Find the Inverse f(x)=5 cube root of x+4-2

$f (x) = 5 \sqrt[3]{x + 4} - 2$

Step 1

Write $f (x) = 5 \sqrt[3]{x + 4} - 2$ as an equation.

$y = 5 \sqrt[3]{x + 4} - 2$

Step 2

Interchange the variables.

$x = 5 \sqrt[3]{y + 4} - 2$

Step 3

Solve for $y$ .

Rewrite the equation as $5 \sqrt[3]{y + 4} - 2 = x$ .

$5 \sqrt[3]{y + 4} - 2 = x$

Add $2$ to both sides of the equation.

$5 \sqrt[3]{y + 4} = x + 2$

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

${(5 \sqrt[3]{y + 4})}^{3} = {(x + 2)}^{3}$

Simplify each side of the equation.

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

${(5 {(y + 4)}^{\frac{1}{3}})}^{3} = {(x + 2)}^{3}$

Simplify the left side.

Simplify ${(5 {(y + 4)}^{\frac{1}{3}})}^{3}$ .

Apply the product rule to $5 {(y + 4)}^{\frac{1}{3}}$ .

$5^{3} {({(y + 4)}^{\frac{1}{3}})}^{3} = {(x + 2)}^{3}$

Raise $5$ to the power of $3$ .

$125 {({(y + 4)}^{\frac{1}{3}})}^{3} = {(x + 2)}^{3}$

Multiply the exponents in ${({(y + 4)}^{\frac{1}{3}})}^{3}$ .

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

$125 {(y + 4)}^{\frac{1}{3} \cdot 3} = {(x + 2)}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$125 {(y + 4)}^{\frac{1}{3} \cdot 3} = {(x + 2)}^{3}$

Rewrite the expression.

$125 {(y + 4)}^{1} = {(x + 2)}^{3}$

Simplify.

$125 (y + 4) = {(x + 2)}^{3}$

Apply the distributive property.

$125 y + 125 \cdot 4 = {(x + 2)}^{3}$

Multiply $125$ by $4$ .

$125 y + 500 = {(x + 2)}^{3}$

Simplify the right side.

Simplify ${(x + 2)}^{3}$ .

Use the Binomial Theorem.

$125 y + 500 = x^{3} + 3 x^{2} \cdot 2 + 3 x \cdot 2^{2} + 2^{3}$

Simplify each term.

Multiply $2$ by $3$ .

$125 y + 500 = x^{3} + 6 x^{2} + 3 x \cdot 2^{2} + 2^{3}$

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

$125 y + 500 = x^{3} + 6 x^{2} + 3 x \cdot 4 + 2^{3}$

Multiply $4$ by $3$ .

$125 y + 500 = x^{3} + 6 x^{2} + 12 x + 2^{3}$

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

$125 y + 500 = x^{3} + 6 x^{2} + 12 x + 8$

Solve for $y$ .

Move all terms not containing $y$ to the right side of the equation.

Subtract $500$ from both sides of the equation.

$125 y = x^{3} + 6 x^{2} + 12 x + 8 - 500$

Subtract $500$ from $8$ .

$125 y = x^{3} + 6 x^{2} + 12 x - 492$

Divide each term in $125 y = x^{3} + 6 x^{2} + 12 x - 492$ by $125$ and simplify.

Divide each term in $125 y = x^{3} + 6 x^{2} + 12 x - 492$ by $125$ .

$\frac{125 y}{125} = \frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} + \frac{- 492}{125}$

Simplify the left side.

Cancel the common factor of $125$ .

Cancel the common factor.

$\frac{125 y}{125} = \frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} + \frac{- 492}{125}$

Divide $y$ by $1$ .

$y = \frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} + \frac{- 492}{125}$

Simplify the right side.

Move the negative in front of the fraction.

$y = \frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}$

Step 4

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

$f^{- 1} (x) = \frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}$

Step 5

Verify if $f^{- 1} (x) = \frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}$ is the inverse of $f (x) = 5 \sqrt[3]{x + 4} - 2$ .

