Find the Inverse f(x)=((x^3+7)^(1/5))/7

$f (x) = \frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}$

Step 1

Write $f (x) = \frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}$ as an equation.

$y = \frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}$

Step 2

Interchange the variables.

$x = \frac{{(y^{3} + 7)}^{\frac{1}{5}}}{7}$

Step 3

Solve for $y$ .

Rewrite the equation as $\frac{{(y^{3} + 7)}^{\frac{1}{5}}}{7} = x$ .

$\frac{{(y^{3} + 7)}^{\frac{1}{5}}}{7} = x$

Multiply both sides of the equation by $7$ .

$7 \frac{{(y^{3} + 7)}^{\frac{1}{5}}}{7} = 7 x$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$7 \frac{{(y^{3} + 7)}^{\frac{1}{5}}}{7} = 7 x$

Rewrite the expression.

${(y^{3} + 7)}^{\frac{1}{5}} = 7 x$

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

${({(y^{3} + 7)}^{\frac{1}{5}})}^{5} = {(7 x)}^{5}$

Simplify the exponent.

Simplify the left side.

Simplify ${({(y^{3} + 7)}^{\frac{1}{5}})}^{5}$ .

Multiply the exponents in ${({(y^{3} + 7)}^{\frac{1}{5}})}^{5}$ .

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

${(y^{3} + 7)}^{\frac{1}{5} \cdot 5} = {(7 x)}^{5}$

Cancel the common factor of $5$ .

Cancel the common factor.

${(y^{3} + 7)}^{\frac{1}{5} \cdot 5} = {(7 x)}^{5}$

Rewrite the expression.

${(y^{3} + 7)}^{1} = {(7 x)}^{5}$

Simplify.

$y^{3} + 7 = {(7 x)}^{5}$

Simplify the right side.

Simplify ${(7 x)}^{5}$ .

Apply the product rule to $7 x$ .

$y^{3} + 7 = 7^{5} x^{5}$

Raise $7$ to the power of $5$ .

$y^{3} + 7 = 16807 x^{5}$

Solve for $y$ .

Subtract $7$ from both sides of the equation.

$y^{3} = 16807 x^{5} - 7$

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

$y = \sqrt[3]{16807 x^{5} - 7}$

Factor $7$ out of $16807 x^{5} - 7$ .

Factor $7$ out of $16807 x^{5}$ .

$y = \sqrt[3]{7 (2401 x^{5}) - 7}$

Factor $7$ out of $- 7$ .

$y = \sqrt[3]{7 (2401 x^{5}) + 7 (- 1)}$

Factor $7$ out of $7 (2401 x^{5}) + 7 (- 1)$ .

$y = \sqrt[3]{7 (2401 x^{5} - 1)}$

Step 4

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

$f^{- 1} (x) = \sqrt[3]{7 (2401 x^{5} - 1)}$

Step 5

Verify if $f^{- 1} (x) = \sqrt[3]{7 (2401 x^{5} - 1)}$ is the inverse of $f (x) = \frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}$ .

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 $f^{- 1} (\frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}) = \sqrt[3]{7 (2401 {(\frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7})}^{5} - 1)}$

Apply the product rule to $\frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}$ .

$f^{- 1} (\frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}) = \sqrt[3]{7 (2401 (\frac{{({(x^{3} + 7)}^{\frac{1}{5}})}^{5}}{7^{5}}) - 1)}$

Simplify the numerator.

Multiply the exponents in ${({(x^{3} + 7)}^{\frac{1}{5}})}^{5}$ .

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

$f^{- 1} (\frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}) = \sqrt[3]{7 (2401 (\frac{{(x^{3} + 7)}^{\frac{1}{5} \cdot 5}}{7^{5}}) - 1)}$

Cancel the common factor of $5$ .

Cancel the common factor.

$f^{- 1} (\frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}) = \sqrt[3]{7 (2401 (\frac{{(x^{3} + 7)}^{\frac{1}{5} \cdot 5}}{7^{5}}) - 1)}$

Rewrite the expression.

$f^{- 1} (\frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}) = \sqrt[3]{7 (2401 (\frac{x^{3} + 7}{7^{5}}) - 1)}$

Simplify.

