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

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

Step 1

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

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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

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

Subtract $5$ from both sides of the equation.

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

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

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

Simplify each side of the equation.

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

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

Simplify the left side.

Simplify ${(\frac{1}{2} \cdot {(y - 3)}^{\frac{1}{3}})}^{3}$ .

Combine fractions.

Combine $\frac{1}{2}$ and ${(y - 3)}^{\frac{1}{3}}$ .

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

Apply the product rule to $\frac{{(y - 3)}^{\frac{1}{3}}}{2}$ .

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

Simplify the numerator.

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Raise $2$ to the power of $3$ .

$\frac{y - 3}{8} = {(x - 5)}^{3}$

Simplify the right side.

Simplify ${(x - 5)}^{3}$ .

Use the Binomial Theorem.

$\frac{y - 3}{8} = x^{3} + 3 x^{2} \cdot - 5 + 3 x {(- 5)}^{2} + {(- 5)}^{3}$

Simplify each term.

Multiply $- 5$ by $3$ .

$\frac{y - 3}{8} = x^{3} - 15 x^{2} + 3 x {(- 5)}^{2} + {(- 5)}^{3}$

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

$\frac{y - 3}{8} = x^{3} - 15 x^{2} + 3 x \cdot 25 + {(- 5)}^{3}$

Multiply $25$ by $3$ .

$\frac{y - 3}{8} = x^{3} - 15 x^{2} + 75 x + {(- 5)}^{3}$

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

$\frac{y - 3}{8} = x^{3} - 15 x^{2} + 75 x - 125$

Solve for $y$ .

Multiply both sides of the equation by $8$ .

$8 \frac{y - 3}{8} = 8 (x^{3} - 15 x^{2} + 75 x - 125)$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$8 \frac{y - 3}{8} = 8 (x^{3} - 15 x^{2} + 75 x - 125)$

Rewrite the expression.

$y - 3 = 8 (x^{3} - 15 x^{2} + 75 x - 125)$

Simplify the right side.

Simplify $8 (x^{3} - 15 x^{2} + 75 x - 125)$ .

Apply the distributive property.

$y - 3 = 8 x^{3} + 8 (- 15 x^{2}) + 8 (75 x) + 8 \cdot - 125$

Simplify.

Multiply $- 15$ by $8$ .

$y - 3 = 8 x^{3} - 120 x^{2} + 8 (75 x) + 8 \cdot - 125$

Multiply $75$ by $8$ .

$y - 3 = 8 x^{3} - 120 x^{2} + 600 x + 8 \cdot - 125$

Multiply $8$ by $- 125$ .

$y - 3 = 8 x^{3} - 120 x^{2} + 600 x - 1000$

Move all terms not containing $y$ to the right side of the equation.

Add $3$ to both sides of the equation.

$y = 8 x^{3} - 120 x^{2} + 600 x - 1000 + 3$

Add $- 1000$ and $3$ .

$y = 8 x^{3} - 120 x^{2} + 600 x - 997$

Step 4

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

$f^{- 1} (x) = 8 x^{3} - 120 x^{2} + 600 x - 997$

Step 5

Verify if $f^{- 1} (x) = 8 x^{3} - 120 x^{2} + 600 x - 997$ is the inverse of $f (x) = \frac{1}{2} \cdot \sqrt[3]{x - 3} + 5$ .

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

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = 8 {(\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5)}^{3} - 120 {(\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5)}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Simplify each term.

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

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = 8 {(\frac{\sqrt[3]{x - 3}}{2} + 5)}^{3} - 120 {(\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5)}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

To write $5$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = 8 {(\frac{\sqrt[3]{x - 3}}{2} + 5 \cdot \frac{2}{2})}^{3} - 120 {(\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5)}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Combine $5$ and $\frac{2}{2}$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = 8 {(\frac{\sqrt[3]{x - 3}}{2} + \frac{5 \cdot 2}{2})}^{3} - 120 {(\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5)}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = 8 {(\frac{\sqrt[3]{x - 3} + 5 \cdot 2}{2})}^{3} - 120 {(\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5)}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Multiply $5$ by $2$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = 8 {(\frac{\sqrt[3]{x - 3} + 10}{2})}^{3} - 120 {(\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5)}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Apply the product rule to $\frac{\sqrt[3]{x - 3} + 10}{2}$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = 8 (\frac{{(\sqrt[3]{x - 3} + 10)}^{3}}{2^{3}}) - 120 {(\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5)}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Raise $2$ to the power of $3$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = 8 (\frac{{(\sqrt[3]{x - 3} + 10)}^{3}}{8}) - 120 {(\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5)}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Cancel the common factor of $8$ .

