Find the Inverse square root of -2+x

$\sqrt{- 2 + x}$

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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

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

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

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

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Add $2$ to both sides of the equation.

$y = x^{2} + 2$

Step 3

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

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

Step 4

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

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

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

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

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

Add $- 2$ and $2$ .

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

Add $x$ and $0$ .

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

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 $f$ .

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

Remove parentheses.

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

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

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

Find the Inverse square root of -2+x

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