Find the Inverse f(x)=((x+2)/7)^(1/7)

$f (x) = {(\frac{x + 2}{7})}^{\frac{1}{7}}$

Write $f (x) = {(\frac{x + 2}{7})}^{\frac{1}{7}}$ as an equation.

$y = {(\frac{x + 2}{7})}^{\frac{1}{7}}$

Interchange the variables.

$x = {(\frac{y + 2}{7})}^{\frac{1}{7}}$

Solve for $y$ .

Rewrite the equation as ${(\frac{y + 2}{7})}^{\frac{1}{7}} = x$ .

${(\frac{y + 2}{7})}^{\frac{1}{7}} = x$

Raise each side of the equation to the power of $7$ to eliminate the fractional exponent on the left side.

${({(\frac{y + 2}{7})}^{\frac{1}{7}})}^{7} = x^{7}$

Simplify the left side.

Simplify ${({(\frac{y + 2}{7})}^{\frac{1}{7}})}^{7}$ .

Multiply the exponents in ${({(\frac{y + 2}{7})}^{\frac{1}{7}})}^{7}$ .

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

${(\frac{y + 2}{7})}^{\frac{1}{7} \cdot 7} = x^{7}$

Cancel the common factor of $7$ .

Cancel the common factor.

${(\frac{y + 2}{7})}^{\frac{1}{7} \cdot 7} = x^{7}$

Rewrite the expression.

${(\frac{y + 2}{7})}^{1} = x^{7}$

Split the fraction $\frac{y + 2}{7}$ into two fractions.

${(\frac{y}{7} + \frac{2}{7})}^{1} = x^{7}$

Simplify.

$\frac{y}{7} + \frac{2}{7} = x^{7}$

Solve for $y$ .

Subtract $\frac{2}{7}$ from both sides of the equation.

$\frac{y}{7} = x^{7} - \frac{2}{7}$

Multiply both sides of the equation by $7$ .

$7 \frac{y}{7} = 7 (x^{7} - \frac{2}{7})$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$7 \frac{y}{7} = 7 (x^{7} - \frac{2}{7})$

Rewrite the expression.

$y = 7 (x^{7} - \frac{2}{7})$

Simplify the right side.

Simplify $7 (x^{7} - \frac{2}{7})$ .

Apply the distributive property.

$y = 7 x^{7} + 7 (- \frac{2}{7})$

Cancel the common factor of $7$ .

Move the leading negative in $- \frac{2}{7}$ into the numerator.

$y = 7 x^{7} + 7 (\frac{- 2}{7})$

Cancel the common factor.

$y = 7 x^{7} + 7 (\frac{- 2}{7})$

Rewrite the expression.

$y = 7 x^{7} - 2$

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

$f^{- 1} (x) = 7 x^{7} - 2$

Verify if $f^{- 1} (x) = 7 x^{7} - 2$ is the inverse of $f (x) = {(\frac{x + 2}{7})}^{\frac{1}{7}}$ .

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 $f^{- 1} ({(\frac{x + 2}{7})}^{\frac{1}{7}})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} ({(\frac{x + 2}{7})}^{\frac{1}{7}}) = 7 {({(\frac{x + 2}{7})}^{\frac{1}{7}})}^{7} - 2$

Simplify each term.

Multiply the exponents in ${({(\frac{x + 2}{7})}^{\frac{1}{7}})}^{7}$ .

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

$f^{- 1} ({(\frac{x + 2}{7})}^{\frac{1}{7}}) = 7 {(\frac{x + 2}{7})}^{\frac{1}{7} \cdot 7} - 2$

Cancel the common factor of $7$ .

Cancel the common factor.

$f^{- 1} ({(\frac{x + 2}{7})}^{\frac{1}{7}}) = 7 {(\frac{x + 2}{7})}^{\frac{1}{7} \cdot 7} - 2$

Rewrite the expression.

$f^{- 1} ({(\frac{x + 2}{7})}^{\frac{1}{7}}) = 7 (\frac{x + 2}{7}) - 2$

Simplify.

$f^{- 1} ({(\frac{x + 2}{7})}^{\frac{1}{7}}) = 7 (\frac{x + 2}{7}) - 2$

Cancel the common factor of $7$ .

Cancel the common factor.

$f^{- 1} ({(\frac{x + 2}{7})}^{\frac{1}{7}}) = 7 (\frac{x + 2}{7}) - 2$

Rewrite the expression.

$f^{- 1} ({(\frac{x + 2}{7})}^{\frac{1}{7}}) = x + 2 - 2$

Combine the opposite terms in $x + 2 - 2$ .

Subtract $2$ from $2$ .

$f^{- 1} ({(\frac{x + 2}{7})}^{\frac{1}{7}}) = x + 0$

Add $x$ and $0$ .

$f^{- 1} ({(\frac{x + 2}{7})}^{\frac{1}{7}}) = x$

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

Set up the composite result function.

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

Evaluate $f (7 x^{7} - 2)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (7 x^{7} - 2) = {(\frac{(7 x^{7} - 2) + 2}{7})}^{\frac{1}{7}}$

Simplify the numerator.

Add $- 2$ and $2$ .

$f (7 x^{7} - 2) = {(\frac{7 x^{7} + 0}{7})}^{\frac{1}{7}}$

Add $7 x^{7}$ and $0$ .

$f (7 x^{7} - 2) = {(\frac{7 x^{7}}{7})}^{\frac{1}{7}}$

Reduce the expression by cancelling the common factors.

Cancel the common factor of $7$ .

Cancel the common factor.

$f (7 x^{7} - 2) = {(\frac{7 x^{7}}{7})}^{\frac{1}{7}}$

Divide $x^{7}$ by $1$ .

$f (7 x^{7} - 2) = {(x^{7})}^{\frac{1}{7}}$

Multiply the exponents in ${(x^{7})}^{\frac{1}{7}}$ .

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

$f (7 x^{7} - 2) = x^{7 (\frac{1}{7})}$

Cancel the common factor of $7$ .

Cancel the common factor.

$f (7 x^{7} - 2) = x^{7 (\frac{1}{7})}$

Rewrite the expression.

$f (7 x^{7} - 2) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = 7 x^{7} - 2$ is the inverse of $f (x) = {(\frac{x + 2}{7})}^{\frac{1}{7}}$ .

$f^{- 1} (x) = 7 x^{7} - 2$

Find the Inverse f(x)=((x+2)/7)^(1/7)

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