Find the Inverse V(r)=4/3pir^3

$V (r) = \frac{4}{3} π r^{3}$

Write $V (r) = \frac{4}{3} π r^{3}$ as an equation.

$y = \frac{4}{3} π r^{3}$

Interchange the variables.

$r = \frac{4}{3} \cdot (π y^{3})$

Solve for $y$ .

Rewrite the equation as $\frac{4}{3} \cdot (π y^{3}) = r$ .

$\frac{4}{3} \cdot (π y^{3}) = r$

Multiply both sides of the equation by $\frac{1}{\frac{4}{3} π}$ .

$\frac{1}{\frac{4}{3} π} (\frac{4}{3} \cdot (π y^{3})) = \frac{1}{\frac{4}{3} π} r$

Simplify both sides of the equation.

Simplify the left side.

Simplify $\frac{1}{\frac{4}{3} π} (\frac{4}{3} \cdot (π y^{3}))$ .

Combine $\frac{4}{3}$ and $π$ .

$\frac{1}{\frac{4 π}{3}} (\frac{4}{3} \cdot (π y^{3})) = \frac{1}{\frac{4}{3} π} r$

Multiply the numerator by the reciprocal of the denominator.

$1 \frac{3}{4 π} (\frac{4}{3} \cdot (π y^{3})) = \frac{1}{\frac{4}{3} π} r$

Multiply $\frac{3}{4 π}$ by $1$ .

$\frac{3}{4 π} (\frac{4}{3} \cdot (π y^{3})) = \frac{1}{\frac{4}{3} π} r$

Cancel the common factor of $π$ .

Factor $π$ out of $4 π$ .

$\frac{3}{π \cdot 4} (\frac{4}{3} \cdot (π y^{3})) = \frac{1}{\frac{4}{3} π} r$

Factor $π$ out of $\frac{4}{3} \cdot (π y^{3})$ .

$\frac{3}{π \cdot 4} (π (\frac{4}{3} \cdot (y^{3}))) = \frac{1}{\frac{4}{3} π} r$

Cancel the common factor.

$\frac{3}{π \cdot 4} (π (\frac{4}{3} \cdot (y^{3}))) = \frac{1}{\frac{4}{3} π} r$

Rewrite the expression.

$\frac{3}{4} (\frac{4}{3} \cdot (y^{3})) = \frac{1}{\frac{4}{3} π} r$

Combine $\frac{4}{3}$ and $y^{3}$ .

$\frac{3}{4} \cdot \frac{4 y^{3}}{3} = \frac{1}{\frac{4}{3} π} r$

Multiply $\frac{3}{4}$ by $\frac{4 y^{3}}{3}$ .

$\frac{3 (4 y^{3})}{4 \cdot 3} = \frac{1}{\frac{4}{3} π} r$

Multiply.

Multiply $4$ by $3$ .

$\frac{12 y^{3}}{4 \cdot 3} = \frac{1}{\frac{4}{3} π} r$

Multiply $4$ by $3$ .

$\frac{12 y^{3}}{12} = \frac{1}{\frac{4}{3} π} r$

Cancel the common factor of $12$ .

Cancel the common factor.

$\frac{12 y^{3}}{12} = \frac{1}{\frac{4}{3} π} r$

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

$y^{3} = \frac{1}{\frac{4}{3} π} r$

Simplify the right side.

Simplify $\frac{1}{\frac{4}{3} π} r$ .

Combine $\frac{4}{3}$ and $π$ .

$y^{3} = \frac{1}{\frac{4 π}{3}} r$

Multiply the numerator by the reciprocal of the denominator.

$y^{3} = 1 \frac{3}{4 π} r$

Multiply $\frac{3}{4 π}$ by $1$ .

$y^{3} = \frac{3}{4 π} r$

Combine $\frac{3}{4 π}$ and $r$ .

$y^{3} = \frac{3 r}{4 π}$

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

$y = \sqrt[3]{\frac{3 r}{4 π}}$

Simplify $\sqrt[3]{\frac{3 r}{4 π}}$ .

Rewrite $\sqrt[3]{\frac{3 r}{4 π}}$ as $\frac{\sqrt[3]{3 r}}{\sqrt[3]{4 π}}$ .

