Find the Inverse f(x)=-12x^3

$f (x) = - 12 x^{3}$

Step 1

Write $f (x) = - 12 x^{3}$ as an equation.

$y = - 12 x^{3}$

Step 2

Interchange the variables.

$x = - 12 y^{3}$

Step 3

Solve for $y$ .

Rewrite the equation as $- 12 y^{3} = x$ .

$- 12 y^{3} = x$

Divide each term in $- 12 y^{3} = x$ by $- 12$ and simplify.

Divide each term in $- 12 y^{3} = x$ by $- 12$ .

$\frac{- 12 y^{3}}{- 12} = \frac{x}{- 12}$

Simplify the left side.

Cancel the common factor of $- 12$ .

Cancel the common factor.

$\frac{- 12 y^{3}}{- 12} = \frac{x}{- 12}$

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

$y^{3} = \frac{x}{- 12}$

Simplify the right side.

Move the negative in front of the fraction.

$y^{3} = - \frac{x}{12}$

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

$y = \sqrt[3]{- \frac{x}{12}}$

Simplify $\sqrt[3]{- \frac{x}{12}}$ .

Rewrite $- \frac{x}{12}$ as ${({(- 1)}^{3})}^{3} \frac{x}{12}$ .

Rewrite $- 1$ as ${(- 1)}^{3}$ .

$y = \sqrt[3]{{(- 1)}^{3} \frac{x}{12}}$

Rewrite $- 1$ as ${(- 1)}^{3}$ .

$y = \sqrt[3]{{({(- 1)}^{3})}^{3} \frac{x}{12}}$

Pull terms out from under the radical.

$y = {(- 1)}^{3} \sqrt[3]{\frac{x}{12}}$

Raise $- 1$ to the power of $3$ .

$y = - \sqrt[3]{\frac{x}{12}}$

Rewrite $\sqrt[3]{\frac{x}{12}}$ as $\frac{\sqrt[3]{x}}{\sqrt[3]{12}}$ .

$y = - \frac{\sqrt[3]{x}}{\sqrt[3]{12}}$

Multiply $\frac{\sqrt[3]{x}}{\sqrt[3]{12}}$ by $\frac{{\sqrt[3]{12}}^{2}}{{\sqrt[3]{12}}^{2}}$ .

$y = - (\frac{\sqrt[3]{x}}{\sqrt[3]{12}} \cdot \frac{{\sqrt[3]{12}}^{2}}{{\sqrt[3]{12}}^{2}})$

Combine and simplify the denominator.

Multiply $\frac{\sqrt[3]{x}}{\sqrt[3]{12}}$ by $\frac{{\sqrt[3]{12}}^{2}}{{\sqrt[3]{12}}^{2}}$ .

$y = - \frac{\sqrt[3]{x} {\sqrt[3]{12}}^{2}}{\sqrt[3]{12} {\sqrt[3]{12}}^{2}}$

Raise $\sqrt[3]{12}$ to the power of $1$ .

$y = - \frac{\sqrt[3]{x} {\sqrt[3]{12}}^{2}}{{\sqrt[3]{12}}^{1} {\sqrt[3]{12}}^{2}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y = - \frac{\sqrt[3]{x} {\sqrt[3]{12}}^{2}}{{\sqrt[3]{12}}^{1 + 2}}$

Add $1$ and $2$ .

$y = - \frac{\sqrt[3]{x} {\sqrt[3]{12}}^{2}}{{\sqrt[3]{12}}^{3}}$

Rewrite ${\sqrt[3]{12}}^{3}$ as $12$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{12}$ as $12^{\frac{1}{3}}$ .

$y = - \frac{\sqrt[3]{x} {\sqrt[3]{12}}^{2}}{{(12^{\frac{1}{3}})}^{3}}$

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

$y = - \frac{\sqrt[3]{x} {\sqrt[3]{12}}^{2}}{12^{\frac{1}{3} \cdot 3}}$

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

$y = - \frac{\sqrt[3]{x} {\sqrt[3]{12}}^{2}}{12^{\frac{3}{3}}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$y = - \frac{\sqrt[3]{x} {\sqrt[3]{12}}^{2}}{12^{\frac{3}{3}}}$

Rewrite the expression.

