Find the Inverse 5x^3-2

$5 x^{3} - 2$

Step 1

Interchange the variables.

$x = 5 y^{3} - 2$

Step 2

Solve for $y$ .

Rewrite the equation as $5 y^{3} - 2 = x$ .

$5 y^{3} - 2 = x$

Add $2$ to both sides of the equation.

$5 y^{3} = x + 2$

Divide each term in $5 y^{3} = x + 2$ by $5$ and simplify.

Divide each term in $5 y^{3} = x + 2$ by $5$ .

$\frac{5 y^{3}}{5} = \frac{x}{5} + \frac{2}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 y^{3}}{5} = \frac{x}{5} + \frac{2}{5}$

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

$y^{3} = \frac{x}{5} + \frac{2}{5}$

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

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

Simplify $\sqrt[3]{\frac{x}{5} + \frac{2}{5}}$ .

Combine the numerators over the common denominator.

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

Rewrite $\sqrt[3]{\frac{x + 2}{5}}$ as $\frac{\sqrt[3]{x + 2}}{\sqrt[3]{5}}$ .

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

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

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

Combine and simplify the denominator.

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

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

Raise $\sqrt[3]{5}$ to the power of $1$ .

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

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

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

Add $1$ and $2$ .

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

Rewrite ${\sqrt[3]{5}}^{3}$ as $5$ .

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

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

Simplify the numerator.

Rewrite ${\sqrt[3]{5}}^{2}$ as $\sqrt[3]{5^{2}}$ .

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

Raise $5$ to the power of $2$ .

$y = \frac{\sqrt[3]{x + 2} \sqrt[3]{25}}{5}$

Simplify with factoring out.

Combine using the product rule for radicals.

$y = \frac{\sqrt[3]{(x + 2) \cdot 25}}{5}$

Reorder factors in $\frac{\sqrt[3]{(x + 2) \cdot 25}}{5}$ .

$y = \frac{\sqrt[3]{25 (x + 2)}}{5}$

Step 3

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

$f^{- 1} (x) = \frac{\sqrt[3]{25 (x + 2)}}{5}$

Step 4

Verify if $f^{- 1} (x) = \frac{\sqrt[3]{25 (x + 2)}}{5}$ is the inverse of $f (x) = 5 x^{3} - 2$ .

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 $f^{- 1} (5 x^{3} - 2)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (5 x^{3} - 2) = \frac{\sqrt[3]{25 ((5 x^{3} - 2) + 2)}}{5}$

Simplify the numerator.

Add $- 2$ and $2$ .

$f^{- 1} (5 x^{3} - 2) = \frac{\sqrt[3]{25 (5 x^{3} + 0)}}{5}$

Add $5 x^{3}$ and $0$ .

$f^{- 1} (5 x^{3} - 2) = \frac{\sqrt[3]{25 \cdot (5 x^{3})}}{5}$

Multiply $25$ by $5$ .

$f^{- 1} (5 x^{3} - 2) = \frac{\sqrt[3]{125 x^{3}}}{5}$

Rewrite $125 x^{3}$ as ${(5 x)}^{3}$ .

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

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

$f^{- 1} (5 x^{3} - 2) = \frac{5 x}{5}$

Cancel the common factor of $5$ .

Cancel the common factor.

$f^{- 1} (5 x^{3} - 2) = \frac{5 x}{5}$

Divide $x$ by $1$ .

$f^{- 1} (5 x^{3} - 2) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{\sqrt[3]{25 (x + 2)}}{5})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\sqrt[3]{25 (x + 2)}}{5}) = 5 {(\frac{\sqrt[3]{25 (x + 2)}}{5})}^{3} - 2$

Simplify each term.

Apply the product rule to $\frac{\sqrt[3]{25 (x + 2)}}{5}$ .

$f (\frac{\sqrt[3]{25 (x + 2)}}{5}) = 5 (\frac{{\sqrt[3]{25 (x + 2)}}^{3}}{5^{3}}) - 2$

Simplify the numerator.