$y = \frac{\sqrt[3]{3 r}}{\sqrt[3]{4 π}}$

Multiply $\frac{\sqrt[3]{3 r}}{\sqrt[3]{4 π}}$ by $\frac{{\sqrt[3]{4 π}}^{2}}{{\sqrt[3]{4 π}}^{2}}$ .

$y = \frac{\sqrt[3]{3 r}}{\sqrt[3]{4 π}} \cdot \frac{{\sqrt[3]{4 π}}^{2}}{{\sqrt[3]{4 π}}^{2}}$

Combine and simplify the denominator.

Multiply $\frac{\sqrt[3]{3 r}}{\sqrt[3]{4 π}}$ by $\frac{{\sqrt[3]{4 π}}^{2}}{{\sqrt[3]{4 π}}^{2}}$ .

$y = \frac{\sqrt[3]{3 r} {\sqrt[3]{4 π}}^{2}}{\sqrt[3]{4 π} {\sqrt[3]{4 π}}^{2}}$

Raise $\sqrt[3]{4 π}$ to the power of $1$ .

$y = \frac{\sqrt[3]{3 r} {\sqrt[3]{4 π}}^{2}}{{\sqrt[3]{4 π}}^{1} {\sqrt[3]{4 π}}^{2}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y = \frac{\sqrt[3]{3 r} {\sqrt[3]{4 π}}^{2}}{{\sqrt[3]{4 π}}^{1 + 2}}$

Add $1$ and $2$ .

$y = \frac{\sqrt[3]{3 r} {\sqrt[3]{4 π}}^{2}}{{\sqrt[3]{4 π}}^{3}}$

Rewrite ${\sqrt[3]{4 π}}^{3}$ as $4 π$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{4 π}$ as ${(4 π)}^{\frac{1}{3}}$ .

$y = \frac{\sqrt[3]{3 r} {\sqrt[3]{4 π}}^{2}}{{({(4 π)}^{\frac{1}{3}})}^{3}}$

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

$y = \frac{\sqrt[3]{3 r} {\sqrt[3]{4 π}}^{2}}{{(4 π)}^{\frac{1}{3} \cdot 3}}$

Combine $\frac{1}{3}$ and $3$ .

$y = \frac{\sqrt[3]{3 r} {\sqrt[3]{4 π}}^{2}}{{(4 π)}^{\frac{3}{3}}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$y = \frac{\sqrt[3]{3 r} {\sqrt[3]{4 π}}^{2}}{{(4 π)}^{\frac{3}{3}}}$

Rewrite the expression.

$y = \frac{\sqrt[3]{3 r} {\sqrt[3]{4 π}}^{2}}{{(4 π)}^{1}}$

Simplify.

$y = \frac{\sqrt[3]{3 r} {\sqrt[3]{4 π}}^{2}}{4 π}$

Simplify the numerator.

Rewrite ${\sqrt[3]{4 π}}^{2}$ as ${({(4 π)}^{2})}^{\frac{1}{3}}$ .

$y = \frac{\sqrt[3]{3 r} \sqrt[3]{{(4 π)}^{2}}}{4 π}$

Apply the product rule to $4 π$ .

$y = \frac{\sqrt[3]{3 r} \sqrt[3]{4^{2} π^{2}}}{4 π}$

Raise $4$ to the power of $2$ .

$y = \frac{\sqrt[3]{3 r} \sqrt[3]{16 π^{2}}}{4 π}$

Rewrite $16 π^{2}$ as $2^{3} \cdot (2 π^{2})$ .

Factor $8$ out of $16$ .

$y = \frac{\sqrt[3]{3 r} \sqrt[3]{8 (2) π^{2}}}{4 π}$

Rewrite $8$ as $2^{3}$ .

$y = \frac{\sqrt[3]{3 r} \sqrt[3]{2^{3} \cdot 2 π^{2}}}{4 π}$

Add parentheses.

$y = \frac{\sqrt[3]{3 r} \sqrt[3]{2^{3} \cdot (2 π^{2})}}{4 π}$

Pull terms out from under the radical.

$y = \frac{\sqrt[3]{3 r} \cdot 2 \sqrt[3]{2 π^{2}}}{4 π}$

Combine exponents.

Combine using the product rule for radicals.

