Find the Inverse f(x)=- cube root of (2x+4)/3

$f (x) = - \sqrt[3]{\frac{2 x + 4}{3}}$

Step 1

Write $f (x) = - \sqrt[3]{\frac{2 x + 4}{3}}$ as an equation.

$y = - \sqrt[3]{\frac{2 x + 4}{3}}$

Step 2

Interchange the variables.

$x = - \sqrt[3]{\frac{2 y + 4}{3}}$

Step 3

Solve for $y$ .

Rewrite the equation as $- \sqrt[3]{\frac{2 y + 4}{3}} = x$ .

$- \sqrt[3]{\frac{2 y + 4}{3}} = x$

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

${(- \sqrt[3]{\frac{2 y + 4}{3}})}^{3} = x^{3}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{\frac{2 y + 4}{3}}$ as ${(\frac{2 y + 4}{3})}^{\frac{1}{3}}$ .

${(- {(\frac{2 y + 4}{3})}^{\frac{1}{3}})}^{3} = x^{3}$

Simplify the left side.

Simplify ${(- {(\frac{2 y + 4}{3})}^{\frac{1}{3}})}^{3}$ .

Factor $2$ out of $2 y + 4$ .

Factor $2$ out of $2 y$ .

${(- {(\frac{2 (y) + 4}{3})}^{\frac{1}{3}})}^{3} = x^{3}$

Factor $2$ out of $4$ .

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

Factor $2$ out of $2 y + 2 \cdot 2$ .

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

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

Apply the product rule to $\frac{2 (y + 2)}{3}$ .

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

Apply the product rule to $2 (y + 2)$ .

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

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

Apply the product rule to $- \frac{2^{\frac{1}{3}} {(y + 2)}^{\frac{1}{3}}}{3^{\frac{1}{3}}}$ .

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

Apply the product rule to $\frac{2^{\frac{1}{3}} {(y + 2)}^{\frac{1}{3}}}{3^{\frac{1}{3}}}$ .

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

Apply the product rule to $2^{\frac{1}{3}} {(y + 2)}^{\frac{1}{3}}$ .

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

Raise $- 1$ to the power of $3$ .

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

Simplify the numerator.

Multiply the exponents in ${(2^{\frac{1}{3}})}^{3}$ .

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

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

Multiply the exponents in ${({(y + 2)}^{\frac{1}{3}})}^{3}$ .

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the denominator.

Multiply the exponents in ${(3^{\frac{1}{3}})}^{3}$ .

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

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

Solve for $y$ .

Multiply both sides of the equation by $- \frac{3}{2}$ .

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

Simplify both sides of the equation.

Simplify the left side.

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

Cancel the common factor of $3$ .

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

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

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

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

Factor $3$ out of $- 3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

Factor $2$ out of $- 2 (y + 2)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply.

Multiply $- 1$ by $- 1$ .

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

Multiply $y + 2$ by $1$ .

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

Simplify the right side.

Simplify $- \frac{3}{2} x^{3}$ .

Combine $x^{3}$ and $\frac{3}{2}$ .

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

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

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

Subtract $2$ from both sides of the equation.

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

Step 4

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

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

Step 5

Verify if $f^{- 1} (x) = - \frac{3 x^{3}}{2} - 2$ is the inverse of $f (x) = - \sqrt[3]{\frac{2 x + 4}{3}}$ .

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

Evaluate $f^{- 1} (f (x))$ .

Set up the composite result function.

$f^{- 1} (f (x))$

Evaluate $f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 {(- \sqrt[3]{\frac{2 x + 4}{3}})}^{3}}{2} - 2$

Simplify each term.

Factor $2$ out of $2 x + 4$ .

