide each term in $- x = - 3$ by $- 1$ and simplify.

Divide each term in $- x = - 3$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

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

Simplify the right side.

Divide $- 3$ by $- 1$ .

$x = 3$

$x = 3$

$x = 3$

$x = 3$

$x = 3$

The final solution is all the values that make $(3 + x) (3 - x) \geq 0$ true.

$x = - 3, 3$

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 3$

$x > 3$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < - 3$ to see if it makes the inequality true.

Choose a value on the interval $x < - 3$ and see if this value makes the original inequality true.

$x = - 6$

Replace $x$ with $- 6$ in the original inequality.

$(3 - 6) (3 - (- 6)) \geq 0$

The left side $- 27$ is less than the right side $0$ , which means that the given statement is false.

False

False

Test a value on the interval $- 3 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$(3 + 0) (3 - (0)) \geq 0$

The left side $9$ is greater than the right side $0$ , which means that the given statement is always true.

True

True

Test a value on the interval $x > 3$ to see if it makes the inequality true.

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

$(3 + 6) (3 - (6)) \geq 0$

The left side $- 27$ is less than the right side $0$ , which means that the given statement is false.

False

False

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 3$ False

$- 3 < x < 3$ True

$x > 3$ False

$x < - 3$ False

$- 3 < x < 3$ True

$x > 3$ False

The solution consists of all of the true intervals.

$- 3 \leq x \leq 3$

$- 3 \leq x \leq 3$

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

$[- 3, 3]$

$[- 3, 3]$

Since the domain of $f^{- 1} (x) = \sqrt{(3 + x) (3 - x)}, - \sqrt{(3 + x) (3 - x)}$ is not equal to the range of $f (x) = \sqrt{9 - x^{2}}$ , then $f^{- 1} (x) = \sqrt{(3 + x) (3 - x)}, - \sqrt{(3 + x) (3 - x)}$ is not an inverse of $f (x) = \sqrt{9 - x^{2}}$ .

There is no inverse

There is no inverse

Step 5

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