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

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

Step 1

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

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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

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

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $5$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

$y - 5 = x^{5}$

Add $5$ to both sides of the equation.

$y = x^{5} + 5$

Step 4

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

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

Step 5

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

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

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

Simplify each term.

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

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

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

Cancel the common factor of $5$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Combine the opposite terms in $x - 5 + 5$ .

Add $- 5$ and $5$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

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

Subtract $5$ from $5$ .

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

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

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

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

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

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

Cancel the common factor of $5$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

Related Questions

Support LCMGCF.com by disabling your adblocker.