$f^{- 1} (\frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}) = \sqrt[3]{7 (2401 (\frac{x^{3} + 7}{7^{5}}) - 1)}$

Reduce the expression by cancelling the common factors.

Raise $7$ to the power of $5$ .

$f^{- 1} (\frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}) = \sqrt[3]{7 (2401 (\frac{x^{3} + 7}{16807}) - 1)}$

Cancel the common factor of $2401$ .

Factor $2401$ out of $16807$ .

$f^{- 1} (\frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}) = \sqrt[3]{7 (2401 (\frac{x^{3} + 7}{2401 (7)}) - 1)}$

Cancel the common factor.

$f^{- 1} (\frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}) = \sqrt[3]{7 (2401 (\frac{x^{3} + 7}{2401 \cdot 7}) - 1)}$

Rewrite the expression.

$f^{- 1} (\frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}) = \sqrt[3]{7 (\frac{x^{3} + 7}{7} - 1)}$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{7}{7}$ .

$f^{- 1} (\frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}) = \sqrt[3]{7 (\frac{x^{3} + 7}{7} - 1 \cdot \frac{7}{7})}$

Combine $- 1$ and $\frac{7}{7}$ .

$f^{- 1} (\frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}) = \sqrt[3]{7 (\frac{x^{3} + 7}{7} + \frac{- 1 \cdot 7}{7})}$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}) = \sqrt[3]{7 (\frac{x^{3} + 7 - 1 \cdot 7}{7})}$

Rewrite $\frac{x^{3} + 7 - 1 \cdot 7}{7}$ in a factored form.

Multiply $- 1$ by $7$ .

$f^{- 1} (\frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}) = \sqrt[3]{7 (\frac{x^{3} + 7 - 7}{7})}$

Subtract $7$ from $7$ .

$f^{- 1} (\frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}) = \sqrt[3]{7 (\frac{x^{3} + 0}{7})}$

Add $x^{3}$ and $0$ .

$f^{- 1} (\frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}) = \sqrt[3]{7 (\frac{x^{3}}{7})}$

Combine $7$ and $\frac{x^{3}}{7}$ .

$f^{- 1} (\frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}) = \sqrt[3]{\frac{7 x^{3}}{7}}$

Reduce the expression by cancelling the common factors.

Reduce the expression $\frac{7 x^{3}}{7}$ by cancelling the common factors.

Cancel the common factor.

$f^{- 1} (\frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}) = \sqrt[3]{\frac{7 x^{3}}{7}}$

Rewrite the expression.

$f^{- 1} (\frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}) = \sqrt[3]{\frac{x^{3}}{1}}$

Divide $x^{3}$ by $1$ .

$f^{- 1} (\frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}) = \sqrt[3]{x^{3}}$

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

$f^{- 1} (\frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}) = x$

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

Set up the composite result function.

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

Evaluate $f (\sqrt[3]{7 (2401 x^{5} - 1)})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\sqrt[3]{7 (2401 x^{5} - 1)}) = \frac{{({(\sqrt[3]{7 (2401 x^{5} - 1)})}^{3} + 7)}^{\frac{1}{5}}}{7}$

Simplify the numerator.

Simplify each term.

Rewrite ${\sqrt[3]{7 (2401 x^{5} - 1)}}^{3}$ as $7 (2401 x^{5} - 1)$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{7 (2401 x^{5} - 1)}$ as ${(7 (2401 x^{5} - 1))}^{\frac{1}{3}}$ .

$f (\sqrt[3]{7 (2401 x^{5} - 1)}) = \frac{{({({(7 (2401 x^{5} - 1))}^{\frac{1}{3}})}^{3} + 7)}^{\frac{1}{5}}}{7}$

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

$f (\sqrt[3]{7 (2401 x^{5} - 1)}) = \frac{{({(7 (2401 x^{5} - 1))}^{\frac{1}{3} \cdot 3} + 7)}^{\frac{1}{5}}}{7}$

Combine $\frac{1}{3}$ and $3$ .

$f (\sqrt[3]{7 (2401 x^{5} - 1)}) = \frac{{({(7 (2401 x^{5} - 1))}^{\frac{3}{3}} + 7)}^{\frac{1}{5}}}{7}$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (\sqrt[3]{7 (2401 x^{5} - 1)}) = \frac{{({(7 (2401 x^{5} - 1))}^{\frac{3}{3}} + 7)}^{\frac{1}{5}}}{7}$

Rewrite the expression.

