Find the Inverse f(x) = cube root of x^5-3

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

Step 1

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

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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

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

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

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

Simplify each side of the equation.

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

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

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 $y$ .

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

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

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

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

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify by adding numbers.

Add $- 3$ and $3$ .

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

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

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

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

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

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 $f$ .

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

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

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

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

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

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

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

Cancel the common factor of $5$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify by subtracting numbers.

Subtract $3$ from $3$ .

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

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

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

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

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

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

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

Find the Inverse f(x) = cube root of x^5-3

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