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 $f^{- 1} (5 \sqrt[3]{x + 4} - 2)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{{(5 \sqrt[3]{x + 4} - 2)}^{3}}{125} + \frac{6 {(5 \sqrt[3]{x + 4} - 2)}^{2}}{125} + \frac{12 (5 \sqrt[3]{x + 4} - 2)}{125} - \frac{492}{125}$

Combine the numerators over the common denominator.

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{{(5 \sqrt[3]{x + 4} - 2)}^{3} + 6 {(5 \sqrt[3]{x + 4} - 2)}^{2} + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Simplify each term.

Use the Binomial Theorem.

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{{(5 \sqrt[3]{x + 4})}^{3} + 3 {(5 \sqrt[3]{x + 4})}^{2} \cdot - 2 + 3 (5 \sqrt[3]{x + 4}) {(- 2)}^{2} + {(- 2)}^{3} + 6 {(5 \sqrt[3]{x + 4} - 2)}^{2} + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Simplify each term.

Apply the product rule to $5 \sqrt[3]{x + 4}$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{5^{3} {\sqrt[3]{x + 4}}^{3} + 3 {(5 \sqrt[3]{x + 4})}^{2} \cdot - 2 + 3 (5 \sqrt[3]{x + 4}) {(- 2)}^{2} + {(- 2)}^{3} + 6 {(5 \sqrt[3]{x + 4} - 2)}^{2} + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Raise $5$ to the power of $3$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 {\sqrt[3]{x + 4}}^{3} + 3 {(5 \sqrt[3]{x + 4})}^{2} \cdot - 2 + 3 (5 \sqrt[3]{x + 4}) {(- 2)}^{2} + {(- 2)}^{3} + 6 {(5 \sqrt[3]{x + 4} - 2)}^{2} + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

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

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

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 {({(x + 4)}^{\frac{1}{3}})}^{3} + 3 {(5 \sqrt[3]{x + 4})}^{2} \cdot - 2 + 3 (5 \sqrt[3]{x + 4}) {(- 2)}^{2} + {(- 2)}^{3} + 6 {(5 \sqrt[3]{x + 4} - 2)}^{2} + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

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

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 {(x + 4)}^{\frac{1}{3} \cdot 3} + 3 {(5 \sqrt[3]{x + 4})}^{2} \cdot - 2 + 3 (5 \sqrt[3]{x + 4}) {(- 2)}^{2} + {(- 2)}^{3} + 6 {(5 \sqrt[3]{x + 4} - 2)}^{2} + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

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

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 {(x + 4)}^{\frac{3}{3}} + 3 {(5 \sqrt[3]{x + 4})}^{2} \cdot - 2 + 3 (5 \sqrt[3]{x + 4}) {(- 2)}^{2} + {(- 2)}^{3} + 6 {(5 \sqrt[3]{x + 4} - 2)}^{2} + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Cancel the common factor of $3$ .

Cancel the common factor.

Rewrite the expression.

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 (x + 4) + 3 {(5 \sqrt[3]{x + 4})}^{2} \cdot - 2 + 3 (5 \sqrt[3]{x + 4}) {(- 2)}^{2} + {(- 2)}^{3} + 6 {(5 \sqrt[3]{x + 4} - 2)}^{2} + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Simplify.

Apply the distributive property.

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 125 \cdot 4 + 3 {(5 \sqrt[3]{x + 4})}^{2} \cdot - 2 + 3 (5 \sqrt[3]{x + 4}) {(- 2)}^{2} + {(- 2)}^{3} + 6 {(5 \sqrt[3]{x + 4} - 2)}^{2} + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Multiply $125$ by $4$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 500 + 3 {(5 \sqrt[3]{x + 4})}^{2} \cdot - 2 + 3 (5 \sqrt[3]{x + 4}) {(- 2)}^{2} + {(- 2)}^{3} + 6 {(5 \sqrt[3]{x + 4} - 2)}^{2} + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Apply the product rule to $5 \sqrt[3]{x + 4}$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 500 + 3 (5^{2} {\sqrt[3]{x + 4}}^{2}) \cdot - 2 + 3 (5 \sqrt[3]{x + 4}) {(- 2)}^{2} + {(- 2)}^{3} + 6 {(5 \sqrt[3]{x + 4} - 2)}^{2} + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Raise $5$ to the power of $2$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 500 + 3 (25 {\sqrt[3]{x + 4}}^{2}) \cdot - 2 + 3 (5 \sqrt[3]{x + 4}) {(- 2)}^{2} + {(- 2)}^{3} + 6 {(5 \sqrt[3]{x + 4} - 2)}^{2} + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

