Find the Inverse f(x)=(3x)^(-2/3)

$f (x) = {(3 x)}^{- \frac{2}{3}}$

Write $f (x) = {(3 x)}^{- \frac{2}{3}}$ as an equation.

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

Interchange the variables.

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

Solve for $y$ .

Rewrite the equation as ${(3 y)}^{- \frac{2}{3}} = x$ .

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

Raise each side of the equation to the power of $- \frac{3}{2}$ to eliminate the fractional exponent on the left side.

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

Simplify the exponent.

Simplify the left side.

Simplify ${({(3 y)}^{- \frac{2}{3}})}^{- \frac{3}{2}}$ .

Multiply the exponents in ${({(3 y)}^{- \frac{2}{3}})}^{- \frac{3}{2}}$ .

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

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{2}{3}$ into the numerator.

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

Move the leading negative in $- \frac{3}{2}$ into the numerator.

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

Factor $2$ out of $- 2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $3$ .

Factor $3$ out of $- 3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 1$ by $- 1$ .

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

Simplify.

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

Simplify the right side.

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

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.

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

Divide each term in $3 y = \frac{1}{x^{\frac{3}{2}}}$ by $3$ and simplify.

Divide each term in $3 y = \frac{1}{x^{\frac{3}{2}}}$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

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

Combine.

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

Simplify the expression.

Multiply $1$ by $1$ .

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

Move $3$ to the left of $x^{\frac{3}{2}}$ .

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

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

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

Divide each term in $3 y = - \frac{1}{x^{\frac{3}{2}}}$ by $3$ and simplify.

Divide each term in $3 y = - \frac{1}{x^{\frac{3}{2}}}$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $\frac{1}{3}$ by $\frac{1}{x^{\frac{3}{2}}}$ .

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

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

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

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

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

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

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

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

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

$f^{- 1} (x) = \frac{1}{3 x^{\frac{3}{2}}}, - \frac{1}{3 x^{\frac{3}{2}}}$

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

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

Find the range of $f (x) = {(3 x)}^{- \frac{2}{3}}$ .

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

Interval Notation:

$(0, \infty)$

Find the domain of $\frac{1}{3 x^{\frac{3}{2}}}$ .

Apply the rule $x^{\frac{m}{n}} = \sqrt[n]{x^{m}}$ to rewrite the exponentiation as a radical.

$\frac{1}{3 \sqrt{x^{3}}}$

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

$x^{3} \geq 0$

Solve for $x$ .

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

$\sqrt[3]{x^{3}} \geq \sqrt[3]{0}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$x \geq \sqrt[3]{0}$

Simplify the right side.

Simplify $\sqrt[3]{0}$ .

Rewrite $0$ as $0^{3}$ .

$x \geq \sqrt[3]{0^{3}}$

Pull terms out from under the radical.

$x \geq 0$

Set the denominator in $\frac{1}{3 \sqrt{x^{3}}}$ equal to $0$ to find where the expression is undefined.

$3 \sqrt{x^{3}} = 0$

Solve for $x$ .

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

${(3 \sqrt{x^{3}})}^{2} = 0^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x^{3}}$ as $x^{\frac{3}{2}}$ .

${(3 x^{\frac{3}{2}})}^{2} = 0^{2}$

Simplify the left side.

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

Apply the product rule to $3 x^{\frac{3}{2}}$ .

$3^{2} {(x^{\frac{3}{2}})}^{2} = 0^{2}$

Raise $3$ to the power of $2$ .

$9 {(x^{\frac{3}{2}})}^{2} = 0^{2}$

Multiply the exponents in ${(x^{\frac{3}{2}})}^{2}$ .

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

$9 x^{\frac{3}{2} \cdot 2} = 0^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$9 x^{\frac{3}{2} \cdot 2} = 0^{2}$

Rewrite the expression.

$9 x^{3} = 0^{2}$

Simplify the right side.

Raising $0$ to any positive power yields $0$ .

$9 x^{3} = 0$

Solve for $x$ .

Divide each term in $9 x^{3} = 0$ by $9$ and simplify.

Divide each term in $9 x^{3} = 0$ by $9$ .

$\frac{9 x^{3}}{9} = \frac{0}{9}$

Simplify the left side.

Cancel the common factor of $9$ .

Cancel the common factor.

$\frac{9 x^{3}}{9} = \frac{0}{9}$

Divide $x^{3}$ by $1$ .

$x^{3} = \frac{0}{9}$

Simplify the right side.

Divide $0$ by $9$ .

$x^{3} = 0$

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

$x = \sqrt[3]{0}$

Simplify $\sqrt[3]{0}$ .

Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

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

$x = 0$

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

$(0, \infty)$

Find the domain of $f (x) = {(3 x)}^{- \frac{2}{3}}$ .

Convert expressions with fractional exponents to radicals.

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{{(3 x)}^{\frac{2}{3}}}$

Apply the rule $x^{\frac{m}{n}} = \sqrt[n]{x^{m}}$ to rewrite the exponentiation as a radical.

$\frac{1}{\sqrt[3]{{(3 x)}^{2}}}$

Set the denominator in $\frac{1}{\sqrt[3]{{(3 x)}^{2}}}$ equal to $0$ to find where the expression is undefined.

$\sqrt[3]{{(3 x)}^{2}} = 0$

Solve for $x$ .

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

${\sqrt[3]{{(3 x)}^{2}}}^{3} = 0^{3}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{{(3 x)}^{2}}$ as ${(3 x)}^{\frac{2}{3}}$ .

${({(3 x)}^{\frac{2}{3}})}^{3} = 0^{3}$

Simplify the left side.

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

Multiply the exponents in ${({(3 x)}^{\frac{2}{3}})}^{3}$ .

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

${(3 x)}^{\frac{2}{3} \cdot 3} = 0^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(3 x)}^{\frac{2}{3} \cdot 3} = 0^{3}$

Rewrite the expression.

${(3 x)}^{2} = 0^{3}$

Apply the product rule to $3 x$ .

$3^{2} x^{2} = 0^{3}$

Raise $3$ to the power of $2$ .

$9 x^{2} = 0^{3}$

Simplify the right side.

Raising $0$ to any positive power yields $0$ .

$9 x^{2} = 0$

Solve for $x$ .

Divide each term in $9 x^{2} = 0$ by $9$ and simplify.

Divide each term in $9 x^{2} = 0$ by $9$ .

$\frac{9 x^{2}}{9} = \frac{0}{9}$

Simplify the left side.

Cancel the common factor of $9$ .

Cancel the common factor.

$\frac{9 x^{2}}{9} = \frac{0}{9}$

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

$x^{2} = \frac{0}{9}$

Simplify the right side.

Divide $0$ by $9$ .

$x^{2} = 0$

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

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

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

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

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

$(- \infty, 0) \cup (0, \infty)$

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

$f^{- 1} (x) = \frac{1}{3 x^{\frac{3}{2}}}, - \frac{1}{3 x^{\frac{3}{2}}}$

Find the Inverse f(x)=(3x)^(-2/3)

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