Factor $2$ out of $2 x$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 {(- \sqrt[3]{\frac{2 (x) + 4}{3}})}^{3}}{2} - 2$

Factor $2$ out of $4$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 {(- \sqrt[3]{\frac{2 x + 2 \cdot 2}{3}})}^{3}}{2} - 2$

Factor $2$ out of $2 x + 2 \cdot 2$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 {(- \sqrt[3]{\frac{2 (x + 2)}{3}})}^{3}}{2} - 2$

Simplify the numerator.

Apply the product rule to $- \sqrt[3]{\frac{2 (x + 2)}{3}}$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 {(- 1)}^{3} {\sqrt[3]{\frac{2 (x + 2)}{3}}}^{3}}{2} - 2$

Raise $- 1$ to the power of $3$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 {\sqrt[3]{\frac{2 (x + 2)}{3}}}^{3})}{2} - 2$

Rewrite $\sqrt[3]{\frac{2 (x + 2)}{3}}$ as $\frac{\sqrt[3]{2 (x + 2)}}{\sqrt[3]{3}}$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 {(\frac{\sqrt[3]{2 (x + 2)}}{\sqrt[3]{3}})}^{3})}{2} - 2$

Multiply $\frac{\sqrt[3]{2 (x + 2)}}{\sqrt[3]{3}}$ by $\frac{{\sqrt[3]{3}}^{2}}{{\sqrt[3]{3}}^{2}}$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 {(\frac{\sqrt[3]{2 (x + 2)}}{\sqrt[3]{3}} \cdot \frac{{\sqrt[3]{3}}^{2}}{{\sqrt[3]{3}}^{2}})}^{3})}{2} - 2$

Combine and simplify the denominator.

Multiply $\frac{\sqrt[3]{2 (x + 2)}}{\sqrt[3]{3}}$ by $\frac{{\sqrt[3]{3}}^{2}}{{\sqrt[3]{3}}^{2}}$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 {(\frac{\sqrt[3]{2 (x + 2)} {\sqrt[3]{3}}^{2}}{\sqrt[3]{3} {\sqrt[3]{3}}^{2}})}^{3})}{2} - 2$

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

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 {(\frac{\sqrt[3]{2 (x + 2)} {\sqrt[3]{3}}^{2}}{\sqrt[3]{3} {\sqrt[3]{3}}^{2}})}^{3})}{2} - 2$

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

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 {(\frac{\sqrt[3]{2 (x + 2)} {\sqrt[3]{3}}^{2}}{{\sqrt[3]{3}}^{1 + 2}})}^{3})}{2} - 2$

Add $1$ and $2$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 {(\frac{\sqrt[3]{2 (x + 2)} {\sqrt[3]{3}}^{2}}{{\sqrt[3]{3}}^{3}})}^{3})}{2} - 2$

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

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

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 {(\frac{\sqrt[3]{2 (x + 2)} {\sqrt[3]{3}}^{2}}{{(3^{\frac{1}{3}})}^{3}})}^{3})}{2} - 2$

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

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 {(\frac{\sqrt[3]{2 (x + 2)} {\sqrt[3]{3}}^{2}}{3^{\frac{1}{3} \cdot 3}})}^{3})}{2} - 2$

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

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 {(\frac{\sqrt[3]{2 (x + 2)} {\sqrt[3]{3}}^{2}}{3^{\frac{3}{3}}})}^{3})}{2} - 2$

Cancel the common factor of $3$ .

Cancel the common factor.

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 {(\frac{\sqrt[3]{2 (x + 2)} {\sqrt[3]{3}}^{2}}{3^{\frac{3}{3}}})}^{3})}{2} - 2$

Rewrite the expression.

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 {(\frac{\sqrt[3]{2 (x + 2)} {\sqrt[3]{3}}^{2}}{3})}^{3})}{2} - 2$

Evaluate the exponent.

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 {(\frac{\sqrt[3]{2 (x + 2)} {\sqrt[3]{3}}^{2}}{3})}^{3})}{2} - 2$

Simplify the numerator.

