Find the Inverse f(x)=-( cube root of 4x)/2

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

Write $f (x) = - \frac{\sqrt[3]{4 x}}{2}$ as an equation.

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

Interchange the variables.

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

Solve for $y$ .

Rewrite the equation as $- \frac{\sqrt[3]{4 y}}{2} = x$ .

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

Multiply both sides of the equation by $- 2$ .

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

Simplify the left side.

Simplify $- 2 (- \frac{\sqrt[3]{4 y}}{2})$ .

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{\sqrt[3]{4 y}}{2}$ into the numerator.

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

Factor $2$ out of $- 2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply.

Multiply $- 1$ by $- 1$ .

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

Multiply $\sqrt[3]{4 y}$ by $1$ .

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

To remove the radical on the left side of the equation, cube both sides of the equation.

${\sqrt[3]{4 y}}^{3} = {(- 2 x)}^{3}$

Simplify each side of the equation.

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

${({(4 y)}^{\frac{1}{3}})}^{3} = {(- 2 x)}^{3}$

Simplify the left side.

Simplify ${({(4 y)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(4 y)}^{\frac{1}{3}})}^{3}$ .

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

${(4 y)}^{\frac{1}{3} \cdot 3} = {(- 2 x)}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(4 y)}^{\frac{1}{3} \cdot 3} = {(- 2 x)}^{3}$

Rewrite the expression.

${(4 y)}^{1} = {(- 2 x)}^{3}$

Simplify.

$4 y = {(- 2 x)}^{3}$

Simplify the right side.

Simplify ${(- 2 x)}^{3}$ .

Apply the product rule to $- 2 x$ .

$4 y = {(- 2)}^{3} x^{3}$

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

$4 y = - 8 x^{3}$

Divide each term in $4 y = - 8 x^{3}$ by $4$ and simplify.

Divide each term in $4 y = - 8 x^{3}$ by $4$ .

$\frac{4 y}{4} = \frac{- 8 x^{3}}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 y}{4} = \frac{- 8 x^{3}}{4}$

Divide $y$ by $1$ .

$y = \frac{- 8 x^{3}}{4}$

Simplify the right side.

Cancel the common factor of $- 8$ and $4$ .

Factor $4$ out of $- 8 x^{3}$ .

$y = \frac{4 (- 2 x^{3})}{4}$

Cancel the common factors.

Factor $4$ out of $4$ .

$y = \frac{4 (- 2 x^{3})}{4 (1)}$

Cancel the common factor.

$y = \frac{4 (- 2 x^{3})}{4 \cdot 1}$

Rewrite the expression.

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

Divide $- 2 x^{3}$ by $1$ .

$y = - 2 x^{3}$

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

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

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

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

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

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

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

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

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

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

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

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

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 ${(4 x)}^{\frac{1}{3}}$ .

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

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

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{4 x}{8}$ into the numerator.

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

Factor $2$ out of $- 2$ .

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

Factor $2$ out of $8$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $- 4$ and $4$ .

Factor $4$ out of $- 4 x$ .

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

Cancel the common factors.

Factor $4$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- x$ by $1$ .

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

Multiply $- 1 (- x)$ .

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Simplify the numerator.

Multiply $- 2$ by $4$ .

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

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

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

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

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

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

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

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

Cancel the common factors.

Factor $2$ out of $2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- x$ by $1$ .

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

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

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

Find the Inverse f(x)=-( cube root of 4x)/2

Related Questions

Support LCMGCF.com by disabling your adblocker.