Algebra Examples

$f (x) = \frac{3}{4} \sqrt{x + 1}$

Step 1

Write $f (x) = \frac{3}{4} \sqrt{x + 1}$ as an equation.

$y = \frac{3}{4} \sqrt{x + 1}$

Step 2

Interchange the variables.

$x = \frac{3}{4} \cdot \sqrt{y + 1}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{3}{4} \cdot \sqrt{y + 1} = x$ .

$\frac{3}{4} \cdot \sqrt{y + 1} = x$

Step 3.2

Multiply both sides of the equation by $\frac{4}{3}$ .

$\frac{4}{3} (\frac{3}{4} \cdot \sqrt{y + 1}) = \frac{4}{3} x$

Step 3.3

Simplify both sides of the equation.

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Step 3.3.1

Simplify the left side.

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Step 3.3.1.1

Simplify $\frac{4}{3} (\frac{3}{4} \cdot \sqrt{y + 1})$ .

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Step 3.3.1.1.1

Combine $\frac{3}{4}$ and $\sqrt{y + 1}$ .

$\frac{4}{3} \cdot \frac{3 \sqrt{y + 1}}{4} = \frac{4}{3} x$

Step 3.3.1.1.2

Cancel the common factor of $4$ .

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Step 3.3.1.1.2.1

Cancel the common factor.

$\frac{4}{3} \cdot \frac{3 \sqrt{y + 1}}{4} = \frac{4}{3} x$

Step 3.3.1.1.2.2

Rewrite the expression.

$\frac{1}{3} (3 \sqrt{y + 1}) = \frac{4}{3} x$

Step 3.3.1.1.3

Cancel the common factor of $3$ .

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Step 3.3.1.1.3.1

Factor $3$ out of $3 \sqrt{y + 1}$ .

$\frac{1}{3} (3 (\sqrt{y + 1})) = \frac{4}{3} x$

Step 3.3.1.1.3.2

Cancel the common factor.

$\frac{1}{3} (3 \sqrt{y + 1}) = \frac{4}{3} x$

Step 3.3.1.1.3.3

Rewrite the expression.

$\sqrt{y + 1} = \frac{4}{3} x$

Step 3.3.2

Simplify the right side.

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Step 3.3.2.1

Combine $\frac{4}{3}$ and $x$ .

$\sqrt{y + 1} = \frac{4 x}{3}$

Step 3.4

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

${\sqrt{y + 1}}^{2} = {(\frac{4 x}{3})}^{2}$

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

${({(y + 1)}^{\frac{1}{2}})}^{2} = {(\frac{4 x}{3})}^{2}$

Step 3.5.2

Simplify the left side.

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Step 3.5.2.1

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

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Step 3.5.2.1.1

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

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Step 3.5.2.1.1.1

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

${(y + 1)}^{\frac{1}{2} \cdot 2} = {(\frac{4 x}{3})}^{2}$

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.

${(y + 1)}^{\frac{1}{2} \cdot 2} = {(\frac{4 x}{3})}^{2}$

Step 3.5.2.1.1.2.2

Rewrite the expression.

${(y + 1)}^{1} = {(\frac{4 x}{3})}^{2}$

Step 3.5.2.1.2

Simplify.

$y + 1 = {(\frac{4 x}{3})}^{2}$

Step 3.5.3

Simplify the right side.

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Step 3.5.3.1

Simplify ${(\frac{4 x}{3})}^{2}$ .

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Step 3.5.3.1.1

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 3.5.3.1.1.1

Apply the product rule to $\frac{4 x}{3}$ .

$y + 1 = \frac{{(4 x)}^{2}}{3^{2}}$

Step 3.5.3.1.1.2

Apply the product rule to $4 x$ .

$y + 1 = \frac{4^{2} x^{2}}{3^{2}}$

Step 3.5.3.1.2

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

$y + 1 = \frac{16 x^{2}}{3^{2}}$

Step 3.5.3.1.3

Raise $3$ to the power of $2$ .

$y + 1 = \frac{16 x^{2}}{9}$

Step 3.6

Subtract $1$ from both sides of the equation.