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

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 500 + 3 (25 \sqrt[3]{{(x + 4)}^{2}}) \cdot - 2 + 3 (5 \sqrt[3]{x + 4}) {(- 2)}^{2} + {(- 2)}^{3} + 6 {(5 \sqrt[3]{x + 4} - 2)}^{2} + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Multiply $25$ by $3$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 500 + 75 \sqrt[3]{{(x + 4)}^{2}} \cdot - 2 + 3 (5 \sqrt[3]{x + 4}) {(- 2)}^{2} + {(- 2)}^{3} + 6 {(5 \sqrt[3]{x + 4} - 2)}^{2} + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Multiply $- 2$ by $75$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 500 - 150 \sqrt[3]{{(x + 4)}^{2}} + 3 (5 \sqrt[3]{x + 4}) {(- 2)}^{2} + {(- 2)}^{3} + 6 {(5 \sqrt[3]{x + 4} - 2)}^{2} + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Multiply $5$ by $3$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 500 - 150 \sqrt[3]{{(x + 4)}^{2}} + 15 \sqrt[3]{x + 4} {(- 2)}^{2} + {(- 2)}^{3} + 6 {(5 \sqrt[3]{x + 4} - 2)}^{2} + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

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

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 500 - 150 \sqrt[3]{{(x + 4)}^{2}} + 15 \sqrt[3]{x + 4} \cdot 4 + {(- 2)}^{3} + 6 {(5 \sqrt[3]{x + 4} - 2)}^{2} + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Multiply $4$ by $15$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 500 - 150 \sqrt[3]{{(x + 4)}^{2}} + 60 \sqrt[3]{x + 4} + {(- 2)}^{3} + 6 {(5 \sqrt[3]{x + 4} - 2)}^{2} + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

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

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 500 - 150 \sqrt[3]{{(x + 4)}^{2}} + 60 \sqrt[3]{x + 4} - 8 + 6 {(5 \sqrt[3]{x + 4} - 2)}^{2} + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Subtract $8$ from $500$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 492 - 150 \sqrt[3]{{(x + 4)}^{2}} + 60 \sqrt[3]{x + 4} + 6 {(5 \sqrt[3]{x + 4} - 2)}^{2} + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Rewrite ${(5 \sqrt[3]{x + 4} - 2)}^{2}$ as $(5 \sqrt[3]{x + 4} - 2) (5 \sqrt[3]{x + 4} - 2)$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 492 - 150 \sqrt[3]{{(x + 4)}^{2}} + 60 \sqrt[3]{x + 4} + 6 ((5 \sqrt[3]{x + 4} - 2) (5 \sqrt[3]{x + 4} - 2)) + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Expand $(5 \sqrt[3]{x + 4} - 2) (5 \sqrt[3]{x + 4} - 2)$ using the FOIL Method.

Apply the distributive property.

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 492 - 150 \sqrt[3]{{(x + 4)}^{2}} + 60 \sqrt[3]{x + 4} + 6 (5 \sqrt[3]{x + 4} (5 \sqrt[3]{x + 4} - 2) - 2 (5 \sqrt[3]{x + 4} - 2)) + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Apply the distributive property.

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 492 - 150 \sqrt[3]{{(x + 4)}^{2}} + 60 \sqrt[3]{x + 4} + 6 (5 \sqrt[3]{x + 4} (5 \sqrt[3]{x + 4}) + 5 \sqrt[3]{x + 4} \cdot - 2 - 2 (5 \sqrt[3]{x + 4} - 2)) + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Apply the distributive property.

Simplify and combine like terms.

Simplify each term.

Multiply $5 \sqrt[3]{x + 4} (5 \sqrt[3]{x + 4})$ .

