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

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

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

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

Interchange the variables.

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

Solve for $y$ .

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

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

Subtract $3$ from both sides of the equation.

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

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

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

Multiply both sides of the equation by $2$ .

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

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the right side.

Simplify $2 (x - 3)$ .

Apply the distributive property.

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

Multiply $2$ by $- 3$ .

${(y - 1)}^{3} = 2 x - 6$

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

$y - 1 = \sqrt[3]{2 x - 6}$

Simplify $\sqrt[3]{2 x - 6}$ .

Rewrite.

$y - 1 = 0 + 0 + \sqrt[3]{2 x - 6}$

Simplify by adding zeros.

$y - 1 = \sqrt[3]{2 x - 6}$

Factor $2$ out of $2 x - 6$ .

Factor $2$ out of $2 x$ .

$y - 1 = \sqrt[3]{2 (x) - 6}$

Factor $2$ out of $- 6$ .

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

Factor $2$ out of $2 x + 2 \cdot - 3$ .

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

Add $1$ to both sides of the equation.

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

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

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

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

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

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

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

Subtract $3$ from $3$ .

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

Add $\frac{1}{2} \cdot {(x - 1)}^{3}$ and $0$ .

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

Simplify each term.

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

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

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

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

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.

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

Rewrite the expression.

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

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

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

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

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

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

Add $- 1$ and $1$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

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

Subtract $1$ from $1$ .

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

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

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

Simplify each term.

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

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Combine the opposite terms in $x - 3 + 3$ .

Add $- 3$ and $3$ .

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

Add $x$ and $0$ .

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

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

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

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

Related Questions

Support LCMGCF.com by disabling your adblocker.