$y = - \frac{\sqrt[3]{x} {\sqrt[3]{12}}^{2}}{12^{1}}$

Evaluate the exponent.

$y = - \frac{\sqrt[3]{x} {\sqrt[3]{12}}^{2}}{12}$

Simplify the numerator.

Rewrite ${\sqrt[3]{12}}^{2}$ as $\sqrt[3]{12^{2}}$ .

$y = - \frac{\sqrt[3]{x} \sqrt[3]{12^{2}}}{12}$

Raise $12$ to the power of $2$ .

$y = - \frac{\sqrt[3]{x} \sqrt[3]{144}}{12}$

Rewrite $144$ as $2^{3} \cdot 18$ .

Factor $8$ out of $144$ .

$y = - \frac{\sqrt[3]{x} \sqrt[3]{8 (18)}}{12}$

Rewrite $8$ as $2^{3}$ .

$y = - \frac{\sqrt[3]{x} \sqrt[3]{2^{3} \cdot 18}}{12}$

Pull terms out from under the radical.

$y = - \frac{\sqrt[3]{x} \cdot 2 \sqrt[3]{18}}{12}$

Combine using the product rule for radicals.

$y = - \frac{2 \sqrt[3]{x \cdot 18}}{12}$

Reduce the expression by cancelling the common factors.

Cancel the common factor of $2$ and $12$ .

Factor $2$ out of $2 \sqrt[3]{x \cdot 18}$ .

$y = - \frac{2 (\sqrt[3]{x \cdot 18})}{12}$

Cancel the common factors.

Factor $2$ out of $12$ .

$y = - \frac{2 \sqrt[3]{x \cdot 18}}{2 \cdot 6}$

Cancel the common factor.

$y = - \frac{2 \sqrt[3]{x \cdot 18}}{2 \cdot 6}$

Rewrite the expression.

$y = - \frac{\sqrt[3]{x \cdot 18}}{6}$

Reorder factors in $- \frac{\sqrt[3]{x \cdot 18}}{6}$ .

$y = - \frac{\sqrt[3]{18 x}}{6}$

Step 4

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

$f^{- 1} (x) = - \frac{\sqrt[3]{18 x}}{6}$

Step 5

Verify if $f^{- 1} (x) = - \frac{\sqrt[3]{18 x}}{6}$ is the inverse of $f (x) = - 12 x^{3}$ .

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 $f^{- 1} (- 12 x^{3})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (- 12 x^{3}) = - \frac{\sqrt[3]{18 (- 12 x^{3})}}{6}$

Simplify the numerator.

Multiply $- 12$ by $18$ .

$f^{- 1} (- 12 x^{3}) = - \frac{\sqrt[3]{- 216 x^{3}}}{6}$

Rewrite $- 216 x^{3}$ as ${(- 6 x)}^{3}$ .

$f^{- 1} (- 12 x^{3}) = - \frac{\sqrt[3]{{(- 6 x)}^{3}}}{6}$

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

$f^{- 1} (- 12 x^{3}) = - \frac{- 6 x}{6}$

Cancel the common factor of $- 6$ and $6$ .

Factor $6$ out of $- 6 x$ .

$f^{- 1} (- 12 x^{3}) = - \frac{6 (- x)}{6}$

Cancel the common factors.

Factor $6$ out of $6$ .

$f^{- 1} (- 12 x^{3}) = - \frac{6 (- x)}{6 (1)}$

Cancel the common factor.

$f^{- 1} (- 12 x^{3}) = - \frac{6 (- x)}{6 \cdot 1}$

Rewrite the expression.

$f^{- 1} (- 12 x^{3}) = - \frac{- x}{1}$

Divide $- x$ by $1$ .

$f^{- 1} (- 12 x^{3}) = x$

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

Set up the composite result function.

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

Evaluate $f (- \frac{\sqrt[3]{18 x}}{6})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (- \frac{\sqrt[3]{18 x}}{6}) = - 12 {(- \frac{\sqrt[3]{18 x}}{6})}^{3}$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{\sqrt[3]{18 x}}{6}$ .