Multiply $5$ by $5$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 492 - 150 \sqrt[3]{{(x + 4)}^{2}} + 60 \sqrt[3]{x + 4} + 6 (25 \sqrt[3]{x + 4} \sqrt[3]{x + 4} + 5 \sqrt[3]{x + 4} \cdot - 2 - 2 (5 \sqrt[3]{x + 4}) - 2 \cdot - 2) + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

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

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 492 - 150 \sqrt[3]{{(x + 4)}^{2}} + 60 \sqrt[3]{x + 4} + 6 (25 (\sqrt[3]{x + 4} \sqrt[3]{x + 4}) + 5 \sqrt[3]{x + 4} \cdot - 2 - 2 (5 \sqrt[3]{x + 4}) - 2 \cdot - 2) + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

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

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

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 492 - 150 \sqrt[3]{{(x + 4)}^{2}} + 60 \sqrt[3]{x + 4} + 6 (25 {\sqrt[3]{x + 4}}^{1 + 1} + 5 \sqrt[3]{x + 4} \cdot - 2 - 2 (5 \sqrt[3]{x + 4}) - 2 \cdot - 2) + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Add $1$ and $1$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 492 - 150 \sqrt[3]{{(x + 4)}^{2}} + 60 \sqrt[3]{x + 4} + 6 (25 {\sqrt[3]{x + 4}}^{2} + 5 \sqrt[3]{x + 4} \cdot - 2 - 2 (5 \sqrt[3]{x + 4}) - 2 \cdot - 2) + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

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

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 492 - 150 \sqrt[3]{{(x + 4)}^{2}} + 60 \sqrt[3]{x + 4} + 6 (25 \sqrt[3]{{(x + 4)}^{2}} + 5 \sqrt[3]{x + 4} \cdot - 2 - 2 (5 \sqrt[3]{x + 4}) - 2 \cdot - 2) + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Multiply $- 2$ by $5$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 492 - 150 \sqrt[3]{{(x + 4)}^{2}} + 60 \sqrt[3]{x + 4} + 6 (25 \sqrt[3]{{(x + 4)}^{2}} - 10 \sqrt[3]{x + 4} - 2 (5 \sqrt[3]{x + 4}) - 2 \cdot - 2) + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Multiply $5$ by $- 2$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 492 - 150 \sqrt[3]{{(x + 4)}^{2}} + 60 \sqrt[3]{x + 4} + 6 (25 \sqrt[3]{{(x + 4)}^{2}} - 10 \sqrt[3]{x + 4} - 10 \sqrt[3]{x + 4} - 2 \cdot - 2) + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Multiply $- 2$ by $- 2$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 492 - 150 \sqrt[3]{{(x + 4)}^{2}} + 60 \sqrt[3]{x + 4} + 6 (25 \sqrt[3]{{(x + 4)}^{2}} - 10 \sqrt[3]{x + 4} - 10 \sqrt[3]{x + 4} + 4) + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Subtract $10 \sqrt[3]{x + 4}$ from $- 10 \sqrt[3]{x + 4}$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 492 - 150 \sqrt[3]{{(x + 4)}^{2}} + 60 \sqrt[3]{x + 4} + 6 (25 \sqrt[3]{{(x + 4)}^{2}} - 20 \sqrt[3]{x + 4} + 4) + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Apply the distributive property.

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 492 - 150 \sqrt[3]{{(x + 4)}^{2}} + 60 \sqrt[3]{x + 4} + 6 (25 \sqrt[3]{{(x + 4)}^{2}}) + 6 (- 20 \sqrt[3]{x + 4}) + 6 \cdot 4 + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Simplify.

Multiply $25$ by $6$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 492 - 150 \sqrt[3]{{(x + 4)}^{2}} + 60 \sqrt[3]{x + 4} + 150 \sqrt[3]{{(x + 4)}^{2}} + 6 (- 20 \sqrt[3]{x + 4}) + 6 \cdot 4 + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Multiply $- 20$ by $6$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 492 - 150 \sqrt[3]{{(x + 4)}^{2}} + 60 \sqrt[3]{x + 4} + 150 \sqrt[3]{{(x + 4)}^{2}} - 120 \sqrt[3]{x + 4} + 6 \cdot 4 + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Multiply $6$ by $4$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 492 - 150 \sqrt[3]{{(x + 4)}^{2}} + 60 \sqrt[3]{x + 4} + 150 \sqrt[3]{{(x + 4)}^{2}} - 120 \sqrt[3]{x + 4} + 24 + 12 (5 \sqrt[3]{x + 4} - 2) - 492}{125}$

Apply the distributive property.