$y = \frac{2 \sqrt[3]{3 r (2 π^{2})}}{4 π}$

Multiply $2$ by $3$ .

$y = \frac{2 \sqrt[3]{6 r π^{2}}}{4 π}$

Cancel the common factor of $2$ and $4$ .

Factor $2$ out of $2 \sqrt[3]{6 r π^{2}}$ .

$y = \frac{2 (\sqrt[3]{6 r π^{2}})}{4 π}$

Cancel the common factors.

Factor $2$ out of $4 π$ .

$y = \frac{2 (\sqrt[3]{6 r π^{2}})}{2 (2 π)}$

Cancel the common factor.

$y = \frac{2 \sqrt[3]{6 r π^{2}}}{2 (2 π)}$

Rewrite the expression.

$y = \frac{\sqrt[3]{6 r π^{2}}}{2 π}$

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

$V^{- 1} (r) = \frac{\sqrt[3]{6 r π^{2}}}{2 π}$

Verify if $V^{- 1} (r) = \frac{\sqrt[3]{6 r π^{2}}}{2 π}$ is the inverse of $V (r) = \frac{4}{3} \cdot (π r^{3})$ .

To verify the inverse, check if $V^{- 1} (V (r)) = r$ and $V (V^{- 1} (r)) = r$ .

Evaluate $V^{- 1} (V (r))$ .

Set up the composite result function.

$V^{- 1} (V (r))$

Evaluate $V^{- 1} (\frac{4}{3} \cdot (π r^{3}))$ by substituting in the value of $V$ into $V^{- 1}$ .

$V^{- 1} (\frac{4}{3} \cdot (π r^{3})) = \frac{\sqrt[3]{6 (\frac{4}{3} \cdot (π r^{3})) \cdot π^{2}}}{2 π}$

Simplify the numerator.

Combine $6$ and $\frac{4}{3}$ .

$V^{- 1} (\frac{4}{3} \cdot (π r^{3})) = \frac{\sqrt[3]{\frac{6 \cdot 4}{3} \cdot (π r^{3} π^{2})}}{2 π}$

Combine $\frac{6 \cdot 4}{3}$ and $π$ .

$V^{- 1} (\frac{4}{3} \cdot (π r^{3})) = \frac{\sqrt[3]{\frac{6 \cdot (4 π)}{3} \cdot (r^{3} π^{2})}}{2 π}$

Combine $\frac{6 \cdot 4 π}{3}$ and $r^{3}$ .

$V^{- 1} (\frac{4}{3} \cdot (π r^{3})) = \frac{\sqrt[3]{\frac{6 \cdot (4 π r^{3})}{3} \cdot π^{2}}}{2 π}$

Combine $\frac{6 \cdot 4 π r^{3}}{3}$ and $π^{2}$ .

$V^{- 1} (\frac{4}{3} \cdot (π r^{3})) = \frac{\sqrt[3]{\frac{6 \cdot (4 π r^{3} π^{2})}{3}}}{2 π}$

Reduce the expression by cancelling the common factors.

Reduce the expression $\frac{6 \cdot 4 π r^{3} π^{2}}{3}$ by cancelling the common factors.

Factor $3$ out of $6 \cdot 4 π r^{3} π^{2}$ .

$V^{- 1} (\frac{4}{3} \cdot (π r^{3})) = \frac{\sqrt[3]{\frac{3 (2 \cdot (4 π r^{3} π^{2}))}{3}}}{2 π}$

Factor $3$ out of $3$ .

$V^{- 1} (\frac{4}{3} \cdot (π r^{3})) = \frac{\sqrt[3]{\frac{3 (2 \cdot (4 π r^{3} π^{2}))}{3 (1)}}}{2 π}$

Cancel the common factor.

$V^{- 1} (\frac{4}{3} \cdot (π r^{3})) = \frac{\sqrt[3]{\frac{3 (2 \cdot (4 π r^{3} π^{2}))}{3 \cdot 1}}}{2 π}$

Rewrite the expression.

$V^{- 1} (\frac{4}{3} \cdot (π r^{3})) = \frac{\sqrt[3]{\frac{2 \cdot (4 π r^{3} π^{2})}{1}}}{2 π}$

Divide $2 \cdot 4 π r^{3} π^{2}$ by $1$ .

