Algebra Examples

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

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

Step 1

Write

$f (x) = - 2 x^{3} + 1$ as an equation.

$y = - 2 x^{3} + 1$

Step 2

Interchange the variables.

$x = - 2 y^{3} + 1$

Step 3

Solve for

$y$ .

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Step 3.1

Rewrite the equation as

$- 2 y^{3} + 1 = x$ .

$- 2 y^{3} + 1 = x$

Step 3.2

Subtract

$1$ from both sides of the equation.

$- 2 y^{3} = x - 1$

Step 3.3

Divide each term in

$- 2 y^{3} = x - 1$ by

$- 2$ and simplify.

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Step 3.3.1

Divide each term in

$- 2 y^{3} = x - 1$ by

$- 2$ .

$\frac{- 2 y^{3}}{- 2} = \frac{x}{- 2} + \frac{- 1}{- 2}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of

$- 2$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{- 2 y^{3}}{- 2} = \frac{x}{- 2} + \frac{- 1}{- 2}$

Step 3.3.2.1.2

Divide

$y^{3}$ by

$1$ .

$y^{3} = \frac{x}{- 2} + \frac{- 1}{- 2}$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Simplify each term.

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Step 3.3.3.1.1

Move the negative in front of the fraction.

$y^{3} = - \frac{x}{2} + \frac{- 1}{- 2}$

Step 3.3.3.1.2

Dividing two negative values results in a positive value.

$y^{3} = - \frac{x}{2} + \frac{1}{2}$

Step 3.4

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

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

Step 3.5

Simplify

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

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Step 3.5.1

Combine the numerators over the common denominator.

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

Step 3.5.2

Rewrite

$\sqrt[3]{\frac{- x + 1}{2}}$ as

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

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

Step 3.5.3

Multiply

$\frac{\sqrt[3]{- x + 1}}{\sqrt[3]{2}}$ by

$\frac{{\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{2}}$ .

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

Step 3.5.4

Combine and simplify the denominator.

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Step 3.5.4.1

Multiply

$\frac{\sqrt[3]{- x + 1}}{\sqrt[3]{2}}$ by

$\frac{{\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{2}}$ .

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

Step 3.5.4.2

Raise

$\sqrt[3]{2}$ to the power of

$1$ .

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

Step 3.5.4.3

Use the power rule

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 3.5.4.4

Add

$1$ and

$2$ .

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

Step 3.5.4.5

Rewrite

${\sqrt[3]{2}}^{3}$ as

$2$ .

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Step 3.5.4.5.1

Use

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

$\sqrt[3]{2}$ as

$2^{\frac{1}{3}}$ .

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

Step 3.5.4.5.2

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

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

Step 3.5.4.5.3

Combine

$\frac{1}{3}$ and

$3$ .

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

Step 3.5.4.5.4

Cancel the common factor of

$3$ .

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Step 3.5.4.5.4.1

Cancel the common factor.

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

Step 3.5.4.5.4.2

Rewrite the expression.

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

Step 3.5.4.5.5

Evaluate the exponent.

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

Step 3.5.5

Simplify the numerator.

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Step 3.5.5.1

Rewrite

${\sqrt[3]{2}}^{2}$ as

$\sqrt[3]{2^{2}}$ .

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

Step 3.5.5.2

Raise

$2$ to the power of

$2$ .

$y = \frac{\sqrt[3]{- x + 1} \sqrt[3]{4}}{2}$

Step 3.5.6

Simplify with factoring out.

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Step 3.5.6.1

Combine using the product rule for radicals.

$y = \frac{\sqrt[3]{(- x + 1) \cdot 4}}{2}$

Step 3.5.6.2

Reorder factors in

$\frac{\sqrt[3]{(- x + 1) \cdot 4}}{2}$ .