Rewrite ${\sqrt[3]{25 (x + 2)}}^{3}$ as $25 (x + 2)$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{25 (x + 2)}$ as ${(25 (x + 2))}^{\frac{1}{3}}$ .

$f (\frac{\sqrt[3]{25 (x + 2)}}{5}) = 5 (\frac{{({(25 (x + 2))}^{\frac{1}{3}})}^{3}}{5^{3}}) - 2$

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

$f (\frac{\sqrt[3]{25 (x + 2)}}{5}) = 5 (\frac{{(25 (x + 2))}^{\frac{1}{3} \cdot 3}}{5^{3}}) - 2$

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

$f (\frac{\sqrt[3]{25 (x + 2)}}{5}) = 5 (\frac{{(25 (x + 2))}^{\frac{3}{3}}}{5^{3}}) - 2$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (\frac{\sqrt[3]{25 (x + 2)}}{5}) = 5 (\frac{{(25 (x + 2))}^{\frac{3}{3}}}{5^{3}}) - 2$

Rewrite the expression.

$f (\frac{\sqrt[3]{25 (x + 2)}}{5}) = 5 (\frac{25 (x + 2)}{5^{3}}) - 2$

Simplify.

$f (\frac{\sqrt[3]{25 (x + 2)}}{5}) = 5 (\frac{25 (x + 2)}{5^{3}}) - 2$

Apply the distributive property.

$f (\frac{\sqrt[3]{25 (x + 2)}}{5}) = 5 (\frac{25 x + 25 \cdot 2}{5^{3}}) - 2$

Multiply $25$ by $2$ .

$f (\frac{\sqrt[3]{25 (x + 2)}}{5}) = 5 (\frac{25 x + 50}{5^{3}}) - 2$

Factor $25$ out of $25 x + 50$ .

Factor $25$ out of $25 x$ .

$f (\frac{\sqrt[3]{25 (x + 2)}}{5}) = 5 (\frac{25 (x) + 50}{5^{3}}) - 2$

Factor $25$ out of $50$ .

$f (\frac{\sqrt[3]{25 (x + 2)}}{5}) = 5 (\frac{25 x + 25 \cdot 2}{5^{3}}) - 2$

Factor $25$ out of $25 x + 25 \cdot 2$ .

$f (\frac{\sqrt[3]{25 (x + 2)}}{5}) = 5 (\frac{25 (x + 2)}{5^{3}}) - 2$

Raise $5$ to the power of $3$ .

$f (\frac{\sqrt[3]{25 (x + 2)}}{5}) = 5 (\frac{25 (x + 2)}{125}) - 2$

Cancel the common factor of $5$ .

Factor $5$ out of $125$ .

$f (\frac{\sqrt[3]{25 (x + 2)}}{5}) = 5 (\frac{25 (x + 2)}{5 (25)}) - 2$

Cancel the common factor.

$f (\frac{\sqrt[3]{25 (x + 2)}}{5}) = 5 (\frac{25 (x + 2)}{5 \cdot 25}) - 2$

Rewrite the expression.

$f (\frac{\sqrt[3]{25 (x + 2)}}{5}) = \frac{25 (x + 2)}{25} - 2$

Cancel the common factor of $25$ .

Cancel the common factor.

$f (\frac{\sqrt[3]{25 (x + 2)}}{5}) = \frac{25 (x + 2)}{25} - 2$

Divide $x + 2$ by $1$ .

$f (\frac{\sqrt[3]{25 (x + 2)}}{5}) = x + 2 - 2$

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

Subtract $2$ from $2$ .

$f (\frac{\sqrt[3]{25 (x + 2)}}{5}) = x + 0$

Add $x$ and $0$ .

$f (\frac{\sqrt[3]{25 (x + 2)}}{5}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\sqrt[3]{25 (x + 2)}}{5}$ is the inverse of $f (x) = 5 x^{3} - 2$ .

$f^{- 1} (x) = \frac{\sqrt[3]{25 (x + 2)}}{5}$

Find the Inverse 5x^3-2

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