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

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 {(\frac{\sqrt[3]{2 (x + 2)} \sqrt[3]{3^{2}}}{3})}^{3})}{2} - 2$

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

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 {(\frac{\sqrt[3]{2 (x + 2)} \sqrt[3]{9}}{3})}^{3})}{2} - 2$

Simplify the numerator.

Combine using the product rule for radicals.

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 {(\frac{\sqrt[3]{2 (x + 2) \cdot 9}}{3})}^{3})}{2} - 2$

Multiply $9$ by $2$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 {(\frac{\sqrt[3]{18 (x + 2)}}{3})}^{3})}{2} - 2$

Apply the product rule to $\frac{\sqrt[3]{18 (x + 2)}}{3}$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 \frac{{\sqrt[3]{18 (x + 2)}}^{3}}{3^{3}})}{2} - 2$

Simplify the numerator.

Rewrite ${\sqrt[3]{18 (x + 2)}}^{3}$ as $18 (x + 2)$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{18 (x + 2)}$ as ${(18 (x + 2))}^{\frac{1}{3}}$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 \frac{{({(18 (x + 2))}^{\frac{1}{3}})}^{3}}{3^{3}})}{2} - 2$

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

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 \frac{{(18 (x + 2))}^{\frac{1}{3} \cdot 3}}{3^{3}})}{2} - 2$

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

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 \frac{{(18 (x + 2))}^{\frac{3}{3}}}{3^{3}})}{2} - 2$

Cancel the common factor of $3$ .

Cancel the common factor.

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 \frac{{(18 (x + 2))}^{\frac{3}{3}}}{3^{3}})}{2} - 2$

Rewrite the expression.

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 \frac{18 (x + 2)}{3^{3}})}{2} - 2$

Simplify.

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 \frac{18 (x + 2)}{3^{3}})}{2} - 2$

Apply the distributive property.

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 \frac{18 x + 18 \cdot 2}{3^{3}})}{2} - 2$

Multiply $18$ by $2$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 \frac{18 x + 36}{3^{3}})}{2} - 2$

Factor $18$ out of $18 x + 36$ .

Factor $18$ out of $18 x$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 \frac{18 (x) + 36}{3^{3}})}{2} - 2$

Factor $18$ out of $36$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 \frac{18 x + 18 \cdot 2}{3^{3}})}{2} - 2$

Factor $18$ out of $18 x + 18 \cdot 2$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 \frac{18 (x + 2)}{3^{3}})}{2} - 2$

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

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 \frac{18 (x + 2)}{27})}{2} - 2$

Cancel the common factor of $18$ and $27$ .

Factor $9$ out of $18 (x + 2)$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 \frac{9 (2 (x + 2))}{27})}{2} - 2$

Cancel the common factors.

Factor $9$ out of $27$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 \frac{9 (2 (x + 2))}{9 (3)})}{2} - 2$

Cancel the common factor.

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 \frac{9 (2 (x + 2))}{9 \cdot 3})}{2} - 2$

Rewrite the expression.

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{3 \cdot (- 1 \frac{2 (x + 2)}{3})}{2} - 2$

Combine exponents.

Factor out negative.

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{- (3 (\frac{2 (x + 2)}{3}))}{2} - 2$

Combine $3$ and $\frac{2 (x + 2)}{3}$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{- \frac{3 (2 (x + 2))}{3}}{2} - 2$

Multiply $2$ by $3$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{- \frac{6 (x + 2)}{3}}{2} - 2$

Reduce the expression by cancelling the common factors.

Reduce the expression $\frac{6 (x + 2)}{3}$ by cancelling the common factors.

Factor $3$ out of $6 (x + 2)$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{- \frac{3 (2 (x + 2))}{3}}{2} - 2$

Factor $3$ out of $3$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{- \frac{3 (2 (x + 2))}{3 (1)}}{2} - 2$

Cancel the common factor.

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{- \frac{3 (2 (x + 2))}{3 \cdot 1}}{2} - 2$

Rewrite the expression.