$y = \frac{\sqrt[3]{4 (- x + 1)}}{2}$

Step 4

Replace

$y$ with

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

$f^{- 1} (x) = \frac{\sqrt[3]{4 (- x + 1)}}{2}$

Step 5

Verify if

$f^{- 1} (x) = \frac{\sqrt[3]{4 (- x + 1)}}{2}$ is the inverse of

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

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Step 5.1

To verify the inverse, check if

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

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

Step 5.2

Evaluate

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

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Step 5.2.1

Set up the composite result function.

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

Step 5.2.2

Evaluate

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

$f$ into

$f^{- 1}$ .

$f^{- 1} (- 2 x^{3} + 1) = \frac{\sqrt[3]{4 (- (- 2 x^{3} + 1) + 1)}}{2}$

Step 5.2.3

Simplify the numerator.

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Step 5.2.3.1

Apply the distributive property.

$f^{- 1} (- 2 x^{3} + 1) = \frac{\sqrt[3]{4 (- (- 2 x^{3}) - 1 \cdot 1 + 1)}}{2}$

Step 5.2.3.2

Multiply

$- 2$ by

$- 1$ .

$f^{- 1} (- 2 x^{3} + 1) = \frac{\sqrt[3]{4 (2 x^{3} - 1 \cdot 1 + 1)}}{2}$

Step 5.2.3.3

Multiply

$- 1$ by

$1$ .

$f^{- 1} (- 2 x^{3} + 1) = \frac{\sqrt[3]{4 (2 x^{3} - 1 + 1)}}{2}$

Step 5.2.3.4

Add

$- 1$ and

$1$ .

$f^{- 1} (- 2 x^{3} + 1) = \frac{\sqrt[3]{4 (2 x^{3} + 0)}}{2}$

Step 5.2.3.5

Add

$2 x^{3}$ and

$0$ .

$f^{- 1} (- 2 x^{3} + 1) = \frac{\sqrt[3]{4 \cdot (2 x^{3})}}{2}$

Step 5.2.3.6

Multiply

$4$ by

$2$ .

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

Step 5.2.3.7

Rewrite

$8 x^{3}$ as

${(2 x)}^{3}$ .

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

Step 5.2.3.8

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

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

Step 5.2.4

Cancel the common factor of

$2$ .

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Step 5.2.4.1

Cancel the common factor.

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

Step 5.2.4.2

Divide

$x$ by

$1$ .

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

Step 5.3

Evaluate

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

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Step 5.3.1

Set up the composite result function.

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

Step 5.3.2

Evaluate

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2})$ by substituting in the value of

$f^{- 1}$ into

$f$ .

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = - 2 {(\frac{\sqrt[3]{4 (- x + 1)}}{2})}^{3} + 1$

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Apply the product rule to

$\frac{\sqrt[3]{4 (- x + 1)}}{2}$ .

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = - 2 \frac{{\sqrt[3]{4 (- x + 1)}}^{3}}{2^{3}} + 1$

Step 5.3.3.2

Simplify the numerator.

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Step 5.3.3.2.1

Rewrite

${\sqrt[3]{4 (- x + 1)}}^{3}$ as

$4 (- x + 1)$ .

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Step 5.3.3.2.1.1

Use

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

$\sqrt[3]{4 (- x + 1)}$ as

${(4 (- x + 1))}^{\frac{1}{3}}$ .

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = - 2 \frac{{({(4 (- x + 1))}^{\frac{1}{3}})}^{3}}{2^{3}} + 1$

Step 5.3.3.2.1.2

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = - 2 \frac{{(4 (- x + 1))}^{\frac{1}{3} \cdot 3}}{2^{3}} + 1$

Step 5.3.3.2.1.3

Combine

$\frac{1}{3}$ and

$3$ .

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = - 2 \frac{{(4 (- x + 1))}^{\frac{3}{3}}}{2^{3}} + 1$

Step 5.3.3.2.1.4

Cancel the common factor of

$3$ .

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Step 5.3.3.2.1.4.1

Cancel the common factor.

