Find the Inverse 5-9x^3

$5 - 9 x^{3}$

Step 1

Interchange the variables.

$x = 5 - 9 y^{3}$

Step 2

Solve for $y$ .

Rewrite the equation as $5 - 9 y^{3} = x$ .

$5 - 9 y^{3} = x$

Subtract $5$ from both sides of the equation.

$- 9 y^{3} = x - 5$

Divide each term in $- 9 y^{3} = x - 5$ by $- 9$ and simplify.

Divide each term in $- 9 y^{3} = x - 5$ by $- 9$ .

$\frac{- 9 y^{3}}{- 9} = \frac{x}{- 9} + \frac{- 5}{- 9}$

Simplify the left side.

Cancel the common factor of $- 9$ .

Cancel the common factor.

$\frac{- 9 y^{3}}{- 9} = \frac{x}{- 9} + \frac{- 5}{- 9}$

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

$y^{3} = \frac{x}{- 9} + \frac{- 5}{- 9}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$y^{3} = - \frac{x}{9} + \frac{- 5}{- 9}$

Dividing two negative values results in a positive value.

$y^{3} = - \frac{x}{9} + \frac{5}{9}$

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

$y = \sqrt[3]{- \frac{x}{9} + \frac{5}{9}}$

Simplify $\sqrt[3]{- \frac{x}{9} + \frac{5}{9}}$ .

Combine the numerators over the common denominator.

$y = \sqrt[3]{\frac{- x + 5}{9}}$

Rewrite $\sqrt[3]{\frac{- x + 5}{9}}$ as $\frac{\sqrt[3]{- x + 5}}{\sqrt[3]{9}}$ .

$y = \frac{\sqrt[3]{- x + 5}}{\sqrt[3]{9}}$

Multiply $\frac{\sqrt[3]{- x + 5}}{\sqrt[3]{9}}$ by $\frac{{\sqrt[3]{9}}^{2}}{{\sqrt[3]{9}}^{2}}$ .

$y = \frac{\sqrt[3]{- x + 5}}{\sqrt[3]{9}} \cdot \frac{{\sqrt[3]{9}}^{2}}{{\sqrt[3]{9}}^{2}}$

Combine and simplify the denominator.

Multiply $\frac{\sqrt[3]{- x + 5}}{\sqrt[3]{9}}$ by $\frac{{\sqrt[3]{9}}^{2}}{{\sqrt[3]{9}}^{2}}$ .

$y = \frac{\sqrt[3]{- x + 5} {\sqrt[3]{9}}^{2}}{\sqrt[3]{9} {\sqrt[3]{9}}^{2}}$

Raise $\sqrt[3]{9}$ to the power of $1$ .

$y = \frac{\sqrt[3]{- x + 5} {\sqrt[3]{9}}^{2}}{{\sqrt[3]{9}}^{1} {\sqrt[3]{9}}^{2}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y = \frac{\sqrt[3]{- x + 5} {\sqrt[3]{9}}^{2}}{{\sqrt[3]{9}}^{1 + 2}}$

Add $1$ and $2$ .

$y = \frac{\sqrt[3]{- x + 5} {\sqrt[3]{9}}^{2}}{{\sqrt[3]{9}}^{3}}$

Rewrite ${\sqrt[3]{9}}^{3}$ as $9$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{9}$ as $9^{\frac{1}{3}}$ .

$y = \frac{\sqrt[3]{- x + 5} {\sqrt[3]{9}}^{2}}{{(9^{\frac{1}{3}})}^{3}}$

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

$y = \frac{\sqrt[3]{- x + 5} {\sqrt[3]{9}}^{2}}{9^{\frac{1}{3} \cdot 3}}$

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

$y = \frac{\sqrt[3]{- x + 5} {\sqrt[3]{9}}^{2}}{9^{\frac{3}{3}}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$y = \frac{\sqrt[3]{- x + 5} {\sqrt[3]{9}}^{2}}{9^{\frac{3}{3}}}$

Rewrite the expression.

$y = \frac{\sqrt[3]{- x + 5} {\sqrt[3]{9}}^{2}}{9^{1}}$

Evaluate the exponent.

$y = \frac{\sqrt[3]{- x + 5} {\sqrt[3]{9}}^{2}}{9}$

Simplify the numerator.

