Find the Inverse g(x)=1/3* cube root of 2x+5

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

Step 1

Write $g (x) = \frac{1}{3} \cdot \sqrt[3]{2 x + 5}$ as an equation.

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

Rewrite the equation as $\frac{1}{3} \cdot \sqrt[3]{2 y + 5} = x$ .

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

Multiply both sides of the equation by $3$ .

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

Simplify the left side.

Simplify $3 (\frac{1}{3} \cdot \sqrt[3]{2 y + 5})$ .

Combine $\frac{1}{3}$ and $\sqrt[3]{2 y + 5}$ .

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

To remove the radical on the left side of the equation, cube both sides of the equation.

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

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{2 y + 5}$ as ${(2 y + 5)}^{\frac{1}{3}}$ .

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

Simplify the left side.

Simplify ${({(2 y + 5)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(2 y + 5)}^{\frac{1}{3}})}^{3}$ .

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

${(2 y + 5)}^{1} = {(3 x)}^{3}$

Simplify.

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

Simplify the right side.

Simplify ${(3 x)}^{3}$ .

Apply the product rule to $3 x$ .

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

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

$2 y + 5 = 27 x^{3}$

Solve for $y$ .

Subtract $5$ from both sides of the equation.

$2 y = 27 x^{3} - 5$

Divide each term in $2 y = 27 x^{3} - 5$ by $2$ and simplify.

Divide each term in $2 y = 27 x^{3} - 5$ by $2$ .

$\frac{2 y}{2} = \frac{27 x^{3}}{2} + \frac{- 5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y}{2} = \frac{27 x^{3}}{2} + \frac{- 5}{2}$

Divide $y$ by $1$ .

$y = \frac{27 x^{3}}{2} + \frac{- 5}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$y = \frac{27 x^{3}}{2} - \frac{5}{2}$

Step 4

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

$g^{- 1} (x) = \frac{27 x^{3}}{2} - \frac{5}{2}$

Step 5

Verify if $g^{- 1} (x) = \frac{27 x^{3}}{2} - \frac{5}{2}$ is the inverse of $g (x) = \frac{1}{3} \cdot \sqrt[3]{2 x + 5}$ .

To verify the inverse, check if $g^{- 1} (g (x)) = x$ and $g (g^{- 1} (x)) = x$ .

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

Set up the composite result function.

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

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

$g^{- 1} (\frac{1}{3} \cdot \sqrt[3]{2 x + 5}) = \frac{27 {(\frac{1}{3} \cdot \sqrt[3]{2 x + 5})}^{3}}{2} - \frac{5}{2}$

Combine the numerators over the common denominator.

$g^{- 1} (\frac{1}{3} \cdot \sqrt[3]{2 x + 5}) = \frac{27 {(\frac{1}{3} \cdot \sqrt[3]{2 x + 5})}^{3} - 5}{2}$

Simplify each term.

Combine $\frac{1}{3}$ and $\sqrt[3]{2 x + 5}$ .

$g^{- 1} (\frac{1}{3} \cdot \sqrt[3]{2 x + 5}) = \frac{27 {(\frac{\sqrt[3]{2 x + 5}}{3})}^{3} - 5}{2}$

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

$g^{- 1} (\frac{1}{3} \cdot \sqrt[3]{2 x + 5}) = \frac{27 (\frac{{\sqrt[3]{2 x + 5}}^{3}}{3^{3}}) - 5}{2}$

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

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{2 x + 5}$ as ${(2 x + 5)}^{\frac{1}{3}}$ .

$g^{- 1} (\frac{1}{3} \cdot \sqrt[3]{2 x + 5}) = \frac{27 (\frac{{({(2 x + 5)}^{\frac{1}{3}})}^{3}}{3^{3}}) - 5}{2}$

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

$g^{- 1} (\frac{1}{3} \cdot \sqrt[3]{2 x + 5}) = \frac{27 (\frac{{(2 x + 5)}^{\frac{1}{3} \cdot 3}}{3^{3}}) - 5}{2}$

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

$g^{- 1} (\frac{1}{3} \cdot \sqrt[3]{2 x + 5}) = \frac{27 (\frac{{(2 x + 5)}^{\frac{3}{3}}}{3^{3}}) - 5}{2}$

Cancel the common factor of $3$ .

Cancel the common factor.

$g^{- 1} (\frac{1}{3} \cdot \sqrt[3]{2 x + 5}) = \frac{27 (\frac{{(2 x + 5)}^{\frac{3}{3}}}{3^{3}}) - 5}{2}$

Rewrite the expression.

