Algebra Examples

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

Step 1

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

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $2 y^{2 + 3} = x$ .

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

Step 3.2

Add $2$ and $3$ .

$2 y^{5} = x$

Step 3.3

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

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Step 3.3.1

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

$\frac{2 y^{5}}{2} = \frac{x}{2}$

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.

$\frac{2 y^{5}}{2} = \frac{x}{2}$

Step 3.3.2.1.2

Divide $y^{5}$ by $1$ .

$y^{5} = \frac{x}{2}$

Step 3.4

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

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

Step 3.5

Simplify $\sqrt[5]{\frac{x}{2}}$ .

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Step 3.5.1

Rewrite $\sqrt[5]{\frac{x}{2}}$ as $\frac{\sqrt[5]{x}}{\sqrt[5]{2}}$ .

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

Step 3.5.2

Multiply $\frac{\sqrt[5]{x}}{\sqrt[5]{2}}$ by $\frac{{\sqrt[5]{2}}^{4}}{{\sqrt[5]{2}}^{4}}$ .

$y = \frac{\sqrt[5]{x}}{\sqrt[5]{2}} \cdot \frac{{\sqrt[5]{2}}^{4}}{{\sqrt[5]{2}}^{4}}$

Step 3.5.3

Combine and simplify the denominator.

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Step 3.5.3.1

Multiply $\frac{\sqrt[5]{x}}{\sqrt[5]{2}}$ by $\frac{{\sqrt[5]{2}}^{4}}{{\sqrt[5]{2}}^{4}}$ .

$y = \frac{\sqrt[5]{x} {\sqrt[5]{2}}^{4}}{\sqrt[5]{2} {\sqrt[5]{2}}^{4}}$

Step 3.5.3.2

Raise $\sqrt[5]{2}$ to the power of $1$ .

$y = \frac{\sqrt[5]{x} {\sqrt[5]{2}}^{4}}{{\sqrt[5]{2}}^{1} {\sqrt[5]{2}}^{4}}$

Step 3.5.3.3

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

$y = \frac{\sqrt[5]{x} {\sqrt[5]{2}}^{4}}{{\sqrt[5]{2}}^{1 + 4}}$

Step 3.5.3.4

Add $1$ and $4$ .

$y = \frac{\sqrt[5]{x} {\sqrt[5]{2}}^{4}}{{\sqrt[5]{2}}^{5}}$

Step 3.5.3.5

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

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Step 3.5.3.5.1

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

$y = \frac{\sqrt[5]{x} {\sqrt[5]{2}}^{4}}{{(2^{\frac{1}{5}})}^{5}}$

Step 3.5.3.5.2

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

$y = \frac{\sqrt[5]{x} {\sqrt[5]{2}}^{4}}{2^{\frac{1}{5} \cdot 5}}$

Step 3.5.3.5.3

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

$y = \frac{\sqrt[5]{x} {\sqrt[5]{2}}^{4}}{2^{\frac{5}{5}}}$

Step 3.5.3.5.4

Cancel the common factor of $5$ .

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Step 3.5.3.5.4.1

Cancel the common factor.

$y = \frac{\sqrt[5]{x} {\sqrt[5]{2}}^{4}}{2^{\frac{5}{5}}}$

Step 3.5.3.5.4.2

Rewrite the expression.

$y = \frac{\sqrt[5]{x} {\sqrt[5]{2}}^{4}}{2^{1}}$

Step 3.5.3.5.5

Evaluate the exponent.

$y = \frac{\sqrt[5]{x} {\sqrt[5]{2}}^{4}}{2}$

Step 3.5.4

Simplify the numerator.

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Step 3.5.4.1

Rewrite ${\sqrt[5]{2}}^{4}$ as $\sqrt[5]{2^{4}}$ .

$y = \frac{\sqrt[5]{x} \sqrt[5]{2^{4}}}{2}$

Step 3.5.4.2

Raise $2$ to the power of $4$ .

$y = \frac{\sqrt[5]{x} \sqrt[5]{16}}{2}$

Step 3.5.5

Simplify with factoring out.

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Step 3.5.5.1

Combine using the product rule for radicals.

$y = \frac{\sqrt[5]{x \cdot 16}}{2}$

Step 3.5.5.2

Reorder factors in $\frac{\sqrt[5]{x \cdot 16}}{2}$ .

$y = \frac{\sqrt[5]{16 x}}{2}$

Step 4

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

$f^{- 1} (x) = \frac{\sqrt[5]{16 x}}{2}$

Step 5

Verify if $f^{- 1} (x) = \frac{\sqrt[5]{16 x}}{2}$ is the inverse of $f (x) = 2 x^{2 + 3}$ .

