Find the Inverse square root of x-5

$\sqrt{x - 5}$

Interchange the variables.

$x = \sqrt{y - 5}$

Solve for $y$ .

Rewrite the equation as $\sqrt{y - 5} = x$ .

$\sqrt{y - 5} = x$

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

${\sqrt{y - 5}}^{2} = x^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{y - 5}$ as ${(y - 5)}^{\frac{1}{2}}$ .

${({(y - 5)}^{\frac{1}{2}})}^{2} = x^{2}$

Simplify the left side.

Simplify ${({(y - 5)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(y - 5)}^{\frac{1}{2}})}^{2}$ .

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

${(y - 5)}^{\frac{1}{2} \cdot 2} = x^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(y - 5)}^{\frac{1}{2} \cdot 2} = x^{2}$

Rewrite the expression.

${(y - 5)}^{1} = x^{2}$

Simplify.

$y - 5 = x^{2}$

Add $5$ to both sides of the equation.

$y = x^{2} + 5$

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

$f^{- 1} (x) = x^{2} + 5$

Verify if $f^{- 1} (x) = x^{2} + 5$ is the inverse of $f (x) = \sqrt{x - 5}$ .

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 $f^{- 1} (\sqrt{x - 5})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\sqrt{x - 5}) = {(\sqrt{x - 5})}^{2} + 5$

Rewrite ${\sqrt{x - 5}}^{2}$ as $x - 5$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x - 5}$ as ${(x - 5)}^{\frac{1}{2}}$ .

$f^{- 1} (\sqrt{x - 5}) = {({(x - 5)}^{\frac{1}{2}})}^{2} + 5$

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

$f^{- 1} (\sqrt{x - 5}) = {(x - 5)}^{\frac{1}{2} \cdot 2} + 5$

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

$f^{- 1} (\sqrt{x - 5}) = {(x - 5)}^{\frac{2}{2}} + 5$

Cancel the common factor of $2$ .

Cancel the common factor.

$f^{- 1} (\sqrt{x - 5}) = {(x - 5)}^{\frac{2}{2}} + 5$

Rewrite the expression.

$f^{- 1} (\sqrt{x - 5}) = (x - 5) + 5$

Simplify.

$f^{- 1} (\sqrt{x - 5}) = x - 5 + 5$

Combine the opposite terms in $x - 5 + 5$ .

Add $- 5$ and $5$ .

$f^{- 1} (\sqrt{x - 5}) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (\sqrt{x - 5}) = x$

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

Set up the composite result function.

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

Evaluate $f (x^{2} + 5)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (x^{2} + 5) = \sqrt{(x^{2} + 5) - 5}$

Subtract $5$ from $5$ .

$f (x^{2} + 5) = \sqrt{x^{2} + 0}$

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

$f (x^{2} + 5) = \sqrt{x^{2}}$

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

$f (x^{2} + 5) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = x^{2} + 5$ is the inverse of $f (x) = \sqrt{x - 5}$ .

$f^{- 1} (x) = x^{2} + 5$

Find the Inverse square root of x-5

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