Find the Inverse square root of 2x+2

$\sqrt{2 x + 2}$

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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

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

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

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

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Solve for $y$ .

Subtract $2$ from both sides of the equation.

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

Divide each term in $2 y = x^{2} - 2$ by $2$ and simplify.

Divide each term in $2 y = x^{2} - 2$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Divide $- 2$ by $2$ .

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

Step 3

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

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

Step 4

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

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

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

Simplify each term.

Simplify the numerator.

Factor $2$ out of $2 x + 2$ .

Factor $2$ out of $2 x$ .

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

Factor $2$ out of $2$ .

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

Factor $2$ out of $2 (x) + 2 (1)$ .

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

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Apply the distributive property.

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

Multiply $2$ by $1$ .

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

Factor $2$ out of $2 x + 2$ .

Factor $2$ out of $2 x$ .

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

Factor $2$ out of $2$ .

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

Factor $2$ out of $2 (x) + 2 (1)$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x + 1$ by $1$ .

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

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

Subtract $1$ from $1$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Factor $2$ out of $2 (\frac{x^{2}}{2} - 1) + 2$ .

Factor $2$ out of $2$ .

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

Factor $2$ out of $2 (\frac{x^{2}}{2} - 1) + 2 (1)$ .

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

Simplify by adding numbers.

Add $- 1$ and $1$ .

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

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

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

Combine $2$ and $\frac{x^{2}}{2}$ .

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

Reduce the expression by cancelling the common factors.

Reduce the expression $\frac{2 x^{2}}{2}$ by cancelling the common factors.

Cancel the common factor.

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

Rewrite the expression.

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

Divide $x^{2}$ by $1$ .

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

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

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

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

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

Find the Inverse square root of 2x+2

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