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

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

Step 1

Write $f (x) = \frac{1}{2} {(x - 1)}^{3}$ as an equation.

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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

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

Multiply both sides of the equation by $2$ .

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

Simplify the left side.

Simplify $2 (\frac{1}{2} \cdot {(y - 1)}^{3})$ .

Combine $\frac{1}{2}$ and ${(y - 1)}^{3}$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Add $1$ to both sides of the equation.

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

Step 4

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

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

Step 5

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

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

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

Simplify each term.

Combine $2$ and $\frac{1}{2}$ .

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

Combine $\frac{2}{2}$ and ${(x - 1)}^{3}$ .

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

Reduce the expression by cancelling the common factors.

Reduce the expression $\frac{2 {(x - 1)}^{3}}{2}$ by cancelling the common factors.

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

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

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

Add $- 1$ and $1$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

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

Subtract $1$ from $1$ .

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

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

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

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

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Cancel the common factor of $2$ .

Factor $2$ out of $2 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

Related Questions

Support LCMGCF.com by disabling your adblocker.