Cancel the common factor.

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = 8 (\frac{{(\sqrt[3]{x - 3} + 10)}^{3}}{8}) - 120 {(\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5)}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Rewrite the expression.

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = {(\sqrt[3]{x - 3} + 10)}^{3} - 120 {(\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5)}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Use the Binomial Theorem.

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = {\sqrt[3]{x - 3}}^{3} + 3 {\sqrt[3]{x - 3}}^{2} \cdot 10 + 3 \sqrt[3]{x - 3} \cdot 10^{2} + 10^{3} - 120 {(\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5)}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Simplify each term.

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

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

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = {({(x - 3)}^{\frac{1}{3}})}^{3} + 3 {\sqrt[3]{x - 3}}^{2} \cdot 10 + 3 \sqrt[3]{x - 3} \cdot 10^{2} + 10^{3} - 120 {(\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5)}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

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

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = {(x - 3)}^{\frac{1}{3} \cdot 3} + 3 {\sqrt[3]{x - 3}}^{2} \cdot 10 + 3 \sqrt[3]{x - 3} \cdot 10^{2} + 10^{3} - 120 {(\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5)}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

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

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = {(x - 3)}^{\frac{3}{3}} + 3 {\sqrt[3]{x - 3}}^{2} \cdot 10 + 3 \sqrt[3]{x - 3} \cdot 10^{2} + 10^{3} - 120 {(\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5)}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Cancel the common factor of $3$ .

Cancel the common factor.

Rewrite the expression.

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = (x - 3) + 3 {\sqrt[3]{x - 3}}^{2} \cdot 10 + 3 \sqrt[3]{x - 3} \cdot 10^{2} + 10^{3} - 120 {(\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5)}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Simplify.

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x - 3 + 3 {\sqrt[3]{x - 3}}^{2} \cdot 10 + 3 \sqrt[3]{x - 3} \cdot 10^{2} + 10^{3} - 120 {(\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5)}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Rewrite ${\sqrt[3]{x - 3}}^{2}$ as ${({(x - 3)}^{2})}^{\frac{1}{3}}$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x - 3 + 3 \sqrt[3]{{(x - 3)}^{2}} \cdot 10 + 3 \sqrt[3]{x - 3} \cdot 10^{2} + 10^{3} - 120 {(\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5)}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Multiply $10$ by $3$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x - 3 + 30 \sqrt[3]{{(x - 3)}^{2}} + 3 \sqrt[3]{x - 3} \cdot 10^{2} + 10^{3} - 120 {(\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5)}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Raise $10$ to the power of $2$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x - 3 + 30 \sqrt[3]{{(x - 3)}^{2}} + 3 \sqrt[3]{x - 3} \cdot 100 + 10^{3} - 120 {(\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5)}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Multiply $100$ by $3$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x - 3 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} + 10^{3} - 120 {(\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5)}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Raise $10$ to the power of $3$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x - 3 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} + 1000 - 120 {(\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5)}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Add $- 3$ and $1000$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 120 {(\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5)}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

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

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 120 {(\frac{\sqrt[3]{x - 3}}{2} + 5)}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

To write $5$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 120 {(\frac{\sqrt[3]{x - 3}}{2} + 5 \cdot \frac{2}{2})}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Combine $5$ and $\frac{2}{2}$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 120 {(\frac{\sqrt[3]{x - 3}}{2} + \frac{5 \cdot 2}{2})}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 120 {(\frac{\sqrt[3]{x - 3} + 5 \cdot 2}{2})}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Multiply $5$ by $2$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 120 {(\frac{\sqrt[3]{x - 3} + 10}{2})}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Apply the product rule to $\frac{\sqrt[3]{x - 3} + 10}{2}$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 120 \frac{{(\sqrt[3]{x - 3} + 10)}^{2}}{2^{2}} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

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

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 120 \frac{{(\sqrt[3]{x - 3} + 10)}^{2}}{4} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Cancel the common factor of $4$ .

