Algebra Examples

$- 9 x^{2}$

Step 1

Interchange the variables.

$x = - 9 y^{2}$

Step 2

Solve for $y$ .

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Step 2.1

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

$- 9 y^{2} = x$

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of $- 9$ .

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Step 2.2.2.1.1

Cancel the common factor.

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

Step 2.2.2.1.2

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

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

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Move the negative in front of the fraction.

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

Step 2.3

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

$y = \pm \sqrt{- \frac{x}{9}}$

Step 2.4

Simplify $\pm \sqrt{- \frac{x}{9}}$ .

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Step 2.4.1

Rewrite $- \frac{x}{9}$ as ${(\frac{1^{2}}{3})}^{2} \cdot (- x)$ .

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Step 2.4.1.1

Factor the perfect power $1^{2}$ out of $x$ .

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

Step 2.4.1.2

Factor the perfect power $3^{2}$ out of $9$ .

$y = \pm \sqrt{- \frac{1^{2} x}{3^{2} \cdot 1}}$

Step 2.4.1.3

Rearrange the fraction $\frac{1^{2} x}{3^{2} \cdot 1}$ .

$y = \pm \sqrt{- ({(\frac{1}{3})}^{2} x)}$

Step 2.4.1.4

Reorder $- 1$ and ${(\frac{1}{3})}^{2}$ .

$y = \pm \sqrt{{(\frac{1}{3})}^{2} \cdot - 1 x}$

Step 2.4.1.5

Rewrite $1$ as $1^{2}$ .

$y = \pm \sqrt{{(\frac{1^{2}}{3})}^{2} \cdot - 1 x}$

Step 2.4.1.6

Add parentheses.

$y = \pm \sqrt{{(\frac{1^{2}}{3})}^{2} \cdot (- x)}$

Step 2.4.2

Pull terms out from under the radical.

$y = \pm \frac{1^{2}}{3} \sqrt{- x}$

Step 2.4.3

One to any power is one.

$y = \pm \frac{1}{3} \sqrt{- x}$

Step 2.4.4

Combine $\frac{1}{3}$ and $\sqrt{- x}$ .

$y = \pm \frac{\sqrt{- x}}{3}$

Step 2.5

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

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Step 2.5.1

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

$y = \frac{\sqrt{- x}}{3}$

Step 2.5.2

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

$y = - \frac{\sqrt{- x}}{3}$

Step 2.5.3

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

$y = \frac{\sqrt{- x}}{3}$

$y = - \frac{\sqrt{- x}}{3}$

$y = \frac{\sqrt{- x}}{3}$

$y = - \frac{\sqrt{- x}}{3}$

$y = \frac{\sqrt{- x}}{3}$

$y = - \frac{\sqrt{- x}}{3}$

Step 3

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

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

Step 4

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

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Step 4.1

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

Step 4.2

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

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Step 4.2.1

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

Interval Notation:

$(- \infty, 0]$

Step 4.3

Find the domain of $\frac{\sqrt{- x}}{3}$ .

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Step 4.3.1

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

$- x \geq 0$

Step 4.3.2

Divide each term in $- x \geq 0$ by $- 1$ and simplify.

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Step 4.3.2.1

Divide each term in $- x \geq 0$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

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

Step 4.3.2.2

Simplify the left side.

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Step 4.3.2.2.1

Dividing two negative values results in a positive value.

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

Step 4.3.2.2.2

Divide $x$ by $1$ .

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

Step 4.3.2.3

Simplify the right side.

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Step 4.3.2.3.1

Divide $0$ by $- 1$ .

$x \leq 0$

Step 4.3.3

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

$(- \infty, 0]$

Step 4.4

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

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Step 4.4.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

$(- \infty, \infty)$

Step 4.5

Since the domain of $f^{- 1} (x) = \frac{\sqrt{- x}}{3}, - \frac{\sqrt{- x}}{3}$ is the range of $f (x) = - 9 x^{2}$ and the range of $f^{- 1} (x) = \frac{\sqrt{- x}}{3}, - \frac{\sqrt{- x}}{3}$ is the domain of $f (x) = - 9 x^{2}$ , then $f^{- 1} (x) = \frac{\sqrt{- x}}{3}, - \frac{\sqrt{- x}}{3}$ is the inverse of $f (x) = - 9 x^{2}$ .

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

Step 5