Find the Inverse y=1/3(x+7)^3-5

$y = \frac{1}{3} {(x + 7)}^{3} - 5$

Interchange the variables.

$x = \frac{1}{3} \cdot {(y + 7)}^{3} - 5$

Solve for $y$ .

Rewrite the equation as $\frac{1}{3} \cdot {(y + 7)}^{3} - 5 = x$ .

$\frac{1}{3} \cdot {(y + 7)}^{3} - 5 = x$

Add $5$ to both sides of the equation.

$\frac{1}{3} \cdot {(y + 7)}^{3} = x + 5$

Combine $\frac{1}{3}$ and ${(y + 7)}^{3}$ .

$\frac{{(y + 7)}^{3}}{3} = x + 5$

Multiply both sides of the equation by $3$ .

$3 \frac{{(y + 7)}^{3}}{3} = 3 (x + 5)$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$3 \frac{{(y + 7)}^{3}}{3} = 3 (x + 5)$

Rewrite the expression.

${(y + 7)}^{3} = 3 (x + 5)$

Simplify the right side.

Simplify $3 (x + 5)$ .

Apply the distributive property.

${(y + 7)}^{3} = 3 x + 3 \cdot 5$

Multiply $3$ by $5$ .

${(y + 7)}^{3} = 3 x + 15$

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

$y + 7 = \sqrt[3]{3 x + 15}$

Simplify $\sqrt[3]{3 x + 15}$ .

Rewrite.

$y + 7 = 0 + 0 + \sqrt[3]{3 x + 15}$

Simplify by adding zeros.

$y + 7 = \sqrt[3]{3 x + 15}$

Factor $3$ out of $3 x + 15$ .

Factor $3$ out of $3 x$ .

$y + 7 = \sqrt[3]{3 (x) + 15}$

Factor $3$ out of $15$ .

$y + 7 = \sqrt[3]{3 x + 3 \cdot 5}$

Factor $3$ out of $3 x + 3 \cdot 5$ .

$y + 7 = \sqrt[3]{3 (x + 5)}$

Subtract $7$ from both sides of the equation.

$y = \sqrt[3]{3 (x + 5)} - 7$

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

$f^{- 1} (x) = \sqrt[3]{3 (x + 5)} - 7$

Verify if $f^{- 1} (x) = \sqrt[3]{3 (x + 5)} - 7$ is the inverse of $f (x) = \frac{1}{3} \cdot {(x + 7)}^{3} - 5$ .

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 $f^{- 1} (\frac{1}{3} \cdot {(x + 7)}^{3} - 5)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{1}{3} \cdot {(x + 7)}^{3} - 5) = \sqrt[3]{3 ((\frac{1}{3} \cdot {(x + 7)}^{3} - 5) + 5)} - 7$

Combine the opposite terms in $\sqrt[3]{3 ((\frac{1}{3} \cdot {(x + 7)}^{3} - 5) + 5)} - 7$ .

Add $- 5$ and $5$ .

$f^{- 1} (\frac{1}{3} \cdot {(x + 7)}^{3} - 5) = \sqrt[3]{3 (\frac{1}{3} \cdot {(x + 7)}^{3} + 0)} - 7$

Add $\frac{1}{3} \cdot {(x + 7)}^{3}$ and $0$ .

$f^{- 1} (\frac{1}{3} \cdot {(x + 7)}^{3} - 5) = \sqrt[3]{3 (\frac{1}{3} \cdot {(x + 7)}^{3})} - 7$

Simplify each term.

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

$f^{- 1} (\frac{1}{3} \cdot {(x + 7)}^{3} - 5) = \sqrt[3]{\frac{3}{3} \cdot {(x + 7)}^{3}} - 7$

Combine $\frac{3}{3}$ and ${(x + 7)}^{3}$ .

$f^{- 1} (\frac{1}{3} \cdot {(x + 7)}^{3} - 5) = \sqrt[3]{\frac{3 {(x + 7)}^{3}}{3}} - 7$

Reduce the expression by cancelling the common factors.