Factor $4$ out of $- 120$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} + 4 (- 30) (\frac{{(\sqrt[3]{x - 3} + 10)}^{2}}{4}) + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Cancel the common factor.

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} + 4 \cdot (- 30 \frac{{(\sqrt[3]{x - 3} + 10)}^{2}}{4}) + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Rewrite the expression.

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 30 {(\sqrt[3]{x - 3} + 10)}^{2} + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Rewrite ${(\sqrt[3]{x - 3} + 10)}^{2}$ as $(\sqrt[3]{x - 3} + 10) (\sqrt[3]{x - 3} + 10)$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 30 ((\sqrt[3]{x - 3} + 10) (\sqrt[3]{x - 3} + 10)) + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Expand $(\sqrt[3]{x - 3} + 10) (\sqrt[3]{x - 3} + 10)$ using the FOIL Method.

Apply the distributive property.

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 30 (\sqrt[3]{x - 3} (\sqrt[3]{x - 3} + 10) + 10 (\sqrt[3]{x - 3} + 10)) + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Apply the distributive property.

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 30 (\sqrt[3]{x - 3} \sqrt[3]{x - 3} + \sqrt[3]{x - 3} \cdot 10 + 10 (\sqrt[3]{x - 3} + 10)) + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Apply the distributive property.

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 30 (\sqrt[3]{x - 3} \sqrt[3]{x - 3} + \sqrt[3]{x - 3} \cdot 10 + 10 \sqrt[3]{x - 3} + 10 \cdot 10) + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Simplify and combine like terms.

Simplify each term.

Multiply $\sqrt[3]{x - 3} \sqrt[3]{x - 3}$ .

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

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

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 30 ({\sqrt[3]{x - 3}}^{1 + 1} + \sqrt[3]{x - 3} \cdot 10 + 10 \sqrt[3]{x - 3} + 10 \cdot 10) + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Add $1$ and $1$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 30 ({\sqrt[3]{x - 3}}^{2} + \sqrt[3]{x - 3} \cdot 10 + 10 \sqrt[3]{x - 3} + 10 \cdot 10) + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Rewrite ${\sqrt[3]{x - 3}}^{2}$ as ${({(x - 3)}^{2})}^{\frac{1}{3}}$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 30 (\sqrt[3]{{(x - 3)}^{2}} + \sqrt[3]{x - 3} \cdot 10 + 10 \sqrt[3]{x - 3} + 10 \cdot 10) + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Move $10$ to the left of $\sqrt[3]{x - 3}$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 30 (\sqrt[3]{{(x - 3)}^{2}} + 10 \cdot \sqrt[3]{x - 3} + 10 \sqrt[3]{x - 3} + 10 \cdot 10) + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Multiply $10$ by $10$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 30 (\sqrt[3]{{(x - 3)}^{2}} + 10 \sqrt[3]{x - 3} + 10 \sqrt[3]{x - 3} + 100) + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Add $10 \sqrt[3]{x - 3}$ and $10 \sqrt[3]{x - 3}$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 30 (\sqrt[3]{{(x - 3)}^{2}} + 20 \sqrt[3]{x - 3} + 100) + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Apply the distributive property.

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 30 \sqrt[3]{{(x - 3)}^{2}} - 30 (20 \sqrt[3]{x - 3}) - 30 \cdot 100 + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Simplify.

Multiply $20$ by $- 30$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 30 \sqrt[3]{{(x - 3)}^{2}} - 600 \sqrt[3]{x - 3} - 30 \cdot 100 + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

Multiply $- 30$ by $100$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 30 \sqrt[3]{{(x - 3)}^{2}} - 600 \sqrt[3]{x - 3} - 3000 + 600 (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) - 997$

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

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 30 \sqrt[3]{{(x - 3)}^{2}} - 600 \sqrt[3]{x - 3} - 3000 + 600 (\frac{\sqrt[3]{x - 3}}{2} + 5) - 997$

To write $5$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

Combine $5$ and $\frac{2}{2}$ .

Combine the numerators over the common denominator.

Multiply $5$ by $2$ .

Cancel the common factor of $2$ .

Factor $2$ out of $600$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 30 \sqrt[3]{{(x - 3)}^{2}} - 600 \sqrt[3]{x - 3} - 3000 + 2 (300) (\frac{\sqrt[3]{x - 3} + 10}{2}) - 997$

Cancel the common factor.

