Algebra Examples

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

Step 1

Write $f (x) = \sqrt{9 x^{2}}$ as an equation.

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

Simplify each side of the equation.

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Step 3.3.1

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

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

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

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

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Step 3.3.2.1.1

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

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Step 3.3.2.1.1.1

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

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

Step 3.3.2.1.1.2

Cancel the common factor of $2$ .

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Step 3.3.2.1.1.2.1

Cancel the common factor.

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

Step 3.3.2.1.1.2.2

Rewrite the expression.

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

Step 3.3.2.1.2

Simplify.

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

Step 3.4

Solve for $y$ .

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Step 3.4.1

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

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Step 3.4.1.1

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

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

Step 3.4.1.2

Simplify the left side.

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Step 3.4.1.2.1

Cancel the common factor of $9$ .

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Step 3.4.1.2.1.1

Cancel the common factor.

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

Step 3.4.1.2.1.2

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

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

Step 3.4.2

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

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

Step 3.4.3

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

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Step 3.4.3.1

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

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

Step 3.4.3.2

Rewrite $\frac{x^{2}}{3^{2}}$ as ${(\frac{x}{3})}^{2}$ .

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

Step 3.4.3.3

Pull terms out from under the radical, assuming positive real numbers.

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

Step 3.4.4

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

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Step 3.4.4.1

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

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

Step 3.4.4.2

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

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

Step 3.4.4.3

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

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

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

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

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

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

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

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

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

Step 4

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

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

Step 5

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

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Step 5.1

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

Step 5.2

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

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Step 5.2.1

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

Interval Notation:

$(- \infty, \infty)$

Step 5.3

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

$(- \infty, \infty)$

Step 5.4

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

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

Step 6