$V^{- 1} (\frac{4}{3} \cdot (π r^{3})) = \frac{\sqrt[3]{2 \cdot (4 π r^{3} π^{2})}}{2 π}$

Combine exponents.

Multiply $2$ by $4$ .

$V^{- 1} (\frac{4}{3} \cdot (π r^{3})) = \frac{\sqrt[3]{8 π r^{3} π^{2}}}{2 π}$

Multiply $π$ by $π^{2}$ by adding the exponents.

Move $π^{2}$ .

$V^{- 1} (\frac{4}{3} \cdot (π r^{3})) = \frac{\sqrt[3]{8 (π^{2} π) r^{3}}}{2 π}$

Multiply $π^{2}$ by $π$ .

Raise $π$ to the power of $1$ .

$V^{- 1} (\frac{4}{3} \cdot (π r^{3})) = \frac{\sqrt[3]{8 (π^{2} π) r^{3}}}{2 π}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$V^{- 1} (\frac{4}{3} \cdot (π r^{3})) = \frac{\sqrt[3]{8 π^{2 + 1} r^{3}}}{2 π}$

Add $2$ and $1$ .

$V^{- 1} (\frac{4}{3} \cdot (π r^{3})) = \frac{\sqrt[3]{8 π^{3} r^{3}}}{2 π}$

Rewrite $8 π^{3} r^{3}$ as ${(2 π r)}^{3}$ .

$V^{- 1} (\frac{4}{3} \cdot (π r^{3})) = \frac{\sqrt[3]{{(2 π r)}^{3}}}{2 π}$

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

$V^{- 1} (\frac{4}{3} \cdot (π r^{3})) = \frac{2 π r}{2 π}$

Reduce the expression by cancelling the common factors.

Cancel the common factor of $2$ .

Cancel the common factor.

$V^{- 1} (\frac{4}{3} \cdot (π r^{3})) = \frac{2 π r}{2 π}$

Rewrite the expression.

$V^{- 1} (\frac{4}{3} \cdot (π r^{3})) = \frac{π r}{π}$

Cancel the common factor of $π$ .

Cancel the common factor.

$V^{- 1} (\frac{4}{3} \cdot (π r^{3})) = \frac{π r}{π}$

Divide $r$ by $1$ .

$V^{- 1} (\frac{4}{3} \cdot (π r^{3})) = r$

Evaluate $V (V^{- 1} (r))$ .

Set up the composite result function.

$V (V^{- 1} (r))$

Evaluate $V (\frac{\sqrt[3]{6 r π^{2}}}{2 π})$ by substituting in the value of $V^{- 1}$ into $V$ .

$V (\frac{\sqrt[3]{6 r π^{2}}}{2 π}) = \frac{4}{3} \cdot (π {(\frac{\sqrt[3]{6 r π^{2}}}{2 π})}^{3})$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $\frac{\sqrt[3]{6 r π^{2}}}{2 π}$ .

$V (\frac{\sqrt[3]{6 r π^{2}}}{2 π}) = \frac{4}{3} \cdot (π (\frac{{\sqrt[3]{6 r π^{2}}}^{3}}{{(2 π)}^{3}}))$

Apply the product rule to $2 π$ .

$V (\frac{\sqrt[3]{6 r π^{2}}}{2 π}) = \frac{4}{3} \cdot (π (\frac{{\sqrt[3]{6 r π^{2}}}^{3}}{2^{3} π^{3}}))$

Rewrite ${\sqrt[3]{6 r π^{2}}}^{3}$ as $6 r π^{2}$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{6 r π^{2}}$ as ${(6 r π^{2})}^{\frac{1}{3}}$ .

$V (\frac{\sqrt[3]{6 r π^{2}}}{2 π}) = \frac{4}{3} \cdot (π (\frac{{({(6 r π^{2})}^{\frac{1}{3}})}^{3}}{2^{3} π^{3}}))$

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

$V (\frac{\sqrt[3]{6 r π^{2}}}{2 π}) = \frac{4}{3} \cdot (π (\frac{{(6 r π^{2})}^{\frac{1}{3} \cdot 3}}{2^{3} π^{3}}))$

Combine $\frac{1}{3}$ and $3$ .

