Find the Inverse y=2(x-1)^3

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

Rewrite the equation as $2 {(y - 1)}^{3} = x$ .

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

Divide each term in $2 {(y - 1)}^{3} = x$ by $2$ and simplify.

Divide each term in $2 {(y - 1)}^{3} = x$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide ${(y - 1)}^{3}$ by $1$ .

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

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

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

Simplify $\sqrt[3]{\frac{x}{2}}$ .

Rewrite $\sqrt[3]{\frac{x}{2}}$ as $\frac{\sqrt[3]{x}}{\sqrt[3]{2}}$ .

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

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

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

Combine and simplify the denominator.

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

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

Raise $\sqrt[3]{2}$ to the power of $1$ .

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

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

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

Add $1$ and $2$ .

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

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

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

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

Simplify the numerator.

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

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

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

$y - 1 = \frac{\sqrt[3]{x} \sqrt[3]{4}}{2}$

Simplify with factoring out.

Combine using the product rule for radicals.

$y - 1 = \frac{\sqrt[3]{x \cdot 4}}{2}$

Reorder factors in $\frac{\sqrt[3]{x \cdot 4}}{2}$ .

$y - 1 = \frac{\sqrt[3]{4 x}}{2}$

Add $1$ to both sides of the equation.

$y = \frac{\sqrt[3]{4 x}}{2} + 1$

Step 3

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

$f^{- 1} (x) = \frac{\sqrt[3]{4 x}}{2} + 1$

Step 4

Verify if $f^{- 1} (x) = \frac{\sqrt[3]{4 x}}{2} + 1$ is the inverse of $f (x) = 2 {(x - 1)}^{3}$ .

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 $f^{- 1} (2 {(x - 1)}^{3})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (2 {(x - 1)}^{3}) = \frac{\sqrt[3]{4 (2 {(x - 1)}^{3})}}{2} + 1$

Simplify each term.

Simplify the numerator.

Multiply $4$ by $2$ .

$f^{- 1} (2 {(x - 1)}^{3}) = \frac{\sqrt[3]{8 {(x - 1)}^{3}}}{2} + 1$

Rewrite $8 {(x - 1)}^{3}$ as ${(2 (x - 1))}^{3}$ .

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

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

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

Apply the distributive property.

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

Multiply $2$ by $- 1$ .

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

Factor $2$ out of $2 x - 2$ .

Factor $2$ out of $2 x$ .

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

Factor $2$ out of $- 2$ .

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

Factor $2$ out of $2 (x) + 2 (- 1)$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x - 1$ by $1$ .

$f^{- 1} (2 {(x - 1)}^{3}) = x - 1 + 1$

Combine the opposite terms in $x - 1 + 1$ .

Add $- 1$ and $1$ .

$f^{- 1} (2 {(x - 1)}^{3}) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (2 {(x - 1)}^{3}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{\sqrt[3]{4 x}}{2} + 1)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\sqrt[3]{4 x}}{2} + 1) = 2 {((\frac{\sqrt[3]{4 x}}{2} + 1) - 1)}^{3}$

Combine the opposite terms in $(\frac{\sqrt[3]{4 x}}{2} + 1) - 1$ .

Subtract $1$ from $1$ .

$f (\frac{\sqrt[3]{4 x}}{2} + 1) = 2 {(\frac{\sqrt[3]{4 x}}{2} + 0)}^{3}$

Add $\frac{\sqrt[3]{4 x}}{2}$ and $0$ .

$f (\frac{\sqrt[3]{4 x}}{2} + 1) = 2 {(\frac{\sqrt[3]{4 x}}{2})}^{3}$

Apply the product rule to $\frac{\sqrt[3]{4 x}}{2}$ .

$f (\frac{\sqrt[3]{4 x}}{2} + 1) = 2 (\frac{{\sqrt[3]{4 x}}^{3}}{2^{3}})$

Rewrite ${\sqrt[3]{4 x}}^{3}$ as $4 x$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{4 x}$ as ${(4 x)}^{\frac{1}{3}}$ .

$f (\frac{\sqrt[3]{4 x}}{2} + 1) = 2 (\frac{{({(4 x)}^{\frac{1}{3}})}^{3}}{2^{3}})$

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

$f (\frac{\sqrt[3]{4 x}}{2} + 1) = 2 (\frac{{(4 x)}^{\frac{1}{3} \cdot 3}}{2^{3}})$

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

$f (\frac{\sqrt[3]{4 x}}{2} + 1) = 2 (\frac{{(4 x)}^{\frac{3}{3}}}{2^{3}})$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (\frac{\sqrt[3]{4 x}}{2} + 1) = 2 (\frac{{(4 x)}^{\frac{3}{3}}}{2^{3}})$

Rewrite the expression.

$f (\frac{\sqrt[3]{4 x}}{2} + 1) = 2 (\frac{4 x}{2^{3}})$

Simplify.

$f (\frac{\sqrt[3]{4 x}}{2} + 1) = 2 (\frac{4 x}{2^{3}})$

Raise $2$ to the power of $3$ .

$f (\frac{\sqrt[3]{4 x}}{2} + 1) = 2 (\frac{4 x}{8})$

Cancel the common factor of $2$ .

Factor $2$ out of $8$ .

$f (\frac{\sqrt[3]{4 x}}{2} + 1) = 2 (\frac{4 x}{2 (4)})$

Cancel the common factor.

$f (\frac{\sqrt[3]{4 x}}{2} + 1) = 2 (\frac{4 x}{2 \cdot 4})$

Rewrite the expression.

$f (\frac{\sqrt[3]{4 x}}{2} + 1) = \frac{4 x}{4}$

Cancel the common factor of $4$ .

Cancel the common factor.

$f (\frac{\sqrt[3]{4 x}}{2} + 1) = \frac{4 x}{4}$

Divide $x$ by $1$ .

$f (\frac{\sqrt[3]{4 x}}{2} + 1) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\sqrt[3]{4 x}}{2} + 1$ is the inverse of $f (x) = 2 {(x - 1)}^{3}$ .

$f^{- 1} (x) = \frac{\sqrt[3]{4 x}}{2} + 1$

Find the Inverse y=2(x-1)^3

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