Find the Inverse f(x)=3/4 square root of x+1

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

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

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

Interchange the variables.

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

Solve for $y$ .

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

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

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

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

Simplify both sides of the equation.

Simplify the left side.

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

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

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $3$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the right side.

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

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

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

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

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

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

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

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

Apply the product rule to $4 x$ .

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

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

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

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

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

Subtract $1$ from both sides of the equation.

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

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

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

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

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Evaluate $f^{- 1} (f (x))$ .

Set up the composite result function.

$f^{- 1} (f (x))$

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

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

Simplify each term.

Simplify the numerator.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Combine exponents.

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

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

Multiply $16$ by $9$ .

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

Divide $144$ by $16$ .

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

Cancel the common factor of $9$ .

Cancel the common factor.

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

Divide $x + 1$ by $1$ .

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

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

Subtract $1$ from $1$ .

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

Add $x$ and $0$ .

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

Evaluate $f (f^{- 1} (x))$ .

Set up the composite result function.

$f (f^{- 1} (x))$

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

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

Simplify by adding numbers.

Add $- 1$ and $1$ .

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

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

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

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

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

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $4$ .

Factor $4$ out of $4 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Find the Inverse f(x)=3/4 square root of x+1

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