$V (\frac{\sqrt[3]{6 r π^{2}}}{2 π}) = \frac{4}{3} \cdot (π (\frac{{(6 r π^{2})}^{\frac{3}{3}}}{2^{3} π^{3}}))$

Cancel the common factor of $3$ .

Cancel the common factor.

$V (\frac{\sqrt[3]{6 r π^{2}}}{2 π}) = \frac{4}{3} \cdot (π (\frac{{(6 r π^{2})}^{\frac{3}{3}}}{2^{3} π^{3}}))$

Rewrite the expression.

$V (\frac{\sqrt[3]{6 r π^{2}}}{2 π}) = \frac{4}{3} \cdot (π (\frac{6 r π^{2}}{2^{3} π^{3}}))$

Simplify.

$V (\frac{\sqrt[3]{6 r π^{2}}}{2 π}) = \frac{4}{3} \cdot (π (\frac{6 r π^{2}}{2^{3} π^{3}}))$

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

$V (\frac{\sqrt[3]{6 r π^{2}}}{2 π}) = \frac{4}{3} \cdot (π (\frac{6 r π^{2}}{8 π^{3}}))$

Cancel the common factor of $π$ .

Factor $π$ out of $8 π^{3}$ .

$V (\frac{\sqrt[3]{6 r π^{2}}}{2 π}) = \frac{4}{3} \cdot (π (\frac{6 r π^{2}}{π (8 π^{2})}))$

Cancel the common factor.

$V (\frac{\sqrt[3]{6 r π^{2}}}{2 π}) = \frac{4}{3} \cdot (π (\frac{6 r π^{2}}{π (8 π^{2})}))$

Rewrite the expression.

$V (\frac{\sqrt[3]{6 r π^{2}}}{2 π}) = \frac{4}{3} \cdot \frac{6 r π^{2}}{8 π^{2}}$

Combine.

$V (\frac{\sqrt[3]{6 r π^{2}}}{2 π}) = \frac{4 (6 r π^{2})}{3 (8 π^{2})}$

Cancel the common factors.

Factor $4$ out of $3 (8 π^{2})$ .

$V (\frac{\sqrt[3]{6 r π^{2}}}{2 π}) = \frac{4 (6 r π^{2})}{4 (3 (2 π^{2}))}$

Cancel the common factor.

$V (\frac{\sqrt[3]{6 r π^{2}}}{2 π}) = \frac{4 (6 r π^{2})}{4 (3 (2 π^{2}))}$

Rewrite the expression.

$V (\frac{\sqrt[3]{6 r π^{2}}}{2 π}) = \frac{6 r π^{2}}{3 (2 π^{2})}$

Cancel the common factor of $6$ and $3$ .

Factor $3$ out of $6 r π^{2}$ .

$V (\frac{\sqrt[3]{6 r π^{2}}}{2 π}) = \frac{3 (2 r π^{2})}{3 (2 π^{2})}$

Cancel the common factors.

Cancel the common factor.

$V (\frac{\sqrt[3]{6 r π^{2}}}{2 π}) = \frac{3 (2 r π^{2})}{3 (2 π^{2})}$

Rewrite the expression.

$V (\frac{\sqrt[3]{6 r π^{2}}}{2 π}) = \frac{2 r π^{2}}{2 π^{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$V (\frac{\sqrt[3]{6 r π^{2}}}{2 π}) = \frac{2 r π^{2}}{2 π^{2}}$

Rewrite the expression.

$V (\frac{\sqrt[3]{6 r π^{2}}}{2 π}) = \frac{r π^{2}}{π^{2}}$

Cancel the common factor of $π^{2}$ .

Cancel the common factor.

$V (\frac{\sqrt[3]{6 r π^{2}}}{2 π}) = \frac{r π^{2}}{π^{2}}$

Divide $r$ by $1$ .

$V (\frac{\sqrt[3]{6 r π^{2}}}{2 π}) = r$

Since $V^{- 1} (V (r)) = r$ and $V (V^{- 1} (r)) = r$ , then $V^{- 1} (r) = \frac{\sqrt[3]{6 r π^{2}}}{2 π}$ is the inverse of $V (r) = \frac{4}{3} \cdot (π r^{3})$ .

$V^{- 1} (r) = \frac{\sqrt[3]{6 r π^{2}}}{2 π}$

Find the Inverse V(r)=4/3pir^3

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