Find the Inverse y=5x^3

$y = 5 x^{3}$

Step 1

Interchange the variables.

$x = 5 y^{3}$

Step 2

Solve for $y$ .

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

$5 y^{3} = x$

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

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

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

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

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

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

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

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

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

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

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

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

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

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

Combine and simplify the denominator.

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

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

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

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

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

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

Add $1$ and $2$ .

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

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

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

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

Simplify the numerator.

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

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

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

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

Simplify with factoring out.

Combine using the product rule for radicals.

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

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

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

Step 3

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

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

Step 4

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

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

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

Simplify the numerator.

Multiply $25$ by $5$ .

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

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

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

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

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

Cancel the common factor of $5$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

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

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

Rewrite ${\sqrt[3]{25 x}}^{3}$ as $25 x$ .

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

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

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

Cancel the common factor of $5$ .

Factor $5$ out of $125$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $25$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

Find the Inverse y=5x^3

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