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = - 2 \frac{{(4 (- x + 1))}^{\frac{3}{3}}}{2^{3}} + 1$

Step 5.3.3.2.1.4.2

Rewrite the expression.

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = - 2 \frac{4 (- x + 1)}{2^{3}} + 1$

Step 5.3.3.2.1.5

Simplify.

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = - 2 \frac{4 (- x + 1)}{2^{3}} + 1$

Step 5.3.3.2.2

Apply the distributive property.

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = - 2 \frac{4 (- x) + 4 \cdot 1}{2^{3}} + 1$

Step 5.3.3.2.3

Multiply

$- 1$ by

$4$ .

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = - 2 \frac{- 4 x + 4 \cdot 1}{2^{3}} + 1$

Step 5.3.3.2.4

Multiply

$4$ by

$1$ .

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = - 2 \frac{- 4 x + 4}{2^{3}} + 1$

Step 5.3.3.2.5

Factor

$4$ out of

$- 4 x + 4$ .

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Step 5.3.3.2.5.1

Factor

$4$ out of

$- 4 x$ .

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = - 2 \frac{4 (- x) + 4}{2^{3}} + 1$

Step 5.3.3.2.5.2

Factor

$4$ out of

$4$ .

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = - 2 \frac{4 (- x) + 4 (1)}{2^{3}} + 1$

Step 5.3.3.2.5.3

Factor

$4$ out of

$4 (- x) + 4 (1)$ .

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = - 2 \frac{4 (- x + 1)}{2^{3}} + 1$

Step 5.3.3.3

Raise

$2$ to the power of

$3$ .

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = - 2 \frac{4 (- x + 1)}{8} + 1$

Step 5.3.3.4

Cancel the common factor of

$2$ .

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Step 5.3.3.4.1

Factor

$2$ out of

$- 2$ .

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = 2 (- 1) (\frac{4 (- x + 1)}{8}) + 1$

Step 5.3.3.4.2

Factor

$2$ out of

$8$ .

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = 2 \cdot (- 1 \frac{4 (- x + 1)}{2 \cdot 4}) + 1$

Step 5.3.3.4.3

Cancel the common factor.

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = 2 \cdot (- 1 \frac{4 (- x + 1)}{2 \cdot 4}) + 1$

Step 5.3.3.4.4

Rewrite the expression.

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = - 1 \frac{4 (- x + 1)}{4} + 1$

Step 5.3.3.5

Cancel the common factor of

$4$ .

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Step 5.3.3.5.1

Cancel the common factor.

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = - 1 \frac{4 (- x + 1)}{4} + 1$

Step 5.3.3.5.2

Divide

$- x + 1$ by

$1$ .

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = - 1 (- x + 1) + 1$

Step 5.3.3.6

Apply the distributive property.

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = - 1 (- x) - 1 \cdot 1 + 1$

Step 5.3.3.7

Multiply

$- 1 (- x)$ .

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Step 5.3.3.7.1

Multiply

$- 1$ by

$- 1$ .

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = 1 x - 1 \cdot 1 + 1$

Step 5.3.3.7.2

Multiply

$x$ by

$1$ .

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = x - 1 \cdot 1 + 1$

Step 5.3.3.8

Multiply

$- 1$ by

$1$ .

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = x - 1 + 1$

Step 5.3.4

Combine the opposite terms in

$x - 1 + 1$ .

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Step 5.3.4.1

Add

$- 1$ and

$1$ .

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = x + 0$

Step 5.3.4.2

Add

$x$ and

$0$ .

$f (\frac{\sqrt[3]{4 (- x + 1)}}{2}) = x$

Step 5.4

Since

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

$f (f^{- 1} (x)) = x$ , then

$f^{- 1} (x) = \frac{\sqrt[3]{4 (- x + 1)}}{2}$ is the inverse of

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

$f^{- 1} (x) = \frac{\sqrt[3]{4 (- x + 1)}}{2}$