Find the Inverse f(x)=(x+2)^(1/5)

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

Step 1

Write $f (x) = {(x + 2)}^{\frac{1}{5}}$ as an equation.

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

Rewrite the equation as ${(y + 2)}^{\frac{1}{5}} = x$ .

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

Raise each side of the equation to the power of $5$ to eliminate the fractional exponent on the left side.

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $5$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

$y + 2 = x^{5}$

Subtract $2$ from both sides of the equation.

$y = x^{5} - 2$

Step 4

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

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

Step 5

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

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

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

Simplify each term.

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

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

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

Cancel the common factor of $5$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

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

Subtract $2$ from $2$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Combine the opposite terms in $(x^{5} - 2) + 2$ .

Add $- 2$ and $2$ .

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

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

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

Multiply the exponents in ${(x^{5})}^{\frac{1}{5}}$ .

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

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

Cancel the common factor of $5$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Find the Inverse f(x)=(x+2)^(1/5)

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