Find the Inverse f(x) = fifth root of (x-1)/2

$f (x) = \sqrt[5]{\frac{x - 1}{2}}$

Step 1

Write $f (x) = \sqrt[5]{\frac{x - 1}{2}}$ as an equation.

$y = \sqrt[5]{\frac{x - 1}{2}}$

Step 2

Interchange the variables.

$x = \sqrt[5]{\frac{y - 1}{2}}$

Step 3

Solve for $y$ .

Rewrite the equation as $\sqrt[5]{\frac{y - 1}{2}} = x$ .

$\sqrt[5]{\frac{y - 1}{2}} = x$

To remove the radical on the left side of the equation, raise both sides of the equation to the power of $5$ .

${\sqrt[5]{\frac{y - 1}{2}}}^{5} = x^{5}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[5]{\frac{y - 1}{2}}$ as ${(\frac{y - 1}{2})}^{\frac{1}{5}}$ .

${({(\frac{y - 1}{2})}^{\frac{1}{5}})}^{5} = x^{5}$

Simplify the left side.

Simplify ${({(\frac{y - 1}{2})}^{\frac{1}{5}})}^{5}$ .

Multiply the exponents in ${({(\frac{y - 1}{2})}^{\frac{1}{5}})}^{5}$ .

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

${(\frac{y - 1}{2})}^{\frac{1}{5} \cdot 5} = x^{5}$

Cancel the common factor of $5$ .

Cancel the common factor.

${(\frac{y - 1}{2})}^{\frac{1}{5} \cdot 5} = x^{5}$

Rewrite the expression.

${(\frac{y - 1}{2})}^{1} = x^{5}$

Simplify.

$\frac{y - 1}{2} = x^{5}$

Solve for $y$ .

Multiply both sides of the equation by $2$ .

$2 \frac{y - 1}{2} = 2 x^{5}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$2 \frac{y - 1}{2} = 2 x^{5}$

Rewrite the expression.

$y - 1 = 2 x^{5}$

Add $1$ to both sides of the equation.

$y = 2 x^{5} + 1$

Step 4

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

$f^{- 1} (x) = 2 x^{5} + 1$

Step 5

Verify if $f^{- 1} (x) = 2 x^{5} + 1$ is the inverse of $f (x) = \sqrt[5]{\frac{x - 1}{2}}$ .

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 $f^{- 1} (\sqrt[5]{\frac{x - 1}{2}})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 {(\sqrt[5]{\frac{x - 1}{2}})}^{5} + 1$

Simplify each term.

Rewrite $\sqrt[5]{\frac{x - 1}{2}}$ as $\frac{\sqrt[5]{x - 1}}{\sqrt[5]{2}}$ .

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 {(\frac{\sqrt[5]{x - 1}}{\sqrt[5]{2}})}^{5} + 1$

Multiply $\frac{\sqrt[5]{x - 1}}{\sqrt[5]{2}}$ by $\frac{{\sqrt[5]{2}}^{4}}{{\sqrt[5]{2}}^{4}}$ .

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 {(\frac{\sqrt[5]{x - 1}}{\sqrt[5]{2}} \cdot \frac{{\sqrt[5]{2}}^{4}}{{\sqrt[5]{2}}^{4}})}^{5} + 1$

Combine and simplify the denominator.

Multiply $\frac{\sqrt[5]{x - 1}}{\sqrt[5]{2}}$ by $\frac{{\sqrt[5]{2}}^{4}}{{\sqrt[5]{2}}^{4}}$ .

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 {(\frac{\sqrt[5]{x - 1} {\sqrt[5]{2}}^{4}}{\sqrt[5]{2} {\sqrt[5]{2}}^{4}})}^{5} + 1$

Raise $\sqrt[5]{2}$ to the power of $1$ .

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 {(\frac{\sqrt[5]{x - 1} {\sqrt[5]{2}}^{4}}{\sqrt[5]{2} {\sqrt[5]{2}}^{4}})}^{5} + 1$

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

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 {(\frac{\sqrt[5]{x - 1} {\sqrt[5]{2}}^{4}}{{\sqrt[5]{2}}^{1 + 4}})}^{5} + 1$

Add $1$ and $4$ .

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 {(\frac{\sqrt[5]{x - 1} {\sqrt[5]{2}}^{4}}{{\sqrt[5]{2}}^{5}})}^{5} + 1$

Rewrite ${\sqrt[5]{2}}^{5}$ as $2$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[5]{2}$ as $2^{\frac{1}{5}}$ .

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 {(\frac{\sqrt[5]{x - 1} {\sqrt[5]{2}}^{4}}{{(2^{\frac{1}{5}})}^{5}})}^{5} + 1$

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

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 {(\frac{\sqrt[5]{x - 1} {\sqrt[5]{2}}^{4}}{2^{\frac{1}{5} \cdot 5}})}^{5} + 1$

Combine $\frac{1}{5}$ and $5$ .

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 {(\frac{\sqrt[5]{x - 1} {\sqrt[5]{2}}^{4}}{2^{\frac{5}{5}}})}^{5} + 1$

Cancel the common factor of $5$ .

Cancel the common factor.

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 {(\frac{\sqrt[5]{x - 1} {\sqrt[5]{2}}^{4}}{2^{\frac{5}{5}}})}^{5} + 1$

Rewrite the expression.

