Find the Inverse f(x)=(x/3)^3+6

$f (x) = {(\frac{x}{3})}^{3} + 6$

Step 1

Write $f (x) = {(\frac{x}{3})}^{3} + 6$ as an equation.

$y = {(\frac{x}{3})}^{3} + 6$

Step 2

Interchange the variables.

$x = {(\frac{y}{3})}^{3} + 6$

Step 3

Solve for $y$ .

Rewrite the equation as ${(\frac{y}{3})}^{3} + 6 = x$ .

${(\frac{y}{3})}^{3} + 6 = x$

Simplify each term.

Apply the product rule to $\frac{y}{3}$ .

$\frac{y^{3}}{3^{3}} + 6 = x$

Raise $3$ to the power of $3$ .

$\frac{y^{3}}{27} + 6 = x$

Subtract $6$ from both sides of the equation.

$\frac{y^{3}}{27} = x - 6$

Multiply both sides of the equation by $27$ .

$27 \frac{y^{3}}{27} = 27 (x - 6)$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $27$ .

Cancel the common factor.

$27 \frac{y^{3}}{27} = 27 (x - 6)$

Rewrite the expression.

$y^{3} = 27 (x - 6)$

Simplify the right side.

Simplify $27 (x - 6)$ .

Apply the distributive property.

$y^{3} = 27 x + 27 \cdot - 6$

Multiply $27$ by $- 6$ .

$y^{3} = 27 x - 162$

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

$y = \sqrt[3]{27 x - 162}$

Simplify $\sqrt[3]{27 x - 162}$ .

Factor $27$ out of $27 x - 162$ .

Factor $27$ out of $27 x$ .

$y = \sqrt[3]{27 (x) - 162}$

Factor $27$ out of $- 162$ .

$y = \sqrt[3]{27 x + 27 \cdot - 6}$

Factor $27$ out of $27 x + 27 \cdot - 6$ .

$y = \sqrt[3]{27 (x - 6)}$

Rewrite $27$ as $3^{3}$ .

$y = \sqrt[3]{3^{3} (x - 6)}$

Pull terms out from under the radical.

$y = 3 \sqrt[3]{x - 6}$

Step 4

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

$f^{- 1} (x) = 3 \sqrt[3]{x - 6}$

Step 5

Verify if $f^{- 1} (x) = 3 \sqrt[3]{x - 6}$ is the inverse of $f (x) = {(\frac{x}{3})}^{3} + 6$ .

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 $f^{- 1} ({(\frac{x}{3})}^{3} + 6)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} ({(\frac{x}{3})}^{3} + 6) = 3 \sqrt[3]{({(\frac{x}{3})}^{3} + 6) - 6}$

Apply the product rule to $\frac{x}{3}$ .

$f^{- 1} ({(\frac{x}{3})}^{3} + 6) = 3 \sqrt[3]{\frac{x^{3}}{3^{3}} + 6 - 6}$

Raise $3$ to the power of $3$ .

$f^{- 1} ({(\frac{x}{3})}^{3} + 6) = 3 \sqrt[3]{\frac{x^{3}}{27} + 6 - 6}$

Subtract $6$ from $6$ .

$f^{- 1} ({(\frac{x}{3})}^{3} + 6) = 3 \sqrt[3]{\frac{x^{3}}{27} + 0}$

Add $\frac{x^{3}}{27}$ and $0$ .

$f^{- 1} ({(\frac{x}{3})}^{3} + 6) = 3 \sqrt[3]{\frac{x^{3}}{27}}$

Rewrite $27$ as $3^{3}$ .

$f^{- 1} ({(\frac{x}{3})}^{3} + 6) = 3 \sqrt[3]{\frac{x^{3}}{3^{3}}}$

Rewrite $\frac{x^{3}}{3^{3}}$ as ${(\frac{x}{3})}^{3}$ .

$f^{- 1} ({(\frac{x}{3})}^{3} + 6) = 3 \sqrt[3]{{(\frac{x}{3})}^{3}}$

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

$f^{- 1} ({(\frac{x}{3})}^{3} + 6) = 3 (\frac{x}{3})$

Cancel the common factor of $3$ .

Cancel the common factor.

$f^{- 1} ({(\frac{x}{3})}^{3} + 6) = 3 (\frac{x}{3})$

Rewrite the expression.

$f^{- 1} ({(\frac{x}{3})}^{3} + 6) = x$

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

Set up the composite result function.

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

Evaluate $f (3 \sqrt[3]{x - 6})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (3 \sqrt[3]{x - 6}) = {(\frac{3 \sqrt[3]{x - 6}}{3})}^{3} + 6$

Simplify each term.

Cancel the common factor of $3$ .

Cancel the common factor.

$f (3 \sqrt[3]{x - 6}) = {(\frac{3 \sqrt[3]{x - 6}}{3})}^{3} + 6$

Divide $\sqrt[3]{x - 6}$ by $1$ .

$f (3 \sqrt[3]{x - 6}) = {\sqrt[3]{x - 6}}^{3} + 6$

Rewrite ${\sqrt[3]{x - 6}}^{3}$ as $x - 6$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x - 6}$ as ${(x - 6)}^{\frac{1}{3}}$ .

$f (3 \sqrt[3]{x - 6}) = {({(x - 6)}^{\frac{1}{3}})}^{3} + 6$

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

$f (3 \sqrt[3]{x - 6}) = {(x - 6)}^{\frac{1}{3} \cdot 3} + 6$

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

$f (3 \sqrt[3]{x - 6}) = {(x - 6)}^{\frac{3}{3}} + 6$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (3 \sqrt[3]{x - 6}) = {(x - 6)}^{\frac{3}{3}} + 6$

Rewrite the expression.

$f (3 \sqrt[3]{x - 6}) = (x - 6) + 6$

Simplify.

$f (3 \sqrt[3]{x - 6}) = x - 6 + 6$

Combine the opposite terms in $x - 6 + 6$ .

Add $- 6$ and $6$ .

$f (3 \sqrt[3]{x - 6}) = x + 0$

Add $x$ and $0$ .

$f (3 \sqrt[3]{x - 6}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = 3 \sqrt[3]{x - 6}$ is the inverse of $f (x) = {(\frac{x}{3})}^{3} + 6$ .

$f^{- 1} (x) = 3 \sqrt[3]{x - 6}$

Find the Inverse f(x)=(x/3)^3+6

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