$f (\sqrt[3]{7 (2401 x^{5} - 1)}) = \frac{{((7 (2401 x^{5} - 1)) + 7)}^{\frac{1}{5}}}{7}$

Simplify.

$f (\sqrt[3]{7 (2401 x^{5} - 1)}) = \frac{{(7 (2401 x^{5} - 1) + 7)}^{\frac{1}{5}}}{7}$

Apply the distributive property.

$f (\sqrt[3]{7 (2401 x^{5} - 1)}) = \frac{{(7 (2401 x^{5}) + 7 \cdot - 1 + 7)}^{\frac{1}{5}}}{7}$

Multiply $2401$ by $7$ .

$f (\sqrt[3]{7 (2401 x^{5} - 1)}) = \frac{{(16807 x^{5} + 7 \cdot - 1 + 7)}^{\frac{1}{5}}}{7}$

Multiply $7$ by $- 1$ .

$f (\sqrt[3]{7 (2401 x^{5} - 1)}) = \frac{{(16807 x^{5} - 7 + 7)}^{\frac{1}{5}}}{7}$

Combine the opposite terms in $16807 x^{5} - 7 + 7$ .

Add $- 7$ and $7$ .

$f (\sqrt[3]{7 (2401 x^{5} - 1)}) = \frac{{(16807 x^{5} + 0)}^{\frac{1}{5}}}{7}$

Add $16807 x^{5}$ and $0$ .

$f (\sqrt[3]{7 (2401 x^{5} - 1)}) = \frac{{(16807 x^{5})}^{\frac{1}{5}}}{7}$

Apply the product rule to $16807 x^{5}$ .

$f (\sqrt[3]{7 (2401 x^{5} - 1)}) = \frac{16807^{\frac{1}{5}} {(x^{5})}^{\frac{1}{5}}}{7}$

Rewrite $16807$ as $7^{5}$ .

$f (\sqrt[3]{7 (2401 x^{5} - 1)}) = \frac{{(7^{5})}^{\frac{1}{5}} {(x^{5})}^{\frac{1}{5}}}{7}$

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

$f (\sqrt[3]{7 (2401 x^{5} - 1)}) = \frac{7^{5 (\frac{1}{5})} {(x^{5})}^{\frac{1}{5}}}{7}$

Cancel the common factor of $5$ .

Cancel the common factor.

$f (\sqrt[3]{7 (2401 x^{5} - 1)}) = \frac{7^{5 (\frac{1}{5})} {(x^{5})}^{\frac{1}{5}}}{7}$

Rewrite the expression.

$f (\sqrt[3]{7 (2401 x^{5} - 1)}) = \frac{7 {(x^{5})}^{\frac{1}{5}}}{7}$

Evaluate the exponent.

$f (\sqrt[3]{7 (2401 x^{5} - 1)}) = \frac{7 {(x^{5})}^{\frac{1}{5}}}{7}$

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

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

$f (\sqrt[3]{7 (2401 x^{5} - 1)}) = \frac{7 x^{5 (\frac{1}{5})}}{7}$

Cancel the common factor of $5$ .

Cancel the common factor.

$f (\sqrt[3]{7 (2401 x^{5} - 1)}) = \frac{7 x^{5 (\frac{1}{5})}}{7}$

Rewrite the expression.

$f (\sqrt[3]{7 (2401 x^{5} - 1)}) = \frac{7 x}{7}$

Simplify.

$f (\sqrt[3]{7 (2401 x^{5} - 1)}) = \frac{7 x}{7}$

Cancel the common factor of $7$ .

Cancel the common factor.

$f (\sqrt[3]{7 (2401 x^{5} - 1)}) = \frac{7 x}{7}$

Divide $x$ by $1$ .

$f (\sqrt[3]{7 (2401 x^{5} - 1)}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \sqrt[3]{7 (2401 x^{5} - 1)}$ is the inverse of $f (x) = \frac{{(x^{3} + 7)}^{\frac{1}{5}}}{7}$ .

$f^{- 1} (x) = \sqrt[3]{7 (2401 x^{5} - 1)}$

Find the Inverse f(x)=((x^3+7)^(1/5))/7

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