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 30 \sqrt[3]{{(x - 3)}^{2}} - 600 \sqrt[3]{x - 3} - 3000 + 2 \cdot (300 (\frac{\sqrt[3]{x - 3} + 10}{2})) - 997$

Rewrite the expression.

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 30 \sqrt[3]{{(x - 3)}^{2}} - 600 \sqrt[3]{x - 3} - 3000 + 300 (\sqrt[3]{x - 3} + 10) - 997$

Apply the distributive property.

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 30 \sqrt[3]{{(x - 3)}^{2}} - 600 \sqrt[3]{x - 3} - 3000 + 300 \sqrt[3]{x - 3} + 300 \cdot 10 - 997$

Multiply $300$ by $10$ .

Simplify by adding terms.

Combine the opposite terms in $x + 997 + 30 \sqrt[3]{{(x - 3)}^{2}} + 300 \sqrt[3]{x - 3} - 30 \sqrt[3]{{(x - 3)}^{2}} - 600 \sqrt[3]{x - 3} - 3000 + 300 \sqrt[3]{x - 3} + 3000 - 997$ .

Subtract $30 \sqrt[3]{{(x - 3)}^{2}}$ from $30 \sqrt[3]{{(x - 3)}^{2}}$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 0 + 300 \sqrt[3]{x - 3} - 600 \sqrt[3]{x - 3} - 3000 + 300 \sqrt[3]{x - 3} + 3000 - 997$

Add $x + 997$ and $0$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 300 \sqrt[3]{x - 3} - 600 \sqrt[3]{x - 3} - 3000 + 300 \sqrt[3]{x - 3} + 3000 - 997$

Add $- 3000$ and $3000$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 300 \sqrt[3]{x - 3} - 600 \sqrt[3]{x - 3} + 0 + 300 \sqrt[3]{x - 3} - 997$

Add $x + 997 + 300 \sqrt[3]{x - 3} - 600 \sqrt[3]{x - 3}$ and $0$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 997 + 300 \sqrt[3]{x - 3} - 600 \sqrt[3]{x - 3} + 300 \sqrt[3]{x - 3} - 997$

Subtract $997$ from $997$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 0 + 300 \sqrt[3]{x - 3} - 600 \sqrt[3]{x - 3} + 300 \sqrt[3]{x - 3}$

Add $x$ and $0$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x + 300 \sqrt[3]{x - 3} - 600 \sqrt[3]{x - 3} + 300 \sqrt[3]{x - 3}$

Subtract $600 \sqrt[3]{x - 3}$ from $300 \sqrt[3]{x - 3}$ .

$f^{- 1} (\frac{1}{2} \cdot \sqrt[3]{x - 3} + 5) = x - 300 \sqrt[3]{x - 3} + 300 \sqrt[3]{x - 3}$

Combine the opposite terms in $x - 300 \sqrt[3]{x - 3} + 300 \sqrt[3]{x - 3}$ .

Add $- 300 \sqrt[3]{x - 3}$ and $300 \sqrt[3]{x - 3}$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

Evaluate $f (8 x^{3} - 120 x^{2} + 600 x - 997)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = \frac{1}{2} \cdot \sqrt[3]{(8 x^{3} - 120 x^{2} + 600 x - 997) - 3} + 5$

Simplify each term.

Subtract $3$ from $- 997$ .

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = \frac{1}{2} \cdot \sqrt[3]{8 x^{3} - 120 x^{2} + 600 x - 1000} + 5$

Factor $8$ out of $8 x^{3} - 120 x^{2} + 600 x - 1000$ .

Factor $8$ out of $8 x^{3}$ .

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = \frac{1}{2} \cdot \sqrt[3]{8 (x^{3}) - 120 x^{2} + 600 x - 1000} + 5$

Factor $8$ out of $- 120 x^{2}$ .

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = \frac{1}{2} \cdot \sqrt[3]{8 (x^{3}) + 8 (- 15 x^{2}) + 600 x - 1000} + 5$

Factor $8$ out of $600 x$ .

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = \frac{1}{2} \cdot \sqrt[3]{8 (x^{3}) + 8 (- 15 x^{2}) + 8 (75 x) - 1000} + 5$

Factor $8$ out of $- 1000$ .

