Precalculus Examples

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

Step 1

Write $f (x) = {(2 - x^{3})}^{5}$ as an equation.

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

Subtract $2$ from both sides of the equation.

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

Step 3.4

Divide each term in $- y^{3} = \sqrt[5]{x} - 2$ by $- 1$ and simplify.

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Step 3.4.1

Divide each term in $- y^{3} = \sqrt[5]{x} - 2$ by $- 1$ .

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

Step 3.4.2

Simplify the left side.

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Step 3.4.2.1

Dividing two negative values results in a positive value.

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

Step 3.4.2.2

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

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

Step 3.4.3

Simplify the right side.

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Step 3.4.3.1

Simplify each term.

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Step 3.4.3.1.1

Move the negative one from the denominator of $\frac{\sqrt[5]{x}}{- 1}$ .

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

Step 3.4.3.1.2

Rewrite $- 1 \cdot \sqrt[5]{x}$ as $- \sqrt[5]{x}$ .

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

Step 3.4.3.1.3

Divide $- 2$ by $- 1$ .

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

Step 3.5

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

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

Step 4

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

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

Step 5

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

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

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

Step 5.2.3

Remove parentheses.

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

Step 5.2.4

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

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

Step 5.2.5

Apply the distributive property.

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

Step 5.2.6

Multiply $- 1$ by $2$ .

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

Step 5.2.7

Add $- 2$ and $2$ .

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

Step 5.2.8

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

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

Step 5.2.9

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

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

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

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

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

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

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Step 5.3.3.1.1

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

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

Step 5.3.3.1.2

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

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

Step 5.3.3.1.3

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

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

Step 5.3.3.1.4

Cancel the common factor of $3$ .

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Step 5.3.3.1.4.1

Cancel the common factor.

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

Step 5.3.3.1.4.2

Rewrite the expression.

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

Step 5.3.3.1.5

Simplify.

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

Step 5.3.3.2

Apply the distributive property.

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

Step 5.3.3.3

Multiply $- - \sqrt[5]{x}$ .

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Step 5.3.3.3.1

Multiply $- 1$ by $- 1$ .

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

Step 5.3.3.3.2

Multiply $\sqrt[5]{x}$ by $1$ .

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

Step 5.3.3.4

Multiply $- 1$ by $2$ .

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

Step 5.3.4

Simplify terms.

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Step 5.3.4.1

Combine the opposite terms in $2 + \sqrt[5]{x} - 2$ .

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Step 5.3.4.1.1

Subtract $2$ from $2$ .

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

Step 5.3.4.1.2

Add $0$ and $\sqrt[5]{x}$ .

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

Step 5.3.4.2

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

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Step 5.3.4.2.1

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

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

Step 5.3.4.2.2

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

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

Step 5.3.4.2.3

Combine $\frac{1}{5}$ and $5$ .

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

Step 5.3.4.2.4

Cancel the common factor of $5$ .

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Step 5.3.4.2.4.1

Cancel the common factor.

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

Step 5.3.4.2.4.2

Rewrite the expression.

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

Step 5.3.4.2.5

Simplify.

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

Step 5.4

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

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