Find the Inverse y = square root of x+2

$y = \sqrt{x + 2}$

Interchange the variables.

$x = \sqrt{y + 2}$

Solve for $y$ .

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

$\sqrt{y + 2} = x$

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

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

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

$y + 2 = x^{2}$

Subtract $2$ from both sides of the equation.

$y = x^{2} - 2$

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

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

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

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

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

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Combine the opposite terms in $x + 2 - 2$ .

Subtract $2$ from $2$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Add $- 2$ and $2$ .

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

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

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

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

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

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

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

Find the Inverse y = square root of x+2

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