$y = \frac{16 x^{2}}{9} - 1$

Step 4

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

$f^{- 1} (x) = \frac{16 x^{2}}{9} - 1$

Step 5

Verify if $f^{- 1} (x) = \frac{16 x^{2}}{9} - 1$ is the inverse of $f (x) = \frac{3}{4} \cdot \sqrt{x + 1}$ .

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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 $f^{- 1} (\frac{3}{4} \cdot \sqrt{x + 1})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{3}{4} \cdot \sqrt{x + 1}) = \frac{16 {(\frac{3}{4} \cdot \sqrt{x + 1})}^{2}}{9} - 1$

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

Simplify the numerator.

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Step 5.2.3.1.1

Apply the product rule to $\frac{3}{4} \cdot \sqrt{x + 1}$ .

$f^{- 1} (\frac{3}{4} \cdot \sqrt{x + 1}) = \frac{16 {(\frac{3}{4})}^{2} \cdot {\sqrt{x + 1}}^{2}}{9} - 1$

Step 5.2.3.1.2

Apply the product rule to $\frac{3}{4}$ .

$f^{- 1} (\frac{3}{4} \cdot \sqrt{x + 1}) = \frac{16 (\frac{3^{2}}{4^{2}}) \cdot {\sqrt{x + 1}}^{2}}{9} - 1$

Step 5.2.3.1.3

Raise $3$ to the power of $2$ .

$f^{- 1} (\frac{3}{4} \cdot \sqrt{x + 1}) = \frac{16 (\frac{9}{4^{2}}) \cdot {\sqrt{x + 1}}^{2}}{9} - 1$

Step 5.2.3.1.4

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

$f^{- 1} (\frac{3}{4} \cdot \sqrt{x + 1}) = \frac{16 (\frac{9}{16}) \cdot {\sqrt{x + 1}}^{2}}{9} - 1$

Step 5.2.3.1.5

Rewrite ${\sqrt{x + 1}}^{2}$ as $x + 1$ .

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Step 5.2.3.1.5.1

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

$f^{- 1} (\frac{3}{4} \cdot \sqrt{x + 1}) = \frac{16 (\frac{9}{16}) \cdot {({(x + 1)}^{\frac{1}{2}})}^{2}}{9} - 1$

Step 5.2.3.1.5.2

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

$f^{- 1} (\frac{3}{4} \cdot \sqrt{x + 1}) = \frac{16 (\frac{9}{16}) \cdot {(x + 1)}^{\frac{1}{2} \cdot 2}}{9} - 1$

Step 5.2.3.1.5.3

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

$f^{- 1} (\frac{3}{4} \cdot \sqrt{x + 1}) = \frac{16 (\frac{9}{16}) \cdot {(x + 1)}^{\frac{2}{2}}}{9} - 1$

Step 5.2.3.1.5.4

Cancel the common factor of $2$ .

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Step 5.2.3.1.5.4.1

Cancel the common factor.

$f^{- 1} (\frac{3}{4} \cdot \sqrt{x + 1}) = \frac{16 (\frac{9}{16}) \cdot {(x + 1)}^{\frac{2}{2}}}{9} - 1$

Step 5.2.3.1.5.4.2

Rewrite the expression.

$f^{- 1} (\frac{3}{4} \cdot \sqrt{x + 1}) = \frac{16 (\frac{9}{16}) \cdot (x + 1)}{9} - 1$

Step 5.2.3.1.5.5

Simplify.

$f^{- 1} (\frac{3}{4} \cdot \sqrt{x + 1}) = \frac{16 (\frac{9}{16}) \cdot (x + 1)}{9} - 1$

Step 5.2.3.1.6

Combine exponents.

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Step 5.2.3.1.6.1

Combine $16$ and $\frac{9}{16}$ .

$f^{- 1} (\frac{3}{4} \cdot \sqrt{x + 1}) = \frac{\frac{16 \cdot 9}{16} \cdot (x + 1)}{9} - 1$

Step 5.2.3.1.6.2

Multiply $16$ by $9$ .

