Find the Inverse y=7x^3+2

$y = 7 x^{3} + 2$

Step 1

Interchange the variables.

$x = 7 y^{3} + 2$

Step 2

Solve for $y$ .

Rewrite the equation as $7 y^{3} + 2 = x$ .

$7 y^{3} + 2 = x$

Subtract $2$ from both sides of the equation.

$7 y^{3} = x - 2$

Divide each term in $7 y^{3} = x - 2$ by $7$ and simplify.

Divide each term in $7 y^{3} = x - 2$ by $7$ .

$\frac{7 y^{3}}{7} = \frac{x}{7} + \frac{- 2}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 y^{3}}{7} = \frac{x}{7} + \frac{- 2}{7}$

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

$y^{3} = \frac{x}{7} + \frac{- 2}{7}$

Simplify the right side.

Move the negative in front of the fraction.

$y^{3} = \frac{x}{7} - \frac{2}{7}$

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

$y = \sqrt[3]{\frac{x}{7} - \frac{2}{7}}$

Simplify $\sqrt[3]{\frac{x}{7} - \frac{2}{7}}$ .

Combine the numerators over the common denominator.

$y = \sqrt[3]{\frac{x - 2}{7}}$

Rewrite $\sqrt[3]{\frac{x - 2}{7}}$ as $\frac{\sqrt[3]{x - 2}}{\sqrt[3]{7}}$ .

$y = \frac{\sqrt[3]{x - 2}}{\sqrt[3]{7}}$

Multiply $\frac{\sqrt[3]{x - 2}}{\sqrt[3]{7}}$ by $\frac{{\sqrt[3]{7}}^{2}}{{\sqrt[3]{7}}^{2}}$ .

$y = \frac{\sqrt[3]{x - 2}}{\sqrt[3]{7}} \cdot \frac{{\sqrt[3]{7}}^{2}}{{\sqrt[3]{7}}^{2}}$

Combine and simplify the denominator.

Multiply $\frac{\sqrt[3]{x - 2}}{\sqrt[3]{7}}$ by $\frac{{\sqrt[3]{7}}^{2}}{{\sqrt[3]{7}}^{2}}$ .

$y = \frac{\sqrt[3]{x - 2} {\sqrt[3]{7}}^{2}}{\sqrt[3]{7} {\sqrt[3]{7}}^{2}}$

Raise $\sqrt[3]{7}$ to the power of $1$ .

$y = \frac{\sqrt[3]{x - 2} {\sqrt[3]{7}}^{2}}{{\sqrt[3]{7}}^{1} {\sqrt[3]{7}}^{2}}$

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

$y = \frac{\sqrt[3]{x - 2} {\sqrt[3]{7}}^{2}}{{\sqrt[3]{7}}^{1 + 2}}$

Add $1$ and $2$ .

$y = \frac{\sqrt[3]{x - 2} {\sqrt[3]{7}}^{2}}{{\sqrt[3]{7}}^{3}}$

Rewrite ${\sqrt[3]{7}}^{3}$ as $7$ .

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

$y = \frac{\sqrt[3]{x - 2} {\sqrt[3]{7}}^{2}}{{(7^{\frac{1}{3}})}^{3}}$

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

$y = \frac{\sqrt[3]{x - 2} {\sqrt[3]{7}}^{2}}{7^{\frac{1}{3} \cdot 3}}$

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

$y = \frac{\sqrt[3]{x - 2} {\sqrt[3]{7}}^{2}}{7^{\frac{3}{3}}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$y = \frac{\sqrt[3]{x - 2} {\sqrt[3]{7}}^{2}}{7^{\frac{3}{3}}}$

Rewrite the expression.

$y = \frac{\sqrt[3]{x - 2} {\sqrt[3]{7}}^{2}}{7^{1}}$

Evaluate the exponent.

$y = \frac{\sqrt[3]{x - 2} {\sqrt[3]{7}}^{2}}{7}$

Simplify the numerator.

Rewrite ${\sqrt[3]{7}}^{2}$ as $\sqrt[3]{7^{2}}$ .

$y = \frac{\sqrt[3]{x - 2} \sqrt[3]{7^{2}}}{7}$

Raise $7$ to the power of $2$ .

$y = \frac{\sqrt[3]{x - 2} \sqrt[3]{49}}{7}$

Simplify with factoring out.

Combine using the product rule for radicals.

$y = \frac{\sqrt[3]{(x - 2) \cdot 49}}{7}$

Reorder factors in $\frac{\sqrt[3]{(x - 2) \cdot 49}}{7}$ .

$y = \frac{\sqrt[3]{49 (x - 2)}}{7}$

Step 3

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

$f^{- 1} (x) = \frac{\sqrt[3]{49 (x - 2)}}{7}$

Step 4

Verify if $f^{- 1} (x) = \frac{\sqrt[3]{49 (x - 2)}}{7}$ is the inverse of $f (x) = 7 x^{3} + 2$ .

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 $f^{- 1} (7 x^{3} + 2)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (7 x^{3} + 2) = \frac{\sqrt[3]{49 ((7 x^{3} + 2) - 2)}}{7}$

Simplify the numerator.

Subtract $2$ from $2$ .

$f^{- 1} (7 x^{3} + 2) = \frac{\sqrt[3]{49 (7 x^{3} + 0)}}{7}$

Add $7 x^{3}$ and $0$ .

