Precalculus Examples

$y = \sqrt{x - 2}$

Step 1

Interchange the variables.

$x = \sqrt{y - 2}$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $\sqrt{y - 2} = x$ .

$\sqrt{y - 2} = x$

Step 2.2

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

${\sqrt{y - 2}}^{2} = x^{2}$

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 $\sqrt{y - 2}$ as ${(y - 2)}^{\frac{1}{2}}$ .

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

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Simplify ${({(y - 2)}^{\frac{1}{2}})}^{2}$ .

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Step 2.3.2.1.1

Multiply the exponents in ${({(y - 2)}^{\frac{1}{2}})}^{2}$ .

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Step 2.3.2.1.1.1

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

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

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.

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

Step 2.3.2.1.1.2.2

Rewrite the expression.

${(y - 2)}^{1} = x^{2}$

Step 2.3.2.1.2

Simplify.

$y - 2 = x^{2}$

Step 2.4

Add $2$ to both sides of the equation.

$y = x^{2} + 2$

Step 3

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

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

Step 4

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

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

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

Step 4.2.3

Rewrite ${\sqrt{x - 2}}^{2}$ as $x - 2$ .

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Step 4.2.3.1

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

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

Step 4.2.3.2

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

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

Step 4.2.3.3

Combine $\frac{1}{2}$ and $2$ .

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

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.

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

Step 4.2.3.4.2

Rewrite the expression.

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

Step 4.2.3.5

Simplify.

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

Step 4.2.4

Combine the opposite terms in $x - 2 + 2$ .

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Step 4.2.4.1

Add $- 2$ and $2$ .

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

Step 4.2.4.2

Add $x$ and $0$ .

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

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

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

Step 4.3.3

Subtract $2$ from $2$ .

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

Step 4.3.4

Add $x^{2}$ and $0$ .

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

Step 4.3.5

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

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

Step 4.4

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

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