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{- \frac{2 (x + 2)}{1}}{2} - 2$

Divide $2 (x + 2)$ by $1$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{- (2 (x + 2))}{2} - 2$

Multiply $- 1$ by $2$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{- 2 (x + 2)}{2} - 2$

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

Factor $2$ out of $- 2 (x + 2)$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{2 (- (x + 2))}{2} - 2$

Cancel the common factors.

Factor $2$ out of $2$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{2 (- (x + 2))}{2 (1)} - 2$

Cancel the common factor.

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{2 (- (x + 2))}{2 \cdot 1} - 2$

Rewrite the expression.

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - \frac{- (x + 2)}{1} - 2$

Divide $- (x + 2)$ by $1$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = (x + 2) - 2$

Apply the distributive property.

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - (- x - 1 \cdot 2) - 2$

Multiply $- 1$ by $2$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = - (- x - 2) - 2$

Apply the distributive property.

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = x + 2 - 2$

Multiply $- - x$ .

Multiply $- 1$ by $- 1$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = 1 x + 2 - 2$

Multiply $x$ by $1$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = x + 2 - 2$

Multiply $- 1$ by $- 2$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = x + 2 - 2$

Combine the opposite terms in $x + 2 - 2$ .

Subtract $2$ from $2$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (- \sqrt[3]{\frac{2 x + 4}{3}}) = x$

Evaluate $f (f^{- 1} (x))$ .

Set up the composite result function.

$f (f^{- 1} (x))$

Evaluate $f (- \frac{3 x^{3}}{2} - 2)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (- \frac{3 x^{3}}{2} - 2) = - \sqrt[3]{\frac{2 (- \frac{3 x^{3}}{2} - 2) + 4}{3}}$

Factor $2$ out of $2 (- \frac{3 x^{3}}{2} - 2) + 4$ .

Factor $2$ out of $4$ .

$f (- \frac{3 x^{3}}{2} - 2) = - \sqrt[3]{\frac{2 (- \frac{3 x^{3}}{2} - 2) + 2 \cdot 2}{3}}$

Factor $2$ out of $2 (- \frac{3 x^{3}}{2} - 2) + 2 \cdot 2$ .

$f (- \frac{3 x^{3}}{2} - 2) = - \sqrt[3]{\frac{2 (- \frac{3 x^{3}}{2} - 2 + 2)}{3}}$

Add $- 2$ and $2$ .

$f (- \frac{3 x^{3}}{2} - 2) = - \sqrt[3]{\frac{2 (- \frac{3 x^{3}}{2} + 0)}{3}}$

Add $- \frac{3 x^{3}}{2}$ and $0$ .

$f (- \frac{3 x^{3}}{2} - 2) = - \sqrt[3]{\frac{2 \cdot (- 1 \frac{3 x^{3}}{2})}{3}}$

Combine exponents.

Factor out negative.

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

Combine $2$ and $\frac{3 x^{3}}{2}$ .

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

Multiply $3$ by $2$ .

$f (- \frac{3 x^{3}}{2} - 2) = - \sqrt[3]{\frac{- \frac{6 x^{3}}{2}}{3}}$

Cancel the common factor of $6$ and $2$ .

Factor $2$ out of $6 x^{3}$ .

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

Cancel the common factors.

Factor $2$ out of $2$ .

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

Cancel the common factor.

$f (- \frac{3 x^{3}}{2} - 2) = - \sqrt[3]{\frac{- \frac{2 (3 x^{3})}{2 \cdot 1}}{3}}$

Rewrite the expression.

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $- (x^{3})$ by $1$ .

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

Rewrite $- x^{3}$ as ${(- x)}^{3}$ .

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

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

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

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = - \frac{3 x^{3}}{2} - 2$ is the inverse of $f (x) = - \sqrt[3]{\frac{2 x + 4}{3}}$ .

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

Find the Inverse f(x)=- cube root of (2x+4)/3

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