Find the Inverse f(x)=(x+7)^5

$f (x) = {(x + 7)}^{5}$

Step 1

Write $f (x) = {(x + 7)}^{5}$ as an equation.

$y = {(x + 7)}^{5}$

Step 2

Interchange the variables.

$x = {(y + 7)}^{5}$

Step 3

Solve for $y$ .

Rewrite the equation as ${(y + 7)}^{5} = x$ .

${(y + 7)}^{5} = x$

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y + 7 = \sqrt[5]{x}$

Subtract $7$ from both sides of the equation.

$y = \sqrt[5]{x} - 7$

Step 4

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

$f^{- 1} (x) = \sqrt[5]{x} - 7$

Step 5

Verify if $f^{- 1} (x) = \sqrt[5]{x} - 7$ is the inverse of $f (x) = {(x + 7)}^{5}$ .

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 $f^{- 1} ({(x + 7)}^{5})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} ({(x + 7)}^{5}) = \sqrt[5]{{(x + 7)}^{5}} - 7$

Remove parentheses.

$f^{- 1} ({(x + 7)}^{5}) = \sqrt[5]{{(x + 7)}^{5}} - 7$

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

$f^{- 1} ({(x + 7)}^{5}) = x + 7 - 7$

Combine the opposite terms in $x + 7 - 7$ .

Subtract $7$ from $7$ .

$f^{- 1} ({(x + 7)}^{5}) = x + 0$

Add $x$ and $0$ .

$f^{- 1} ({(x + 7)}^{5}) = x$

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

Set up the composite result function.

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

Evaluate $f (\sqrt[5]{x} - 7)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\sqrt[5]{x} - 7) = {((\sqrt[5]{x} - 7) + 7)}^{5}$

Combine the opposite terms in $(\sqrt[5]{x} - 7) + 7$ .

Add $- 7$ and $7$ .

$f (\sqrt[5]{x} - 7) = {(\sqrt[5]{x} + 0)}^{5}$

Add $\sqrt[5]{x}$ and $0$ .

$f (\sqrt[5]{x} - 7) = {\sqrt[5]{x}}^{5}$

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

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

$f (\sqrt[5]{x} - 7) = {(x^{\frac{1}{5}})}^{5}$

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

$f (\sqrt[5]{x} - 7) = x^{\frac{1}{5} \cdot 5}$

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

$f (\sqrt[5]{x} - 7) = x^{\frac{5}{5}}$

Cancel the common factor of $5$ .

Cancel the common factor.

$f (\sqrt[5]{x} - 7) = x^{\frac{5}{5}}$

Rewrite the expression.

$f (\sqrt[5]{x} - 7) = x$

Simplify.

$f (\sqrt[5]{x} - 7) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \sqrt[5]{x} - 7$ is the inverse of $f (x) = {(x + 7)}^{5}$ .

$f^{- 1} (x) = \sqrt[5]{x} - 7$

Find the Inverse f(x)=(x+7)^5

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