Precalculus Examples

$y = \sqrt{x + 5}$

Step 1

Interchange the variables.

$x = \sqrt{y + 5}$

Step 2

Solve for

$y$ .

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Step 2.1

Rewrite the equation as

$\sqrt{y + 5} = x$ .

$\sqrt{y + 5} = x$

Step 2.2

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

${\sqrt{y + 5}}^{2} = x^{2}$

Step 2.3

Simplify each side of the equation.

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Step 2.3.1

Use

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

$\sqrt{y + 5}$ as

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

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

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Simplify

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

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Step 2.3.2.1.1

Multiply the exponents in

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

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Step 2.3.2.1.1.1

Apply the power rule and multiply exponents,

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

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

Step 2.3.2.1.1.2

Cancel the common factor of

$2$ .

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Step 2.3.2.1.1.2.1

Cancel the common factor.

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

Step 2.3.2.1.1.2.2

Rewrite the expression.

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

Step 2.3.2.1.2

Simplify.

$y + 5 = x^{2}$

Step 2.4

Subtract

$5$ from both sides of the equation.

$y = x^{2} - 5$

Step 3

Replace

$y$ with

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

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

Step 4

Verify if

$f^{- 1} (x) = x^{2} - 5$ is the inverse of

$f (x) = \sqrt{x + 5}$ .

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Step 4.1

To verify the inverse, check if

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

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

Step 4.2

Evaluate

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

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Step 4.2.1

Set up the composite result function.

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

Step 4.2.2

Evaluate

$f^{- 1} (\sqrt{x + 5})$ by substituting in the value of

$f$ into

$f^{- 1}$ .

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

Step 4.2.3

Rewrite

${\sqrt{x + 5}}^{2}$ as

$x + 5$ .

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Step 4.2.3.1

Use

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

$\sqrt{x + 5}$ as

${(x + 5)}^{\frac{1}{2}}$ .

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

Step 4.2.3.2

Apply the power rule and multiply exponents,

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

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

Step 4.2.3.3

Combine

$\frac{1}{2}$ and

$2$ .

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

Step 4.2.3.4

Cancel the common factor of

$2$ .

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Step 4.2.3.4.1

Cancel the common factor.

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

Step 4.2.3.4.2

Rewrite the expression.

$f^{- 1} (\sqrt{x + 5}) = (x + 5) - 5$

Step 4.2.3.5

Simplify.

$f^{- 1} (\sqrt{x + 5}) = x + 5 - 5$

Step 4.2.4

Combine the opposite terms in

$x + 5 - 5$ .

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Step 4.2.4.1

Subtract

$5$ from

$5$ .

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

Step 4.2.4.2

Add

$x$ and

$0$ .

$f^{- 1} (\sqrt{x + 5}) = x$

Step 4.3

Evaluate

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

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Step 4.3.1

Set up the composite result function.

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

Step 4.3.2

Evaluate

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

$f^{- 1}$ into

$f$ .

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

Step 4.3.3

Add

$- 5$ and

$5$ .

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

Step 4.3.4

Add

$x^{2}$ and

$0$ .

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

Step 4.3.5

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

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

Step 4.4

Since

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

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

$f^{- 1} (x) = x^{2} - 5$ is the inverse of

$f (x) = \sqrt{x + 5}$ .

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