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 492 - 150 \sqrt[3]{{(x + 4)}^{2}} + 60 \sqrt[3]{x + 4} + 150 \sqrt[3]{{(x + 4)}^{2}} - 120 \sqrt[3]{x + 4} + 24 + 12 (5 \sqrt[3]{x + 4}) + 12 \cdot - 2 - 492}{125}$

Multiply $5$ by $12$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 492 - 150 \sqrt[3]{{(x + 4)}^{2}} + 60 \sqrt[3]{x + 4} + 150 \sqrt[3]{{(x + 4)}^{2}} - 120 \sqrt[3]{x + 4} + 24 + 60 \sqrt[3]{x + 4} + 12 \cdot - 2 - 492}{125}$

Multiply $12$ by $- 2$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 492 - 150 \sqrt[3]{{(x + 4)}^{2}} + 60 \sqrt[3]{x + 4} + 150 \sqrt[3]{{(x + 4)}^{2}} - 120 \sqrt[3]{x + 4} + 24 + 60 \sqrt[3]{x + 4} - 24 - 492}{125}$

Simplify terms.

Combine the opposite terms in $125 x + 492 - 150 \sqrt[3]{{(x + 4)}^{2}} + 60 \sqrt[3]{x + 4} + 150 \sqrt[3]{{(x + 4)}^{2}} - 120 \sqrt[3]{x + 4} + 24 + 60 \sqrt[3]{x + 4} - 24 - 492$ .

Add $- 150 \sqrt[3]{{(x + 4)}^{2}}$ and $150 \sqrt[3]{{(x + 4)}^{2}}$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 492 + 0 + 60 \sqrt[3]{x + 4} - 120 \sqrt[3]{x + 4} + 24 + 60 \sqrt[3]{x + 4} - 24 - 492}{125}$

Add $125 x + 492$ and $0$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 492 + 60 \sqrt[3]{x + 4} - 120 \sqrt[3]{x + 4} + 24 + 60 \sqrt[3]{x + 4} - 24 - 492}{125}$

Subtract $24$ from $24$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 492 + 60 \sqrt[3]{x + 4} - 120 \sqrt[3]{x + 4} + 0 + 60 \sqrt[3]{x + 4} - 492}{125}$

Add $125 x + 492 + 60 \sqrt[3]{x + 4} - 120 \sqrt[3]{x + 4}$ and $0$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 492 + 60 \sqrt[3]{x + 4} - 120 \sqrt[3]{x + 4} + 60 \sqrt[3]{x + 4} - 492}{125}$

Subtract $492$ from $492$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 0 + 60 \sqrt[3]{x + 4} - 120 \sqrt[3]{x + 4} + 60 \sqrt[3]{x + 4}}{125}$

Add $125 x$ and $0$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 60 \sqrt[3]{x + 4} - 120 \sqrt[3]{x + 4} + 60 \sqrt[3]{x + 4}}{125}$

Subtract $120 \sqrt[3]{x + 4}$ from $60 \sqrt[3]{x + 4}$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x - 60 \sqrt[3]{x + 4} + 60 \sqrt[3]{x + 4}}{125}$

Combine the opposite terms in $125 x - 60 \sqrt[3]{x + 4} + 60 \sqrt[3]{x + 4}$ .

Add $- 60 \sqrt[3]{x + 4}$ and $60 \sqrt[3]{x + 4}$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x + 0}{125}$

Add $125 x$ and $0$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x}{125}$

Cancel the common factor of $125$ .

Cancel the common factor.

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = \frac{125 x}{125}$

Divide $x$ by $1$ .

$f^{- 1} (5 \sqrt[3]{x + 4} - 2) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}) = 5 \sqrt[3]{(\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}) + 4} - 2$

Simplify each term.

To write $4$ as a fraction with a common denominator, multiply by $\frac{125}{125}$ .

$f (\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}) = 5 \sqrt[3]{\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125} + 4 \cdot \frac{125}{125}} - 2$

Combine $4$ and $\frac{125}{125}$ .