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

$f^{- 1} (2 x^{2 + 3}) = \frac{\sqrt[5]{16 (2 x^{2 + 3})}}{2}$

Step 5.2.3

Simplify the numerator.

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Step 5.2.3.1

Multiply $16$ by $2$ .

$f^{- 1} (2 x^{2 + 3}) = \frac{\sqrt[5]{32 x^{2 + 3}}}{2}$

Step 5.2.3.2

Add $2$ and $3$ .

$f^{- 1} (2 x^{2 + 3}) = \frac{\sqrt[5]{32 x^{5}}}{2}$

Step 5.2.3.3

Rewrite $32 x^{5}$ as ${(2 x)}^{5}$ .

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

Step 5.2.3.4

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

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

Step 5.2.4

Cancel the common factor of $2$ .

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Step 5.2.4.1

Cancel the common factor.

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

Step 5.2.4.2

Divide $x$ by $1$ .

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

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 $f (\frac{\sqrt[5]{16 x}}{2})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\sqrt[5]{16 x}}{2}) = 2 {(\frac{\sqrt[5]{16 x}}{2})}^{2 + 3}$

Step 5.3.3

Simplify the expression.

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Step 5.3.3.1

Add $2$ and $3$ .

$f (\frac{\sqrt[5]{16 x}}{2}) = 2 {(\frac{\sqrt[5]{16 x}}{2})}^{5}$

Step 5.3.3.2

Apply the product rule to $\frac{\sqrt[5]{16 x}}{2}$ .

$f (\frac{\sqrt[5]{16 x}}{2}) = 2 (\frac{{\sqrt[5]{16 x}}^{5}}{2^{5}})$

Step 5.3.4

Rewrite ${\sqrt[5]{16 x}}^{5}$ as $16 x$ .

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Step 5.3.4.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[5]{16 x}$ as ${(16 x)}^{\frac{1}{5}}$ .

$f (\frac{\sqrt[5]{16 x}}{2}) = 2 (\frac{{({(16 x)}^{\frac{1}{5}})}^{5}}{2^{5}})$

Step 5.3.4.2

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

$f (\frac{\sqrt[5]{16 x}}{2}) = 2 (\frac{{(16 x)}^{\frac{1}{5} \cdot 5}}{2^{5}})$

Step 5.3.4.3

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

$f (\frac{\sqrt[5]{16 x}}{2}) = 2 (\frac{{(16 x)}^{\frac{5}{5}}}{2^{5}})$

Step 5.3.4.4

Cancel the common factor of $5$ .

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Step 5.3.4.4.1

Cancel the common factor.

$f (\frac{\sqrt[5]{16 x}}{2}) = 2 (\frac{{(16 x)}^{\frac{5}{5}}}{2^{5}})$

Step 5.3.4.4.2

Rewrite the expression.

$f (\frac{\sqrt[5]{16 x}}{2}) = 2 (\frac{16 x}{2^{5}})$

Step 5.3.4.5

Simplify.

$f (\frac{\sqrt[5]{16 x}}{2}) = 2 (\frac{16 x}{2^{5}})$

Step 5.3.5

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

$f (\frac{\sqrt[5]{16 x}}{2}) = 2 (\frac{16 x}{32})$

Step 5.3.6

Cancel the common factor of $2$ .

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Step 5.3.6.1

Factor $2$ out of $32$ .

$f (\frac{\sqrt[5]{16 x}}{2}) = 2 (\frac{16 x}{2 (16)})$

Step 5.3.6.2

Cancel the common factor.

$f (\frac{\sqrt[5]{16 x}}{2}) = 2 (\frac{16 x}{2 \cdot 16})$

Step 5.3.6.3

Rewrite the expression.

$f (\frac{\sqrt[5]{16 x}}{2}) = \frac{16 x}{16}$

Step 5.3.7

Cancel the common factor of $16$ .

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Step 5.3.7.1

Cancel the common factor.

$f (\frac{\sqrt[5]{16 x}}{2}) = \frac{16 x}{16}$

Step 5.3.7.2

Divide $x$ by $1$ .

$f (\frac{\sqrt[5]{16 x}}{2}) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\sqrt[5]{16 x}}{2}$ is the inverse of $f (x) = 2 x^{2 + 3}$ .

$f^{- 1} (x) = \frac{\sqrt[5]{16 x}}{2}$