Find the Inverse f(x)=((x-3)/8)^(1/3)

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

Write $f (x) = {(\frac{x - 3}{8})}^{\frac{1}{3}}$ as an equation.

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

Interchange the variables.

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

Solve for $y$ .

Rewrite the equation as ${(\frac{y - 3}{8})}^{\frac{1}{3}} = x$ .

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

Raise each side of the equation to the power of $3$ to eliminate the fractional exponent on the left side.

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

Simplify the left side.

Simplify ${({(\frac{y - 3}{8})}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(\frac{y - 3}{8})}^{\frac{1}{3}})}^{3}$ .

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

${(\frac{y - 3}{8})}^{\frac{1}{3} \cdot 3} = x^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(\frac{y - 3}{8})}^{\frac{1}{3} \cdot 3} = x^{3}$

Rewrite the expression.

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

Split the fraction $\frac{y - 3}{8}$ into two fractions.

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

Move the negative in front of the fraction.

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

Simplify.

$\frac{y}{8} - \frac{3}{8} = x^{3}$

Solve for $y$ .

Add $\frac{3}{8}$ to both sides of the equation.

$\frac{y}{8} = x^{3} + \frac{3}{8}$

Multiply both sides of the equation by $8$ .

$8 \frac{y}{8} = 8 (x^{3} + \frac{3}{8})$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$8 \frac{y}{8} = 8 (x^{3} + \frac{3}{8})$

Rewrite the expression.

$y = 8 (x^{3} + \frac{3}{8})$

Simplify the right side.

Simplify $8 (x^{3} + \frac{3}{8})$ .

Apply the distributive property.

$y = 8 x^{3} + 8 (\frac{3}{8})$

Cancel the common factor of $8$ .

Cancel the common factor.

$y = 8 x^{3} + 8 (\frac{3}{8})$

Rewrite the expression.

$y = 8 x^{3} + 3$

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

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

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

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

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

Simplify each term.

Multiply the exponents in ${({(\frac{x - 3}{8})}^{\frac{1}{3}})}^{3}$ .

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Cancel the common factor of $8$ .

Cancel the common factor.

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

Rewrite the expression.

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

Combine the opposite terms in $x - 3 + 3$ .

Add $- 3$ and $3$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Simplify the numerator.

Subtract $3$ from $3$ .

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

Add $8 x^{3}$ and $0$ .

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

Reduce the expression by cancelling the common factors.

Cancel the common factor of $8$ .

Cancel the common factor.

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

$f (8 x^{3} + 3) = x$

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

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

Find the Inverse f(x)=((x-3)/8)^(1/3)

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