Rewrite ${\sqrt[3]{9}}^{2}$ as $\sqrt[3]{9^{2}}$ .

$y = \frac{\sqrt[3]{- x + 5} \sqrt[3]{9^{2}}}{9}$

Raise $9$ to the power of $2$ .

$y = \frac{\sqrt[3]{- x + 5} \sqrt[3]{81}}{9}$

Rewrite $81$ as $3^{3} \cdot 3$ .

Factor $27$ out of $81$ .

$y = \frac{\sqrt[3]{- x + 5} \sqrt[3]{27 (3)}}{9}$

Rewrite $27$ as $3^{3}$ .

$y = \frac{\sqrt[3]{- x + 5} \sqrt[3]{3^{3} \cdot 3}}{9}$

Pull terms out from under the radical.

$y = \frac{\sqrt[3]{- x + 5} \cdot 3 \sqrt[3]{3}}{9}$

Combine using the product rule for radicals.

$y = \frac{3 \sqrt[3]{(- x + 5) \cdot 3}}{9}$

Reduce the expression by cancelling the common factors.

Cancel the common factor of $3$ and $9$ .

Factor $3$ out of $3 \sqrt[3]{(- x + 5) \cdot 3}$ .

$y = \frac{3 (\sqrt[3]{(- x + 5) \cdot 3})}{9}$

Cancel the common factors.

Factor $3$ out of $9$ .

$y = \frac{3 \sqrt[3]{(- x + 5) \cdot 3}}{3 \cdot 3}$

Cancel the common factor.

$y = \frac{3 \sqrt[3]{(- x + 5) \cdot 3}}{3 \cdot 3}$

Rewrite the expression.

$y = \frac{\sqrt[3]{(- x + 5) \cdot 3}}{3}$

Reorder factors in $\frac{\sqrt[3]{(- x + 5) \cdot 3}}{3}$ .

$y = \frac{\sqrt[3]{3 (- x + 5)}}{3}$

Step 3

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

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

Step 4

Verify if $f^{- 1} (x) = \frac{\sqrt[3]{3 (- x + 5)}}{3}$ is the inverse of $f (x) = 5 - 9 x^{3}$ .

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 $f^{- 1} (5 - 9 x^{3})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (5 - 9 x^{3}) = \frac{\sqrt[3]{3 (- (5 - 9 x^{3}) + 5)}}{3}$

Simplify the numerator.

Apply the distributive property.

$f^{- 1} (5 - 9 x^{3}) = \frac{\sqrt[3]{3 (- 1 \cdot 5 - (- 9 x^{3}) + 5)}}{3}$

Multiply $- 1$ by $5$ .

$f^{- 1} (5 - 9 x^{3}) = \frac{\sqrt[3]{3 (- 5 - (- 9 x^{3}) + 5)}}{3}$

Multiply $- 9$ by $- 1$ .

$f^{- 1} (5 - 9 x^{3}) = \frac{\sqrt[3]{3 (- 5 + 9 x^{3} + 5)}}{3}$

Add $- 5$ and $5$ .

$f^{- 1} (5 - 9 x^{3}) = \frac{\sqrt[3]{3 (9 x^{3} + 0)}}{3}$

Add $9 x^{3}$ and $0$ .

$f^{- 1} (5 - 9 x^{3}) = \frac{\sqrt[3]{3 \cdot (9 x^{3})}}{3}$

Multiply $3$ by $9$ .

$f^{- 1} (5 - 9 x^{3}) = \frac{\sqrt[3]{27 x^{3}}}{3}$

Rewrite $27 x^{3}$ as ${(3 x)}^{3}$ .

$f^{- 1} (5 - 9 x^{3}) = \frac{\sqrt[3]{{(3 x)}^{3}}}{3}$

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

$f^{- 1} (5 - 9 x^{3}) = \frac{3 x}{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$f^{- 1} (5 - 9 x^{3}) = \frac{3 x}{3}$

Divide $x$ by $1$ .

$f^{- 1} (5 - 9 x^{3}) = x$

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

Set up the composite result function.

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

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

$f (\frac{\sqrt[3]{3 (- x + 5)}}{3}) = 5 - 9 {(\frac{\sqrt[3]{3 (- x + 5)}}{3})}^{3}$

Simplify each term.