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = \frac{1}{2} \cdot \sqrt[3]{8 x^{3} + 8 (- 15 x^{2}) + 8 (75 x) + 8 \cdot - 125} + 5$

Factor $8$ out of $8 x^{3} + 8 (- 15 x^{2})$ .

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = \frac{1}{2} \cdot \sqrt[3]{8 (x^{3} - 15 x^{2}) + 8 (75 x) + 8 \cdot - 125} + 5$

Factor $8$ out of $8 (x^{3} - 15 x^{2}) + 8 (75 x)$ .

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = \frac{1}{2} \cdot \sqrt[3]{8 (x^{3} - 15 x^{2} + 75 x) + 8 \cdot - 125} + 5$

Factor $8$ out of $8 (x^{3} - 15 x^{2} + 75 x) + 8 \cdot - 125$ .

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = \frac{1}{2} \cdot \sqrt[3]{8 (x^{3} - 15 x^{2} + 75 x - 125)} + 5$

Rewrite $8 (x^{3} - 15 x^{2} + 75 x - 125)$ as $2^{3} (x^{3} - 15 x^{2} + 75 x + {(- 5)}^{3})$ .

Rewrite $8$ as $2^{3}$ .

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = \frac{1}{2} \cdot \sqrt[3]{2^{3} (x^{3} - 15 x^{2} + 75 x - 125)} + 5$

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

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = \frac{1}{2} \cdot \sqrt[3]{2^{3} (x^{3} - 15 x^{2} + 75 x + {(- 5)}^{3})} + 5$

Pull terms out from under the radical.

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = \frac{1}{2} \cdot (2 \sqrt[3]{x^{3} - 15 x^{2} + 75 x + {(- 5)}^{3}}) + 5$

Cancel the common factor of $2$ .

Factor $2$ out of $2 \sqrt[3]{x^{3} - 15 x^{2} + 75 x + {(- 5)}^{3}}$ .

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = \frac{1}{2} \cdot (2 (\sqrt[3]{x^{3} - 15 x^{2} + 75 x + {(- 5)}^{3}})) + 5$

Cancel the common factor.

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = \frac{1}{2} \cdot (2 \sqrt[3]{x^{3} - 15 x^{2} + 75 x + {(- 5)}^{3}}) + 5$

Rewrite the expression.

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = \sqrt[3]{x^{3} - 15 x^{2} + 75 x + {(- 5)}^{3}} + 5$

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

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = \sqrt[3]{x^{3} - 15 x^{2} + 75 x - 125} + 5$

Rewrite $x^{3} - 15 x^{2} + 75 x - 125$ in a factored form.

Factor $x^{3} - 15 x^{2} + 75 x - 125$ using the rational roots test.

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 125, \pm 5, \pm 25 q = \pm 1$

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 125, \pm 5, \pm 25$

Substitute $5$ and simplify the expression. In this case, the expression is equal to $0$ so $5$ is a root of the polynomial.

Substitute $5$ into the polynomial.

$5^{3} - 15 \cdot 5^{2} + 75 \cdot 5 - 125$

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

$125 - 15 \cdot 5^{2} + 75 \cdot 5 - 125$

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

$125 - 15 \cdot 25 + 75 \cdot 5 - 125$

Multiply $- 15$ by $25$ .

$125 - 375 + 75 \cdot 5 - 125$

Subtract $375$ from $125$ .

$- 250 + 75 \cdot 5 - 125$

Multiply $75$ by $5$ .

$- 250 + 375 - 125$

Add $- 250$ and $375$ .

$125 - 125$

Subtract $125$ from $125$ .

$0$

Since $5$ is a known root, divide the polynomial by $x - 5$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 15 x^{2} + 75 x - 125}{x - 5}$

Divide $x^{3} - 15 x^{2} + 75 x - 125$ by $x - 5$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$5$

$x^{3}$

$15 x^{2}$

$75 x$

$125$

Divide the highest order term in the dividend $x^{3}$ by the highest order term in divisor $x$ .

					$x^{2}$
	$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			+	$x^{3}$	-	$5 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{3} - 5 x^{2}$

				$x^{2}$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			-	$x^{3}$	+	$5 x^{2}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			-	$x^{3}$	+	$5 x^{2}$
					-	$10 x^{2}$

Pull the next terms from the original dividend down into the current dividend.

