Find the Inverse (7x-8)^3

${(7 x - 8)}^{3}$

Step 1

Interchange the variables.

$x = {(7 y - 8)}^{3}$

Step 2

Solve for $y$ .

Rewrite the equation as ${(7 y - 8)}^{3} = x$ .

${(7 y - 8)}^{3} = x$

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

$7 y - 8 = \sqrt[3]{x}$

Add $8$ to both sides of the equation.

$7 y = \sqrt[3]{x} + 8$

Divide each term in $7 y = \sqrt[3]{x} + 8$ by $7$ and simplify.

Divide each term in $7 y = \sqrt[3]{x} + 8$ by $7$ .

$\frac{7 y}{7} = \frac{\sqrt[3]{x}}{7} + \frac{8}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 y}{7} = \frac{\sqrt[3]{x}}{7} + \frac{8}{7}$

Divide $y$ by $1$ .

$y = \frac{\sqrt[3]{x}}{7} + \frac{8}{7}$

Step 3

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

$f^{- 1} (x) = \frac{\sqrt[3]{x}}{7} + \frac{8}{7}$

Step 4

Verify if $f^{- 1} (x) = \frac{\sqrt[3]{x}}{7} + \frac{8}{7}$ is the inverse of $f (x) = {(7 x - 8)}^{3}$ .

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 $f^{- 1} ({(7 x - 8)}^{3})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} ({(7 x - 8)}^{3}) = \frac{\sqrt[3]{{(7 x - 8)}^{3}}}{7} + \frac{8}{7}$

Combine the numerators over the common denominator.

$f^{- 1} ({(7 x - 8)}^{3}) = \frac{\sqrt[3]{{(7 x - 8)}^{3}} + 8}{7}$

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

$f^{- 1} ({(7 x - 8)}^{3}) = \frac{7 x - 8 + 8}{7}$

Combine the opposite terms in $7 x - 8 + 8$ .

Add $- 8$ and $8$ .

$f^{- 1} ({(7 x - 8)}^{3}) = \frac{7 x + 0}{7}$

Add $7 x$ and $0$ .

$f^{- 1} ({(7 x - 8)}^{3}) = \frac{7 x}{7}$

Cancel the common factor of $7$ .

Cancel the common factor.

$f^{- 1} ({(7 x - 8)}^{3}) = \frac{7 x}{7}$

Divide $x$ by $1$ .

$f^{- 1} ({(7 x - 8)}^{3}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{\sqrt[3]{x}}{7} + \frac{8}{7})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\sqrt[3]{x}}{7} + \frac{8}{7}) = {(7 (\frac{\sqrt[3]{x}}{7} + \frac{8}{7}) - 8)}^{3}$

Simplify each term.

Apply the distributive property.

$f (\frac{\sqrt[3]{x}}{7} + \frac{8}{7}) = {(7 (\frac{\sqrt[3]{x}}{7}) + 7 (\frac{8}{7}) - 8)}^{3}$

Cancel the common factor of $7$ .

Cancel the common factor.

$f (\frac{\sqrt[3]{x}}{7} + \frac{8}{7}) = {(7 (\frac{\sqrt[3]{x}}{7}) + 7 (\frac{8}{7}) - 8)}^{3}$

Rewrite the expression.

$f (\frac{\sqrt[3]{x}}{7} + \frac{8}{7}) = {(\sqrt[3]{x} + 7 (\frac{8}{7}) - 8)}^{3}$

Cancel the common factor of $7$ .

Cancel the common factor.

$f (\frac{\sqrt[3]{x}}{7} + \frac{8}{7}) = {(\sqrt[3]{x} + 7 (\frac{8}{7}) - 8)}^{3}$

Rewrite the expression.

$f (\frac{\sqrt[3]{x}}{7} + \frac{8}{7}) = {(\sqrt[3]{x} + 8 - 8)}^{3}$

Simplify terms.

Combine the opposite terms in $\sqrt[3]{x} + 8 - 8$ .

Subtract $8$ from $8$ .

$f (\frac{\sqrt[3]{x}}{7} + \frac{8}{7}) = {(\sqrt[3]{x} + 0)}^{3}$

Add $\sqrt[3]{x}$ and $0$ .

$f (\frac{\sqrt[3]{x}}{7} + \frac{8}{7}) = {\sqrt[3]{x}}^{3}$

Rewrite ${\sqrt[3]{x}}^{3}$ as $x$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x}$ as $x^{\frac{1}{3}}$ .

$f (\frac{\sqrt[3]{x}}{7} + \frac{8}{7}) = {(x^{\frac{1}{3}})}^{3}$

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

$f (\frac{\sqrt[3]{x}}{7} + \frac{8}{7}) = x^{\frac{1}{3} \cdot 3}$

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

$f (\frac{\sqrt[3]{x}}{7} + \frac{8}{7}) = x^{\frac{3}{3}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (\frac{\sqrt[3]{x}}{7} + \frac{8}{7}) = x^{\frac{3}{3}}$

Rewrite the expression.

$f (\frac{\sqrt[3]{x}}{7} + \frac{8}{7}) = x$

Simplify.

$f (\frac{\sqrt[3]{x}}{7} + \frac{8}{7}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\sqrt[3]{x}}{7} + \frac{8}{7}$ is the inverse of $f (x) = {(7 x - 8)}^{3}$ .

$f^{- 1} (x) = \frac{\sqrt[3]{x}}{7} + \frac{8}{7}$

Find the Inverse (7x-8)^3

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