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

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