Algebra Examples

$y = 2 \sqrt{x + 7}$

Step 1

Interchange the variables.

$x = 2 \sqrt{y + 7}$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $2 \sqrt{y + 7} = x$ .

$2 \sqrt{y + 7} = x$

Step 2.2

Divide each term in $2 \sqrt{y + 7} = x$ by $2$ and simplify.

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Step 2.2.1

Divide each term in $2 \sqrt{y + 7} = x$ by $2$ .

$\frac{2 \sqrt{y + 7}}{2} = \frac{x}{2}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of $2$ .

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Step 2.2.2.1.1

Cancel the common factor.

$\frac{2 \sqrt{y + 7}}{2} = \frac{x}{2}$

Step 2.2.2.1.2

Divide $\sqrt{y + 7}$ by $1$ .

$\sqrt{y + 7} = \frac{x}{2}$

Step 2.3

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

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

Step 2.4

Simplify each side of the equation.

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Step 2.4.1

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

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

Step 2.4.2

Simplify the left side.

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Step 2.4.2.1

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

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Step 2.4.2.1.1

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

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Step 2.4.2.1.1.1

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

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

Step 2.4.2.1.1.2

Cancel the common factor of $2$ .

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Step 2.4.2.1.1.2.1

Cancel the common factor.

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

Step 2.4.2.1.1.2.2

Rewrite the expression.

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

Step 2.4.2.1.2

Simplify.

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

Step 2.4.3

Simplify the right side.

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Step 2.4.3.1

Simplify ${(\frac{x}{2})}^{2}$ .

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Step 2.4.3.1.1

Apply the product rule to $\frac{x}{2}$ .

$y + 7 = \frac{x^{2}}{2^{2}}$

Step 2.4.3.1.2

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

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

Step 2.5

Subtract $7$ from both sides of the equation.

$y = \frac{x^{2}}{4} - 7$

Step 3

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

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

Step 4

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

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

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

Step 4.2.3

Simplify each term.

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Step 4.2.3.1

Simplify the numerator.

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Step 4.2.3.1.1

Apply the product rule to $2 \sqrt{x + 7}$ .

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

Step 4.2.3.1.2

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

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

Step 4.2.3.1.3

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

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Step 4.2.3.1.3.1

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

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

Step 4.2.3.1.3.2

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

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

Step 4.2.3.1.3.3

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

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

Step 4.2.3.1.3.4

Cancel the common factor of $2$ .

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Step 4.2.3.1.3.4.1

Cancel the common factor.

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

Step 4.2.3.1.3.4.2

Rewrite the expression.

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

Step 4.2.3.1.3.5

Simplify.

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

Step 4.2.3.2

Cancel the common factor of $4$ .

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Step 4.2.3.2.1

Cancel the common factor.

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

Step 4.2.3.2.2

Divide $x + 7$ by $1$ .

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

Step 4.2.4

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

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Step 4.2.4.1

Subtract $7$ from $7$ .

$f^{- 1} (2 \sqrt{x + 7}) = x + 0$

Step 4.2.4.2

Add $x$ and $0$ .

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

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 $f (\frac{x^{2}}{4} - 7)$ by substituting in the value of $f^{- 1}$ into $f$ .

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

Step 4.3.3

Add $- 7$ and $7$ .

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

Step 4.3.4

Add $\frac{x^{2}}{4}$ and $0$ .

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

Step 4.3.5

Rewrite $4$ as $2^{2}$ .

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

Step 4.3.6

Rewrite $\frac{x^{2}}{2^{2}}$ as ${(\frac{x}{2})}^{2}$ .

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

Step 4.3.7

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

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

Step 4.3.8

Cancel the common factor of $2$ .

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Step 4.3.8.1

Cancel the common factor.

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

Step 4.3.8.2

Rewrite the expression.

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

Step 4.4

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

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