$f^{- 1} (7 x^{3} + 2) = \frac{\sqrt[3]{49 \cdot (7 x^{3})}}{7}$

Multiply $49$ by $7$ .

$f^{- 1} (7 x^{3} + 2) = \frac{\sqrt[3]{343 x^{3}}}{7}$

Rewrite $343 x^{3}$ as ${(7 x)}^{3}$ .

$f^{- 1} (7 x^{3} + 2) = \frac{\sqrt[3]{{(7 x)}^{3}}}{7}$

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

$f^{- 1} (7 x^{3} + 2) = \frac{7 x}{7}$

Cancel the common factor of $7$ .

Cancel the common factor.

$f^{- 1} (7 x^{3} + 2) = \frac{7 x}{7}$

Divide $x$ by $1$ .

$f^{- 1} (7 x^{3} + 2) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{\sqrt[3]{49 (x - 2)}}{7})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\sqrt[3]{49 (x - 2)}}{7}) = 7 {(\frac{\sqrt[3]{49 (x - 2)}}{7})}^{3} + 2$

Simplify each term.

Apply the product rule to $\frac{\sqrt[3]{49 (x - 2)}}{7}$ .

$f (\frac{\sqrt[3]{49 (x - 2)}}{7}) = 7 (\frac{{\sqrt[3]{49 (x - 2)}}^{3}}{7^{3}}) + 2$

Simplify the numerator.

Rewrite ${\sqrt[3]{49 (x - 2)}}^{3}$ as $49 (x - 2)$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{49 (x - 2)}$ as ${(49 (x - 2))}^{\frac{1}{3}}$ .

$f (\frac{\sqrt[3]{49 (x - 2)}}{7}) = 7 (\frac{{({(49 (x - 2))}^{\frac{1}{3}})}^{3}}{7^{3}}) + 2$

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

$f (\frac{\sqrt[3]{49 (x - 2)}}{7}) = 7 (\frac{{(49 (x - 2))}^{\frac{1}{3} \cdot 3}}{7^{3}}) + 2$

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

$f (\frac{\sqrt[3]{49 (x - 2)}}{7}) = 7 (\frac{{(49 (x - 2))}^{\frac{3}{3}}}{7^{3}}) + 2$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (\frac{\sqrt[3]{49 (x - 2)}}{7}) = 7 (\frac{{(49 (x - 2))}^{\frac{3}{3}}}{7^{3}}) + 2$

Rewrite the expression.

$f (\frac{\sqrt[3]{49 (x - 2)}}{7}) = 7 (\frac{49 (x - 2)}{7^{3}}) + 2$

Simplify.

$f (\frac{\sqrt[3]{49 (x - 2)}}{7}) = 7 (\frac{49 (x - 2)}{7^{3}}) + 2$

Apply the distributive property.

$f (\frac{\sqrt[3]{49 (x - 2)}}{7}) = 7 (\frac{49 x + 49 \cdot - 2}{7^{3}}) + 2$

Multiply $49$ by $- 2$ .

$f (\frac{\sqrt[3]{49 (x - 2)}}{7}) = 7 (\frac{49 x - 98}{7^{3}}) + 2$

Factor $49$ out of $49 x - 98$ .

Factor $49$ out of $49 x$ .

$f (\frac{\sqrt[3]{49 (x - 2)}}{7}) = 7 (\frac{49 (x) - 98}{7^{3}}) + 2$

Factor $49$ out of $- 98$ .

$f (\frac{\sqrt[3]{49 (x - 2)}}{7}) = 7 (\frac{49 x + 49 \cdot - 2}{7^{3}}) + 2$

Factor $49$ out of $49 x + 49 \cdot - 2$ .

$f (\frac{\sqrt[3]{49 (x - 2)}}{7}) = 7 (\frac{49 (x - 2)}{7^{3}}) + 2$

Raise $7$ to the power of $3$ .

$f (\frac{\sqrt[3]{49 (x - 2)}}{7}) = 7 (\frac{49 (x - 2)}{343}) + 2$

Cancel the common factor of $7$ .

Factor $7$ out of $343$ .

$f (\frac{\sqrt[3]{49 (x - 2)}}{7}) = 7 (\frac{49 (x - 2)}{7 (49)}) + 2$

Cancel the common factor.

$f (\frac{\sqrt[3]{49 (x - 2)}}{7}) = 7 (\frac{49 (x - 2)}{7 \cdot 49}) + 2$

Rewrite the expression.

$f (\frac{\sqrt[3]{49 (x - 2)}}{7}) = \frac{49 (x - 2)}{49} + 2$

Cancel the common factor of $49$ .

Cancel the common factor.

$f (\frac{\sqrt[3]{49 (x - 2)}}{7}) = \frac{49 (x - 2)}{49} + 2$

Divide $x - 2$ by $1$ .

$f (\frac{\sqrt[3]{49 (x - 2)}}{7}) = x - 2 + 2$

Combine the opposite terms in $x - 2 + 2$ .

Add $- 2$ and $2$ .

$f (\frac{\sqrt[3]{49 (x - 2)}}{7}) = x + 0$

Add $x$ and $0$ .

$f (\frac{\sqrt[3]{49 (x - 2)}}{7}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\sqrt[3]{49 (x - 2)}}{7}$ is the inverse of $f (x) = 7 x^{3} + 2$ .

$f^{- 1} (x) = \frac{\sqrt[3]{49 (x - 2)}}{7}$

Find the Inverse y=7x^3+2

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