Apply the product rule to $\frac{\sqrt[3]{3 (- x + 5)}}{3}$ .

$f (\frac{\sqrt[3]{3 (- x + 5)}}{3}) = 5 - 9 \frac{{\sqrt[3]{3 (- x + 5)}}^{3}}{3^{3}}$

Simplify the numerator.

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

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

$f (\frac{\sqrt[3]{3 (- x + 5)}}{3}) = 5 - 9 \frac{{({(3 (- x + 5))}^{\frac{1}{3}})}^{3}}{3^{3}}$

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

$f (\frac{\sqrt[3]{3 (- x + 5)}}{3}) = 5 - 9 \frac{{(3 (- x + 5))}^{\frac{1}{3} \cdot 3}}{3^{3}}$

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

$f (\frac{\sqrt[3]{3 (- x + 5)}}{3}) = 5 - 9 \frac{{(3 (- x + 5))}^{\frac{3}{3}}}{3^{3}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (\frac{\sqrt[3]{3 (- x + 5)}}{3}) = 5 - 9 \frac{{(3 (- x + 5))}^{\frac{3}{3}}}{3^{3}}$

Rewrite the expression.

$f (\frac{\sqrt[3]{3 (- x + 5)}}{3}) = 5 - 9 \frac{3 (- x + 5)}{3^{3}}$

Simplify.

$f (\frac{\sqrt[3]{3 (- x + 5)}}{3}) = 5 - 9 \frac{3 (- x + 5)}{3^{3}}$

Apply the distributive property.

$f (\frac{\sqrt[3]{3 (- x + 5)}}{3}) = 5 - 9 \frac{3 (- x) + 3 \cdot 5}{3^{3}}$

Multiply $- 1$ by $3$ .

$f (\frac{\sqrt[3]{3 (- x + 5)}}{3}) = 5 - 9 \frac{- 3 x + 3 \cdot 5}{3^{3}}$

Multiply $3$ by $5$ .

$f (\frac{\sqrt[3]{3 (- x + 5)}}{3}) = 5 - 9 \frac{- 3 x + 15}{3^{3}}$

Factor $3$ out of $- 3 x + 15$ .

Factor $3$ out of $- 3 x$ .

$f (\frac{\sqrt[3]{3 (- x + 5)}}{3}) = 5 - 9 \frac{3 (- x) + 15}{3^{3}}$

Factor $3$ out of $15$ .

$f (\frac{\sqrt[3]{3 (- x + 5)}}{3}) = 5 - 9 \frac{3 (- x) + 3 (5)}{3^{3}}$

Factor $3$ out of $3 (- x) + 3 (5)$ .

$f (\frac{\sqrt[3]{3 (- x + 5)}}{3}) = 5 - 9 \frac{3 (- x + 5)}{3^{3}}$

Raise $3$ to the power of $3$ .

$f (\frac{\sqrt[3]{3 (- x + 5)}}{3}) = 5 - 9 \frac{3 (- x + 5)}{27}$

Cancel the common factor of $9$ .

Factor $9$ out of $- 9$ .

$f (\frac{\sqrt[3]{3 (- x + 5)}}{3}) = 5 + 9 (- 1) (\frac{3 (- x + 5)}{27})$

Factor $9$ out of $27$ .

$f (\frac{\sqrt[3]{3 (- x + 5)}}{3}) = 5 + 9 \cdot (- 1 \frac{3 (- x + 5)}{9 \cdot 3})$

Cancel the common factor.

$f (\frac{\sqrt[3]{3 (- x + 5)}}{3}) = 5 + 9 \cdot (- 1 \frac{3 (- x + 5)}{9 \cdot 3})$

Rewrite the expression.

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $- x + 5$ by $1$ .

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

Apply the distributive property.

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

Multiply $- 1 (- x)$ .

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

Multiply $- 1$ by $5$ .

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

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

Subtract $5$ from $5$ .

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

Add $x$ and $0$ .

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

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\sqrt[3]{3 (- x + 5)}}{3}$ is the inverse of $f (x) = 5 - 9 x^{3}$ .

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

Find the Inverse 5-9x^3

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