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 {(\frac{\sqrt[5]{x - 1} {\sqrt[5]{2}}^{4}}{2})}^{5} + 1$

Evaluate the exponent.

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 {(\frac{\sqrt[5]{x - 1} {\sqrt[5]{2}}^{4}}{2})}^{5} + 1$

Simplify the numerator.

Rewrite ${\sqrt[5]{2}}^{4}$ as ${(2^{4})}^{\frac{1}{5}}$ .

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 {(\frac{\sqrt[5]{x - 1} \sqrt[5]{2^{4}}}{2})}^{5} + 1$

Raise $2$ to the power of $4$ .

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 {(\frac{\sqrt[5]{x - 1} \sqrt[5]{16}}{2})}^{5} + 1$

Combine using the product rule for radicals.

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 {(\frac{\sqrt[5]{(x - 1) \cdot 16}}{2})}^{5} + 1$

Apply the product rule to $\frac{\sqrt[5]{(x - 1) \cdot 16}}{2}$ .

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 (\frac{{\sqrt[5]{(x - 1) \cdot 16}}^{5}}{2^{5}}) + 1$

Simplify the numerator.

Rewrite ${\sqrt[5]{(x - 1) \cdot 16}}^{5}$ as $(x - 1) \cdot 16$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[5]{(x - 1) \cdot 16}$ as ${((x - 1) \cdot 16)}^{\frac{1}{5}}$ .

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 (\frac{{({((x - 1) \cdot 16)}^{\frac{1}{5}})}^{5}}{2^{5}}) + 1$

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

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 (\frac{{((x - 1) \cdot 16)}^{\frac{1}{5} \cdot 5}}{2^{5}}) + 1$

Combine $\frac{1}{5}$ and $5$ .

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 (\frac{{((x - 1) \cdot 16)}^{\frac{5}{5}}}{2^{5}}) + 1$

Cancel the common factor of $5$ .

Cancel the common factor.

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 (\frac{{((x - 1) \cdot 16)}^{\frac{5}{5}}}{2^{5}}) + 1$

Rewrite the expression.

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 (\frac{(x - 1) \cdot 16}{2^{5}}) + 1$

Simplify.

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 (\frac{(x - 1) \cdot 16}{2^{5}}) + 1$

Apply the distributive property.

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 (\frac{x \cdot 16 - 1 \cdot 16}{2^{5}}) + 1$

Move $16$ to the left of $x$ .

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 (\frac{16 \cdot x - 1 \cdot 16}{2^{5}}) + 1$

Multiply $- 1$ by $16$ .

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 (\frac{16 \cdot x - 16}{2^{5}}) + 1$

Factor $16$ out of $16 x - 16$ .

Factor $16$ out of $16 x$ .

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 (\frac{16 (x) - 16}{2^{5}}) + 1$

Factor $16$ out of $- 16$ .

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 (\frac{16 (x) + 16 (- 1)}{2^{5}}) + 1$

Factor $16$ out of $16 (x) + 16 (- 1)$ .

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 (\frac{16 (x - 1)}{2^{5}}) + 1$

Raise $2$ to the power of $5$ .

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 (\frac{16 (x - 1)}{32}) + 1$

Cancel the common factor of $2$ .

Factor $2$ out of $32$ .

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 (\frac{16 (x - 1)}{2 (16)}) + 1$

Cancel the common factor.

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = 2 (\frac{16 (x - 1)}{2 \cdot 16}) + 1$

Rewrite the expression.

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = \frac{16 (x - 1)}{16} + 1$

Cancel the common factor of $16$ .

Cancel the common factor.

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = \frac{16 (x - 1)}{16} + 1$

Divide $x - 1$ by $1$ .

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = x - 1 + 1$

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

Add $- 1$ and $1$ .

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (\sqrt[5]{\frac{x - 1}{2}}) = x$

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

Set up the composite result function.

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

Evaluate $f (2 x^{5} + 1)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (2 x^{5} + 1) = \sqrt[5]{\frac{(2 x^{5} + 1) - 1}{2}}$

Subtract $1$ from $1$ .

$f (2 x^{5} + 1) = \sqrt[5]{\frac{2 x^{5} + 0}{2}}$

Add $2 x^{5}$ and $0$ .

$f (2 x^{5} + 1) = \sqrt[5]{\frac{2 x^{5}}{2}}$

Reduce the expression by cancelling the common factors.

Reduce the expression $\frac{2 x^{5}}{2}$ by cancelling the common factors.

Cancel the common factor.

$f (2 x^{5} + 1) = \sqrt[5]{\frac{2 x^{5}}{2}}$

Rewrite the expression.

$f (2 x^{5} + 1) = \sqrt[5]{\frac{x^{5}}{1}}$

Divide $x^{5}$ by $1$ .

$f (2 x^{5} + 1) = \sqrt[5]{x^{5}}$

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

$f (2 x^{5} + 1) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = 2 x^{5} + 1$ is the inverse of $f (x) = \sqrt[5]{\frac{x - 1}{2}}$ .

$f^{- 1} (x) = 2 x^{5} + 1$

Find the Inverse f(x) = fifth root of (x-1)/2

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