$f (- \frac{\sqrt[3]{18 x}}{6}) = - 12 ({(- 1)}^{3} {(\frac{\sqrt[3]{18 x}}{6})}^{3})$

Apply the product rule to $\frac{\sqrt[3]{18 x}}{6}$ .

$f (- \frac{\sqrt[3]{18 x}}{6}) = - 12 ({(- 1)}^{3} (\frac{{\sqrt[3]{18 x}}^{3}}{6^{3}}))$

Raise $- 1$ to the power of $3$ .

$f (- \frac{\sqrt[3]{18 x}}{6}) = - 12 (- \frac{{\sqrt[3]{18 x}}^{3}}{6^{3}})$

Rewrite ${\sqrt[3]{18 x}}^{3}$ as $18 x$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{18 x}$ as ${(18 x)}^{\frac{1}{3}}$ .

$f (- \frac{\sqrt[3]{18 x}}{6}) = - 12 (- \frac{{({(18 x)}^{\frac{1}{3}})}^{3}}{6^{3}})$

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

$f (- \frac{\sqrt[3]{18 x}}{6}) = - 12 (- \frac{{(18 x)}^{\frac{1}{3} \cdot 3}}{6^{3}})$

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

$f (- \frac{\sqrt[3]{18 x}}{6}) = - 12 (- \frac{{(18 x)}^{\frac{3}{3}}}{6^{3}})$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (- \frac{\sqrt[3]{18 x}}{6}) = - 12 (- \frac{{(18 x)}^{\frac{3}{3}}}{6^{3}})$

Rewrite the expression.

$f (- \frac{\sqrt[3]{18 x}}{6}) = - 12 (- \frac{18 x}{6^{3}})$

Simplify.

$f (- \frac{\sqrt[3]{18 x}}{6}) = - 12 (- \frac{18 x}{6^{3}})$

Raise $6$ to the power of $3$ .

$f (- \frac{\sqrt[3]{18 x}}{6}) = - 12 (- \frac{18 x}{216})$

Cancel the common factor of $12$ .

Move the leading negative in $- \frac{18 x}{216}$ into the numerator.

$f (- \frac{\sqrt[3]{18 x}}{6}) = - 12 \frac{- 18 x}{216}$

Factor $12$ out of $- 12$ .

$f (- \frac{\sqrt[3]{18 x}}{6}) = 12 (- 1) (\frac{- 18 x}{216})$

Factor $12$ out of $216$ .

$f (- \frac{\sqrt[3]{18 x}}{6}) = 12 \cdot (- 1 \frac{- 18 x}{12 \cdot 18})$

Cancel the common factor.

$f (- \frac{\sqrt[3]{18 x}}{6}) = 12 \cdot (- 1 \frac{- 18 x}{12 \cdot 18})$

Rewrite the expression.

$f (- \frac{\sqrt[3]{18 x}}{6}) = - 1 \frac{- 18 x}{18}$

Cancel the common factor of $- 18$ and $18$ .

Factor $18$ out of $- 18 x$ .

$f (- \frac{\sqrt[3]{18 x}}{6}) = - 1 \frac{18 (- x)}{18}$

Cancel the common factors.

Factor $18$ out of $18$ .

$f (- \frac{\sqrt[3]{18 x}}{6}) = - 1 \frac{18 (- x)}{18 (1)}$

Cancel the common factor.

$f (- \frac{\sqrt[3]{18 x}}{6}) = - 1 \frac{18 (- x)}{18 \cdot 1}$

Rewrite the expression.

$f (- \frac{\sqrt[3]{18 x}}{6}) = - 1 \frac{- x}{1}$

Divide $- x$ by $1$ .

$f (- \frac{\sqrt[3]{18 x}}{6}) = - 1 (- x)$

Multiply $- 1 (- x)$ .

Multiply $- 1$ by $- 1$ .

$f (- \frac{\sqrt[3]{18 x}}{6}) = 1 x$

Multiply $x$ by $1$ .

$f (- \frac{\sqrt[3]{18 x}}{6}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = - \frac{\sqrt[3]{18 x}}{6}$ is the inverse of $f (x) = - 12 x^{3}$ .

$f^{- 1} (x) = - \frac{\sqrt[3]{18 x}}{6}$

Find the Inverse f(x)=-12x^3

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