Find the Inverse f(x)=(x^5+3)^(1/3)

$f (x) = {(x^{5} + 3)}^{\frac{1}{3}}$

Step 1

Write $f (x) = {(x^{5} + 3)}^{\frac{1}{3}}$ as an equation.

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

Rewrite the equation as ${(y^{5} + 3)}^{\frac{1}{3}} = x$ .

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

Raise each side of the equation to the power of $3$ to eliminate the fractional exponent on the left side.

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

Simplify the left side.

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

Multiply the exponents in ${({(y^{5} + 3)}^{\frac{1}{3}})}^{3}$ .

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

${(y^{5} + 3)}^{1} = x^{3}$

Simplify.

$y^{5} + 3 = x^{3}$

Solve for $y$ .

Subtract $3$ from both sides of the equation.

$y^{5} = x^{3} - 3$

Take the 5th root of both sides of the equation to eliminate the exponent on the left side.

$y = \sqrt[5]{x^{3} - 3}$

Step 4

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

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

Step 5

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

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 $f^{- 1} ({(x^{5} + 3)}^{\frac{1}{3}})$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Multiply the exponents in ${({(x^{5} + 3)}^{\frac{1}{3}})}^{3}$ .

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Subtract $3$ from $3$ .

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

Add $x^{5}$ and $0$ .

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

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

$f^{- 1} ({(x^{5} + 3)}^{\frac{1}{3}}) = x$

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

Set up the composite result function.

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

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

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

Rewrite ${\sqrt[5]{x^{3} - 3}}^{5}$ as $x^{3} - 3$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[5]{x^{3} - 3}$ as ${(x^{3} - 3)}^{\frac{1}{5}}$ .

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

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

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

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

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

Cancel the common factor of $5$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify by adding terms.

Combine the opposite terms in $x^{3} - 3 + 3$ .

Add $- 3$ and $3$ .

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

Add $x^{3}$ and $0$ .

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Find the Inverse f(x)=(x^5+3)^(1/3)

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