				$x^{2}$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			-	$x^{3}$	+	$5 x^{2}$
					-	$10 x^{2}$	+	$75 x$

Divide the highest order term in the dividend $- 10 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$10 x$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			-	$x^{3}$	+	$5 x^{2}$
					-	$10 x^{2}$	+	$75 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$10 x$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			-	$x^{3}$	+	$5 x^{2}$
					-	$10 x^{2}$	+	$75 x$
					-	$10 x^{2}$	+	$50 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 10 x^{2} + 50 x$

				$x^{2}$	-	$10 x$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			-	$x^{3}$	+	$5 x^{2}$
					-	$10 x^{2}$	+	$75 x$
					+	$10 x^{2}$	-	$50 x$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$	-	$10 x$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			-	$x^{3}$	+	$5 x^{2}$
					-	$10 x^{2}$	+	$75 x$
					+	$10 x^{2}$	-	$50 x$
							+	$25 x$

Pull the next terms from the original dividend down into the current dividend.

				$x^{2}$	-	$10 x$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			-	$x^{3}$	+	$5 x^{2}$
					-	$10 x^{2}$	+	$75 x$
					+	$10 x^{2}$	-	$50 x$
							+	$25 x$	-	$125$

Divide the highest order term in the dividend $25 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$10 x$	+	$25$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			-	$x^{3}$	+	$5 x^{2}$
					-	$10 x^{2}$	+	$75 x$
					+	$10 x^{2}$	-	$50 x$
							+	$25 x$	-	$125$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$10 x$	+	$25$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			-	$x^{3}$	+	$5 x^{2}$
					-	$10 x^{2}$	+	$75 x$
					+	$10 x^{2}$	-	$50 x$
							+	$25 x$	-	$125$
							+	$25 x$	-	$125$

The expression needs to be subtracted from the dividend, so change all the signs in $25 x - 125$

				$x^{2}$	-	$10 x$	+	$25$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			-	$x^{3}$	+	$5 x^{2}$
					-	$10 x^{2}$	+	$75 x$
					+	$10 x^{2}$	-	$50 x$
							+	$25 x$	-	$125$
							-	$25 x$	+	$125$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$	-	$10 x$	+	$25$
$x$	-	$5$		$x^{3}$	-	$15 x^{2}$	+	$75 x$	-	$125$
			-	$x^{3}$	+	$5 x^{2}$
					-	$10 x^{2}$	+	$75 x$
					+	$10 x^{2}$	-	$50 x$
							+	$25 x$	-	$125$
							-	$25 x$	+	$125$
										$0$

Since the remander is $0$ , the final answer is the quotient.

$x^{2} - 10 x + 25$

Write $x^{3} - 15 x^{2} + 75 x - 125$ as a set of factors.

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = \sqrt[3]{(x - 5) (x^{2} - 10 x + 25)} + 5$

Factor using the perfect square rule.

Rewrite $25$ as $5^{2}$ .

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = \sqrt[3]{(x - 5) (x^{2} - 10 x + 5^{2})} + 5$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$10 x = 2 \cdot x \cdot 5$

Rewrite the polynomial.

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = \sqrt[3]{(x - 5) (x^{2} - 2 \cdot x \cdot 5 + 5^{2})} + 5$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 5$ .

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = \sqrt[3]{(x - 5) {(x - 5)}^{2}} + 5$

Combine like factors.

Raise $x - 5$ to the power of $1$ .

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = \sqrt[3]{(x - 5) {(x - 5)}^{2}} + 5$

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

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = \sqrt[3]{{(x - 5)}^{1 + 2}} + 5$

Add $1$ and $2$ .

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = \sqrt[3]{{(x - 5)}^{3}} + 5$

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

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = x - 5 + 5$

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

Add $- 5$ and $5$ .

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = x + 0$

Add $x$ and $0$ .

$f (8 x^{3} - 120 x^{2} + 600 x - 997) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = 8 x^{3} - 120 x^{2} + 600 x - 997$ is the inverse of $f (x) = \frac{1}{2} \cdot \sqrt[3]{x - 3} + 5$ .

$f^{- 1} (x) = 8 x^{3} - 120 x^{2} + 600 x - 997$

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

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