$g^{- 1} (\frac{1}{3} \cdot \sqrt[3]{2 x + 5}) = \frac{27 (\frac{2 x + 5}{3^{3}}) - 5}{2}$

Simplify.

$g^{- 1} (\frac{1}{3} \cdot \sqrt[3]{2 x + 5}) = \frac{27 (\frac{2 x + 5}{3^{3}}) - 5}{2}$

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

$g^{- 1} (\frac{1}{3} \cdot \sqrt[3]{2 x + 5}) = \frac{27 (\frac{2 x + 5}{27}) - 5}{2}$

Cancel the common factor of $27$ .

Cancel the common factor.

$g^{- 1} (\frac{1}{3} \cdot \sqrt[3]{2 x + 5}) = \frac{27 (\frac{2 x + 5}{27}) - 5}{2}$

Rewrite the expression.

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

Simplify terms.

Combine the opposite terms in $2 x + 5 - 5$ .

Subtract $5$ from $5$ .

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

Add $2 x$ and $0$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

Evaluate $g (\frac{27 x^{3}}{2} - \frac{5}{2})$ by substituting in the value of $g^{- 1}$ into $g$ .

$g (\frac{27 x^{3}}{2} - \frac{5}{2}) = \frac{1}{3} \cdot \sqrt[3]{2 (\frac{27 x^{3}}{2} - \frac{5}{2}) + 5}$

Apply the distributive property.

$g (\frac{27 x^{3}}{2} - \frac{5}{2}) = \frac{1}{3} \cdot \sqrt[3]{2 (\frac{27 x^{3}}{2}) + 2 (- \frac{5}{2}) + 5}$

Cancel the common factor of $2$ .

Cancel the common factor.

$g (\frac{27 x^{3}}{2} - \frac{5}{2}) = \frac{1}{3} \cdot \sqrt[3]{2 (\frac{27 x^{3}}{2}) + 2 (- \frac{5}{2}) + 5}$

Rewrite the expression.

$g (\frac{27 x^{3}}{2} - \frac{5}{2}) = \frac{1}{3} \cdot \sqrt[3]{27 x^{3} + 2 (- \frac{5}{2}) + 5}$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{5}{2}$ into the numerator.

$g (\frac{27 x^{3}}{2} - \frac{5}{2}) = \frac{1}{3} \cdot \sqrt[3]{27 x^{3} + 2 (\frac{- 5}{2}) + 5}$

Cancel the common factor.

$g (\frac{27 x^{3}}{2} - \frac{5}{2}) = \frac{1}{3} \cdot \sqrt[3]{27 x^{3} + 2 (\frac{- 5}{2}) + 5}$

Rewrite the expression.

$g (\frac{27 x^{3}}{2} - \frac{5}{2}) = \frac{1}{3} \cdot \sqrt[3]{27 x^{3} - 5 + 5}$

Simplify by adding numbers.

Add $- 5$ and $5$ .

$g (\frac{27 x^{3}}{2} - \frac{5}{2}) = \frac{1}{3} \cdot \sqrt[3]{27 x^{3} + 0}$

Add $27 x^{3}$ and $0$ .

$g (\frac{27 x^{3}}{2} - \frac{5}{2}) = \frac{1}{3} \cdot \sqrt[3]{27 x^{3}}$

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

$g (\frac{27 x^{3}}{2} - \frac{5}{2}) = \frac{1}{3} \cdot \sqrt[3]{{(3 x)}^{3}}$

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

$g (\frac{27 x^{3}}{2} - \frac{5}{2}) = \frac{1}{3} \cdot (3 x)$

Cancel the common factor of $3$ .

Factor $3$ out of $3 x$ .

$g (\frac{27 x^{3}}{2} - \frac{5}{2}) = \frac{1}{3} \cdot (3 (x))$

Cancel the common factor.

$g (\frac{27 x^{3}}{2} - \frac{5}{2}) = \frac{1}{3} \cdot (3 x)$

Rewrite the expression.

$g (\frac{27 x^{3}}{2} - \frac{5}{2}) = x$

Since $g^{- 1} (g (x)) = x$ and $g (g^{- 1} (x)) = x$ , then $g^{- 1} (x) = \frac{27 x^{3}}{2} - \frac{5}{2}$ is the inverse of $g (x) = \frac{1}{3} \cdot \sqrt[3]{2 x + 5}$ .

$g^{- 1} (x) = \frac{27 x^{3}}{2} - \frac{5}{2}$

Find the Inverse g(x)=1/3* cube root of 2x+5

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