Find the Inverse f(x) = square root of x^3+5

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

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

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

Interchange the variables.

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

Solve for $y$ .

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

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

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

${\sqrt{y^{3} + 5}}^{2} = x^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{y^{3} + 5}$ as ${(y^{3} + 5)}^{\frac{1}{2}}$ .

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

Simplify the left side.

Simplify ${({(y^{3} + 5)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(y^{3} + 5)}^{\frac{1}{2}})}^{2}$ .

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

$y^{3} + 5 = x^{2}$

Solve for $y$ .

Subtract $5$ from both sides of the equation.

$y^{3} = x^{2} - 5$

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

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

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

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

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

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

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

Rewrite ${\sqrt{x^{3} + 5}}^{2}$ as $x^{3} + 5$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x^{3} + 5}$ as ${(x^{3} + 5)}^{\frac{1}{2}}$ .

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify by subtracting numbers.

Subtract $5$ from $5$ .

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

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

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

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

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

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

Set up the composite result function.

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

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

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

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

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x^{2} - 5}$ as ${(x^{2} - 5)}^{\frac{1}{3}}$ .

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify by adding numbers.

Add $- 5$ and $5$ .

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

Add $x^{2}$ and $0$ .

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

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

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

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

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

Find the Inverse f(x) = square root of x^3+5

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