Combine the numerators over the common denominator.

$f (\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}) = 5 \sqrt[3]{\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} + \frac{- 492 + 4 \cdot 125}{125}} - 2$

Simplify the numerator.

Multiply $4$ by $125$ .

$f (\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}) = 5 \sqrt[3]{\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} + \frac{- 492 + 500}{125}} - 2$

Add $- 492$ and $500$ .

$f (\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}) = 5 \sqrt[3]{\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} + \frac{8}{125}} - 2$

Combine the numerators over the common denominator.

$f (\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}) = 5 \sqrt[3]{\frac{x^{3} + 6 x^{2} + 12 x + 8}{125}} - 2$

Rewrite $\frac{x^{3} + 6 x^{2} + 12 x + 8}{125}$ as ${(\frac{1}{5})}^{3} (x^{3} + 6 x^{2} + 12 x + 8)$ .

Factor the perfect power $1^{3}$ out of $x^{3} + 6 x^{2} + 12 x + 8$ .

$f (\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}) = 5 \sqrt[3]{\frac{1^{3} (x^{3} + 6 x^{2} + 12 x + 8)}{125}} - 2$

Factor the perfect power $5^{3}$ out of $125$ .

$f (\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}) = 5 \sqrt[3]{\frac{1^{3} (x^{3} + 6 x^{2} + 12 x + 8)}{5^{3} \cdot 1}} - 2$

Rearrange the fraction $\frac{1^{3} (x^{3} + 6 x^{2} + 12 x + 8)}{5^{3} \cdot 1}$ .

$f (\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}) = 5 \sqrt[3]{{(\frac{1}{5})}^{3} (x^{3} + 6 x^{2} + 12 x + 8)} - 2$

Pull terms out from under the radical.

$f (\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}) = 5 (\frac{1}{5} \cdot \sqrt[3]{x^{3} + 6 x^{2} + 12 x + 8}) - 2$

Make each term match the terms from the binomial theorem formula.

$f (\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}) = 5 (\frac{1}{5} \cdot \sqrt[3]{x^{3} + 3 \cdot (2 x^{2}) + 3 \cdot (2^{2} x) + 2^{3}}) - 2$

Factor $x^{3} + 3 \cdot 2 x^{2} + 3 \cdot 2^{2} x + 2^{3}$ using the binomial theorem.

$f (\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}) = 5 (\frac{1}{5} \cdot \sqrt[3]{{(x + 2)}^{3}}) - 2$

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

$f (\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}) = 5 (\frac{1}{5} \cdot (x + 2)) - 2$

Apply the distributive property.

$f (\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}) = 5 (\frac{1}{5} x + \frac{1}{5} \cdot 2) - 2$

Combine $\frac{1}{5}$ and $x$ .

$f (\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}) = 5 (\frac{x}{5} + \frac{1}{5} \cdot 2) - 2$

Combine $\frac{1}{5}$ and $2$ .

$f (\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}) = 5 (\frac{x}{5} + \frac{2}{5}) - 2$

Apply the distributive property.

$f (\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}) = 5 (\frac{x}{5}) + 5 (\frac{2}{5}) - 2$

Cancel the common factor of $5$ .

Cancel the common factor.

$f (\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}) = 5 (\frac{x}{5}) + 5 (\frac{2}{5}) - 2$

Rewrite the expression.

$f (\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}) = x + 5 (\frac{2}{5}) - 2$

Cancel the common factor of $5$ .

Cancel the common factor.

$f (\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}) = x + 5 (\frac{2}{5}) - 2$

Rewrite the expression.

$f (\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}) = x + 2 - 2$

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

Subtract $2$ from $2$ .

$f (\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}) = x + 0$

Add $x$ and $0$ .

$f (\frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}$ is the inverse of $f (x) = 5 \sqrt[3]{x + 4} - 2$ .

$f^{- 1} (x) = \frac{x^{3}}{125} + \frac{6 x^{2}}{125} + \frac{12 x}{125} - \frac{492}{125}$

Find the Inverse f(x)=5 cube root of x+4-2

Related Questions

Support LCMGCF.com by disabling your adblocker.