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

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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

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

Multiply both sides of the equation by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Simplify each side of the equation.

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

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

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, ${(a^{m})}^{n} = a^{m n}$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

Simplify ${(3 x)}^{2}$ .

Apply the product rule to $3 x$ .

$y + 1 = 3^{2} x^{2}$

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

$y + 1 = 9 x^{2}$

Subtract $1$ from both sides of the equation.

$y = 9 x^{2} - 1$

Step 3

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

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

Step 4

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

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

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

Simplify each term.

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

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

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

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

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

Cancel the common factor of $9$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

Subtract $1$ from $1$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Simplify the numerator.

Add $- 1$ and $1$ .

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

Add $9 x^{2}$ and $0$ .

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

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

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

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