Reduce the expression $\frac{3 {(x + 7)}^{3}}{3}$ by cancelling the common factors.

Cancel the common factor.

$f^{- 1} (\frac{1}{3} \cdot {(x + 7)}^{3} - 5) = \sqrt[3]{\frac{3 {(x + 7)}^{3}}{3}} - 7$

Rewrite the expression.

$f^{- 1} (\frac{1}{3} \cdot {(x + 7)}^{3} - 5) = \sqrt[3]{\frac{{(x + 7)}^{3}}{1}} - 7$

Divide ${(x + 7)}^{3}$ by $1$ .

$f^{- 1} (\frac{1}{3} \cdot {(x + 7)}^{3} - 5) = \sqrt[3]{{(x + 7)}^{3}} - 7$

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

$f^{- 1} (\frac{1}{3} \cdot {(x + 7)}^{3} - 5) = x + 7 - 7$

Combine the opposite terms in $x + 7 - 7$ .

Subtract $7$ from $7$ .

$f^{- 1} (\frac{1}{3} \cdot {(x + 7)}^{3} - 5) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (\frac{1}{3} \cdot {(x + 7)}^{3} - 5) = x$

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

Set up the composite result function.

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

Evaluate $f (\sqrt[3]{3 (x + 5)} - 7)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\sqrt[3]{3 (x + 5)} - 7) = \frac{1}{3} \cdot {((\sqrt[3]{3 (x + 5)} - 7) + 7)}^{3} - 5$

Combine the opposite terms in $\frac{1}{3} \cdot {((\sqrt[3]{3 (x + 5)} - 7) + 7)}^{3} - 5$ .

Add $- 7$ and $7$ .

$f (\sqrt[3]{3 (x + 5)} - 7) = \frac{1}{3} \cdot {(\sqrt[3]{3 (x + 5)} + 0)}^{3} - 5$

Add $\sqrt[3]{3 (x + 5)}$ and $0$ .

$f (\sqrt[3]{3 (x + 5)} - 7) = \frac{1}{3} \cdot {\sqrt[3]{3 (x + 5)}}^{3} - 5$

Simplify each term.

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

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

$f (\sqrt[3]{3 (x + 5)} - 7) = \frac{1}{3} \cdot {({(3 (x + 5))}^{\frac{1}{3}})}^{3} - 5$

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

$f (\sqrt[3]{3 (x + 5)} - 7) = \frac{1}{3} \cdot {(3 (x + 5))}^{\frac{1}{3} \cdot 3} - 5$

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

$f (\sqrt[3]{3 (x + 5)} - 7) = \frac{1}{3} \cdot {(3 (x + 5))}^{\frac{3}{3}} - 5$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (\sqrt[3]{3 (x + 5)} - 7) = \frac{1}{3} \cdot {(3 (x + 5))}^{\frac{3}{3}} - 5$

Rewrite the expression.

$f (\sqrt[3]{3 (x + 5)} - 7) = \frac{1}{3} \cdot (3 (x + 5)) - 5$

Simplify.

$f (\sqrt[3]{3 (x + 5)} - 7) = \frac{1}{3} \cdot (3 (x + 5)) - 5$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (\sqrt[3]{3 (x + 5)} - 7) = \frac{1}{3} \cdot (3 (x + 5)) - 5$

Rewrite the expression.

$f (\sqrt[3]{3 (x + 5)} - 7) = x + 5 - 5$

Combine the opposite terms in $x + 5 - 5$ .

Subtract $5$ from $5$ .

$f (\sqrt[3]{3 (x + 5)} - 7) = x + 0$

Add $x$ and $0$ .

$f (\sqrt[3]{3 (x + 5)} - 7) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \sqrt[3]{3 (x + 5)} - 7$ is the inverse of $f (x) = \frac{1}{3} \cdot {(x + 7)}^{3} - 5$ .

$f^{- 1} (x) = \sqrt[3]{3 (x + 5)} - 7$

Find the Inverse y=1/3(x+7)^3-5

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