$f^{- 1} (\frac{3}{4} \cdot \sqrt{x + 1}) = \frac{\frac{144}{16} \cdot (x + 1)}{9} - 1$

Step 5.2.3.1.7

Divide $144$ by $16$ .

$f^{- 1} (\frac{3}{4} \cdot \sqrt{x + 1}) = \frac{9 \cdot (x + 1)}{9} - 1$

Step 5.2.3.2

Cancel the common factor of $9$ .

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Step 5.2.3.2.1

Cancel the common factor.

$f^{- 1} (\frac{3}{4} \cdot \sqrt{x + 1}) = \frac{9 \cdot (x + 1)}{9} - 1$

Step 5.2.3.2.2

Divide $x + 1$ by $1$ .

$f^{- 1} (\frac{3}{4} \cdot \sqrt{x + 1}) = x + 1 - 1$

Step 5.2.4

Combine the opposite terms in $x + 1 - 1$ .

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Step 5.2.4.1

Subtract $1$ from $1$ .

$f^{- 1} (\frac{3}{4} \cdot \sqrt{x + 1}) = x + 0$

Step 5.2.4.2

Add $x$ and $0$ .

$f^{- 1} (\frac{3}{4} \cdot \sqrt{x + 1}) = x$

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 $f (\frac{16 x^{2}}{9} - 1)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{16 x^{2}}{9} - 1) = \frac{3}{4} \cdot \sqrt{(\frac{16 x^{2}}{9} - 1) + 1}$

Step 5.3.3

Simplify by adding numbers.

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Step 5.3.3.1

Add $- 1$ and $1$ .

$f (\frac{16 x^{2}}{9} - 1) = \frac{3}{4} \cdot \sqrt{\frac{16 x^{2}}{9} + 0}$

Step 5.3.3.2

Add $\frac{16 x^{2}}{9}$ and $0$ .

$f (\frac{16 x^{2}}{9} - 1) = \frac{3}{4} \cdot \sqrt{\frac{16 x^{2}}{9}}$

Step 5.3.4

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

$f (\frac{16 x^{2}}{9} - 1) = \frac{3}{4} \cdot \sqrt{\frac{{(4 x)}^{2}}{9}}$

Step 5.3.5

Rewrite $9$ as $3^{2}$ .

$f (\frac{16 x^{2}}{9} - 1) = \frac{3}{4} \cdot \sqrt{\frac{{(4 x)}^{2}}{3^{2}}}$

Step 5.3.6

Rewrite $\frac{{(4 x)}^{2}}{3^{2}}$ as ${(\frac{4 x}{3})}^{2}$ .

$f (\frac{16 x^{2}}{9} - 1) = \frac{3}{4} \cdot \sqrt{{(\frac{4 x}{3})}^{2}}$

Step 5.3.7

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

$f (\frac{16 x^{2}}{9} - 1) = \frac{3}{4} \cdot \frac{4 x}{3}$

Step 5.3.8

Cancel the common factor of $3$ .

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Step 5.3.8.1

Cancel the common factor.

$f (\frac{16 x^{2}}{9} - 1) = \frac{3}{4} \cdot \frac{4 x}{3}$

Step 5.3.8.2

Rewrite the expression.

$f (\frac{16 x^{2}}{9} - 1) = \frac{1}{4} \cdot (4 x)$

Step 5.3.9

Cancel the common factor of $4$ .

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Step 5.3.9.1

Factor $4$ out of $4 x$ .

$f (\frac{16 x^{2}}{9} - 1) = \frac{1}{4} \cdot (4 (x))$

Step 5.3.9.2

Cancel the common factor.

$f (\frac{16 x^{2}}{9} - 1) = \frac{1}{4} \cdot (4 x)$

Step 5.3.9.3

Rewrite the expression.

$f (\frac{16 x^{2}}{9} - 1) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{16 x^{2}}{9} - 1$ is the inverse of $f (x) = \frac{3}{4} \cdot \sqrt{x + 1}$ .

$f^{- 1} (x) = \frac{16 x^{2}}{9} - 1$