Precalculus Examples

$f (x) = \frac{\sqrt{x + 7}}{4}$

Step 1

Write $f (x) = \frac{\sqrt{x + 7}}{4}$ as an equation.

$y = \frac{\sqrt{x + 7}}{4}$

Step 2

Interchange the variables.

$x = \frac{\sqrt{y + 7}}{4}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{\sqrt{y + 7}}{4} = x$ .

$\frac{\sqrt{y + 7}}{4} = x$

Step 3.2

Multiply both sides of the equation by $4$ .

$4 \frac{\sqrt{y + 7}}{4} = 4 x$

Step 3.3

Simplify the left side.

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Step 3.3.1

Cancel the common factor of $4$ .

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Step 3.3.1.1

Cancel the common factor.

$4 \frac{\sqrt{y + 7}}{4} = 4 x$

Step 3.3.1.2

Rewrite the expression.

$\sqrt{y + 7} = 4 x$

Step 3.4

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

${\sqrt{y + 7}}^{2} = {(4 x)}^{2}$

Step 3.5

Simplify each side of the equation.

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Step 3.5.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{y + 7}$ as ${(y + 7)}^{\frac{1}{2}}$ .

${({(y + 7)}^{\frac{1}{2}})}^{2} = {(4 x)}^{2}$

Step 3.5.2

Simplify the left side.

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Step 3.5.2.1

Simplify ${({(y + 7)}^{\frac{1}{2}})}^{2}$ .

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Step 3.5.2.1.1

Multiply the exponents in ${({(y + 7)}^{\frac{1}{2}})}^{2}$ .

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Step 3.5.2.1.1.1

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

${(y + 7)}^{\frac{1}{2} \cdot 2} = {(4 x)}^{2}$

Step 3.5.2.1.1.2

Cancel the common factor of $2$ .

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Step 3.5.2.1.1.2.1

Cancel the common factor.

${(y + 7)}^{\frac{1}{2} \cdot 2} = {(4 x)}^{2}$

Step 3.5.2.1.1.2.2

Rewrite the expression.

${(y + 7)}^{1} = {(4 x)}^{2}$

Step 3.5.2.1.2

Simplify.

$y + 7 = {(4 x)}^{2}$

Step 3.5.3

Simplify the right side.

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Step 3.5.3.1

Simplify ${(4 x)}^{2}$ .

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Step 3.5.3.1.1

Apply the product rule to $4 x$ .

$y + 7 = 4^{2} x^{2}$

Step 3.5.3.1.2

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

$y + 7 = 16 x^{2}$

Step 3.6

Subtract $7$ from both sides of the equation.

$y = 16 x^{2} - 7$

Step 4

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

$f^{- 1} (x) = 16 x^{2} - 7$

Step 5

Verify if $f^{- 1} (x) = 16 x^{2} - 7$ is the inverse of $f (x) = \frac{\sqrt{x + 7}}{4}$ .

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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 $f^{- 1} (\frac{\sqrt{x + 7}}{4})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{\sqrt{x + 7}}{4}) = 16 {(\frac{\sqrt{x + 7}}{4})}^{2} - 7$

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

Apply the product rule to $\frac{\sqrt{x + 7}}{4}$ .

$f^{- 1} (\frac{\sqrt{x + 7}}{4}) = 16 (\frac{{\sqrt{x + 7}}^{2}}{4^{2}}) - 7$

Step 5.2.3.2

Rewrite ${\sqrt{x + 7}}^{2}$ as $x + 7$ .

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Step 5.2.3.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 7}$ as ${(x + 7)}^{\frac{1}{2}}$ .

$f^{- 1} (\frac{\sqrt{x + 7}}{4}) = 16 (\frac{{({(x + 7)}^{\frac{1}{2}})}^{2}}{4^{2}}) - 7$

Step 5.2.3.2.2

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

$f^{- 1} (\frac{\sqrt{x + 7}}{4}) = 16 (\frac{{(x + 7)}^{\frac{1}{2} \cdot 2}}{4^{2}}) - 7$

Step 5.2.3.2.3

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

$f^{- 1} (\frac{\sqrt{x + 7}}{4}) = 16 (\frac{{(x + 7)}^{\frac{2}{2}}}{4^{2}}) - 7$

Step 5.2.3.2.4

Cancel the common factor of $2$ .

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Step 5.2.3.2.4.1

Cancel the common factor.

$f^{- 1} (\frac{\sqrt{x + 7}}{4}) = 16 (\frac{{(x + 7)}^{\frac{2}{2}}}{4^{2}}) - 7$

Step 5.2.3.2.4.2

Rewrite the expression.

$f^{- 1} (\frac{\sqrt{x + 7}}{4}) = 16 (\frac{x + 7}{4^{2}}) - 7$

Step 5.2.3.2.5

Simplify.

$f^{- 1} (\frac{\sqrt{x + 7}}{4}) = 16 (\frac{x + 7}{4^{2}}) - 7$

Step 5.2.3.3

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

$f^{- 1} (\frac{\sqrt{x + 7}}{4}) = 16 (\frac{x + 7}{16}) - 7$

Step 5.2.3.4

Cancel the common factor of $16$ .

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Step 5.2.3.4.1

Cancel the common factor.

$f^{- 1} (\frac{\sqrt{x + 7}}{4}) = 16 (\frac{x + 7}{16}) - 7$

Step 5.2.3.4.2

Rewrite the expression.

$f^{- 1} (\frac{\sqrt{x + 7}}{4}) = x + 7 - 7$

Step 5.2.4

Combine the opposite terms in $x + 7 - 7$ .

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Step 5.2.4.1

Subtract $7$ from $7$ .

$f^{- 1} (\frac{\sqrt{x + 7}}{4}) = x + 0$

Step 5.2.4.2

Add $x$ and $0$ .

$f^{- 1} (\frac{\sqrt{x + 7}}{4}) = x$

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 $f (16 x^{2} - 7)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (16 x^{2} - 7) = \frac{\sqrt{(16 x^{2} - 7) + 7}}{4}$

Step 5.3.3

Simplify the numerator.

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Step 5.3.3.1

Add $- 7$ and $7$ .

$f (16 x^{2} - 7) = \frac{\sqrt{16 x^{2} + 0}}{4}$

Step 5.3.3.2

Add $16 x^{2}$ and $0$ .

$f (16 x^{2} - 7) = \frac{\sqrt{16 x^{2}}}{4}$

Step 5.3.3.3

Rewrite $16 x^{2}$ as ${(4 x)}^{2}$ .

$f (16 x^{2} - 7) = \frac{\sqrt{{(4 x)}^{2}}}{4}$

Step 5.3.3.4

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

$f (16 x^{2} - 7) = \frac{4 x}{4}$

Step 5.3.4

Cancel the common factor of $4$ .

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Step 5.3.4.1

Cancel the common factor.

$f (16 x^{2} - 7) = \frac{4 x}{4}$

Step 5.3.4.2

Divide $x$ by $1$ .

$f (16 x^{2} - 7) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = 16 x^{2} - 7$ is the inverse of $f (x) = \frac{\sqrt{x + 7}}{4}$ .

$f^{- 1} (x) = 16 x^{2} - 7$