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

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

Step 1

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

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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

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

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

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

Subtract $\frac{1}{2}$ from both sides of the equation.

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

Multiply both sides of the equation by $2$ .

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

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the right side.

Simplify $2 (x - \frac{1}{2})$ .

Apply the distributive property.

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

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

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

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

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

Apply the distributive property.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify by subtracting numbers.

Subtract $1$ from $1$ .

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

Add $x^{3}$ and $0$ .

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

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

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

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 $f$ .

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

Simplify each term.

Rewrite ${\sqrt[3]{2 x - 1}}^{3}$ as $2 x - 1$ .

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Apply the distributive property.

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

Cancel the common factor of $2$ .

Factor $2$ out of $2 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Move the negative in front of the fraction.

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

Combine the opposite terms in $x - \frac{1}{2} + \frac{1}{2}$ .

Add $- \frac{1}{2}$ and $\frac{1}{2}$ .

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

Add $x$ and $0$ .

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

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

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

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

Related Questions

Support LCMGCF.com by disabling your adblocker.