Find the Inverse -x^3+4

$- x^{3} + 4$

Step 1

Interchange the variables.

$x = - y^{3} + 4$

Step 2

Solve for $y$ .

Rewrite the equation as $- y^{3} + 4 = x$ .

$- y^{3} + 4 = x$

Subtract $4$ from both sides of the equation.

$- y^{3} = x - 4$

Divide each term in $- y^{3} = x - 4$ by $- 1$ and simplify.

Divide each term in $- y^{3} = x - 4$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

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

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

Simplify the right side.

Simplify each term.

Move the negative one from the denominator of $\frac{x}{- 1}$ .

$y^{3} = - 1 \cdot x + \frac{- 4}{- 1}$

Rewrite $- 1 \cdot x$ as $- x$ .

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

Divide $- 4$ by $- 1$ .

$y^{3} = - x + 4$

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

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

Step 3

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

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

Step 4

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

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 $f^{- 1} (- x^{3} + 4)$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Apply the distributive property.

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

Multiply $- 1$ by $4$ .

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

Add $- 4$ and $4$ .

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

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

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

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

$f^{- 1} (- x^{3} + 4) = x$

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

Set up the composite result function.

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

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

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

Simplify each term.

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 ${(- x + 4)}^{\frac{1}{3}}$ .

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Apply the distributive property.

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

Multiply $- - x$ .

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

Multiply $- 1$ by $4$ .

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

Combine the opposite terms in $x - 4 + 4$ .

Add $- 4$ and $4$ .

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

Add $x$ and $0$ .

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

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

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

Find the Inverse -x^3+4

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