Find the Inverse square root of 8+x

$\sqrt{8 + x}$

Step 1

Interchange the variables.

$x = \sqrt{8 + y}$

Step 2

Solve for $y$ .

Rewrite the equation as $\sqrt{8 + y} = x$ .

$\sqrt{8 + y} = x$

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

${\sqrt{8 + y}}^{2} = x^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{8 + y}$ as ${(8 + y)}^{\frac{1}{2}}$ .

${({(8 + y)}^{\frac{1}{2}})}^{2} = x^{2}$

Simplify the left side.

Simplify ${({(8 + y)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(8 + y)}^{\frac{1}{2}})}^{2}$ .

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

${(8 + y)}^{\frac{1}{2} \cdot 2} = x^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(8 + y)}^{\frac{1}{2} \cdot 2} = x^{2}$

Rewrite the expression.

${(8 + y)}^{1} = x^{2}$

Simplify.

$8 + y = x^{2}$

Subtract $8$ from both sides of the equation.

$y = x^{2} - 8$

Step 3

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

$f^{- 1} (x) = x^{2} - 8$

Step 4

Verify if $f^{- 1} (x) = x^{2} - 8$ is the inverse of $f (x) = \sqrt{8 + x}$ .

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 $f^{- 1} (\sqrt{8 + x})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\sqrt{8 + x}) = {(\sqrt{8 + x})}^{2} - 8$

Rewrite ${\sqrt{8 + x}}^{2}$ as $8 + x$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{8 + x}$ as ${(8 + x)}^{\frac{1}{2}}$ .

$f^{- 1} (\sqrt{8 + x}) = {({(8 + x)}^{\frac{1}{2}})}^{2} - 8$

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

$f^{- 1} (\sqrt{8 + x}) = {(8 + x)}^{\frac{1}{2} \cdot 2} - 8$

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

$f^{- 1} (\sqrt{8 + x}) = {(8 + x)}^{\frac{2}{2}} - 8$

Cancel the common factor of $2$ .

Cancel the common factor.

$f^{- 1} (\sqrt{8 + x}) = {(8 + x)}^{\frac{2}{2}} - 8$

Rewrite the expression.

$f^{- 1} (\sqrt{8 + x}) = (8 + x) - 8$

Simplify.

$f^{- 1} (\sqrt{8 + x}) = 8 + x - 8$

Combine the opposite terms in $8 + x - 8$ .

Subtract $8$ from $8$ .

$f^{- 1} (\sqrt{8 + x}) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (\sqrt{8 + x}) = x$

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

Set up the composite result function.

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

Evaluate $f (x^{2} - 8)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (x^{2} - 8) = \sqrt{8 + x^{2} - 8}$

Remove parentheses.

$f (x^{2} - 8) = \sqrt{8 + x^{2} - 8}$

Subtract $8$ from $8$ .

$f (x^{2} - 8) = \sqrt{x^{2} + 0}$

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

$f (x^{2} - 8) = \sqrt{x^{2}}$

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

$f (x^{2} - 8) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = x^{2} - 8$ is the inverse of $f (x) = \sqrt{8 + x}$ .

$f^{- 1} (x) = x^{2} - 8$

Find the Inverse square root of 8+x

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