Find the Inverse y = square root of x-6+2

$y = \sqrt{x - 6} + 2$

Step 1

Interchange the variables.

$x = \sqrt{y - 6} + 2$

Step 2

Solve for $y$ .

Rewrite the equation as $\sqrt{y - 6} + 2 = x$ .

$\sqrt{y - 6} + 2 = x$

Subtract $2$ from both sides of the equation.

$\sqrt{y - 6} = x - 2$

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

${\sqrt{y - 6}}^{2} = {(x - 2)}^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{y - 6}$ as ${(y - 6)}^{\frac{1}{2}}$ .

${({(y - 6)}^{\frac{1}{2}})}^{2} = {(x - 2)}^{2}$

Simplify the left side.

Simplify ${({(y - 6)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(y - 6)}^{\frac{1}{2}})}^{2}$ .

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

${(y - 6)}^{\frac{1}{2} \cdot 2} = {(x - 2)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(y - 6)}^{\frac{1}{2} \cdot 2} = {(x - 2)}^{2}$

Rewrite the expression.

${(y - 6)}^{1} = {(x - 2)}^{2}$

Simplify.

$y - 6 = {(x - 2)}^{2}$

Simplify the right side.

Simplify ${(x - 2)}^{2}$ .

Rewrite ${(x - 2)}^{2}$ as $(x - 2) (x - 2)$ .

$y - 6 = (x - 2) (x - 2)$

Expand $(x - 2) (x - 2)$ using the FOIL Method.

Apply the distributive property.

$y - 6 = x (x - 2) - 2 (x - 2)$

Apply the distributive property.

$y - 6 = x \cdot x + x \cdot - 2 - 2 (x - 2)$

Apply the distributive property.

$y - 6 = x \cdot x + x \cdot - 2 - 2 x - 2 \cdot - 2$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$y - 6 = x^{2} + x \cdot - 2 - 2 x - 2 \cdot - 2$

Move $- 2$ to the left of $x$ .

$y - 6 = x^{2} - 2 \cdot x - 2 x - 2 \cdot - 2$

Multiply $- 2$ by $- 2$ .

$y - 6 = x^{2} - 2 x - 2 x + 4$

Subtract $2 x$ from $- 2 x$ .

$y - 6 = x^{2} - 4 x + 4$

Move all terms not containing $y$ to the right side of the equation.

Add $6$ to both sides of the equation.

$y = x^{2} - 4 x + 4 + 6$

Add $4$ and $6$ .

$y = x^{2} - 4 x + 10$

Step 3

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

$f^{- 1} (x) = x^{2} - 4 x + 10$

Step 4

Verify if $f^{- 1} (x) = x^{2} - 4 x + 10$ is the inverse of $f (x) = \sqrt{x - 6} + 2$ .

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 $f^{- 1} (\sqrt{x - 6} + 2)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\sqrt{x - 6} + 2) = {(\sqrt{x - 6} + 2)}^{2} - 4 (\sqrt{x - 6} + 2) + 10$

Simplify each term.

Rewrite ${(\sqrt{x - 6} + 2)}^{2}$ as $(\sqrt{x - 6} + 2) (\sqrt{x - 6} + 2)$ .

$f^{- 1} (\sqrt{x - 6} + 2) = (\sqrt{x - 6} + 2) (\sqrt{x - 6} + 2) - 4 (\sqrt{x - 6} + 2) + 10$

Expand $(\sqrt{x - 6} + 2) (\sqrt{x - 6} + 2)$ using the FOIL Method.

Apply the distributive property.

$f^{- 1} (\sqrt{x - 6} + 2) = \sqrt{x - 6} (\sqrt{x - 6} + 2) + 2 (\sqrt{x - 6} + 2) - 4 (\sqrt{x - 6} + 2) + 10$

Apply the distributive property.

$f^{- 1} (\sqrt{x - 6} + 2) = \sqrt{x - 6} \sqrt{x - 6} + \sqrt{x - 6} \cdot 2 + 2 (\sqrt{x - 6} + 2) - 4 (\sqrt{x - 6} + 2) + 10$

Apply the distributive property.

$f^{- 1} (\sqrt{x - 6} + 2) = \sqrt{x - 6} \sqrt{x - 6} + \sqrt{x - 6} \cdot 2 + 2 \sqrt{x - 6} + 2 \cdot 2 - 4 (\sqrt{x - 6} + 2) + 10$

Simplify and combine like terms.

Simplify each term.

Multiply $\sqrt{x - 6} \sqrt{x - 6}$ .

Raise $\sqrt{x - 6}$ to the power of $1$ .

$f^{- 1} (\sqrt{x - 6} + 2) = \sqrt{x - 6} \sqrt{x - 6} + \sqrt{x - 6} \cdot 2 + 2 \sqrt{x - 6} + 2 \cdot 2 - 4 (\sqrt{x - 6} + 2) + 10$

Raise $\sqrt{x - 6}$ to the power of $1$ .

$f^{- 1} (\sqrt{x - 6} + 2) = \sqrt{x - 6} \sqrt{x - 6} + \sqrt{x - 6} \cdot 2 + 2 \sqrt{x - 6} + 2 \cdot 2 - 4 (\sqrt{x - 6} + 2) + 10$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f^{- 1} (\sqrt{x - 6} + 2) = {\sqrt{x - 6}}^{1 + 1} + \sqrt{x - 6} \cdot 2 + 2 \sqrt{x - 6} + 2 \cdot 2 - 4 (\sqrt{x - 6} + 2) + 10$

Add $1$ and $1$ .

$f^{- 1} (\sqrt{x - 6} + 2) = {\sqrt{x - 6}}^{2} + \sqrt{x - 6} \cdot 2 + 2 \sqrt{x - 6} + 2 \cdot 2 - 4 (\sqrt{x - 6} + 2) + 10$

Rewrite ${\sqrt{x - 6}}^{2}$ as $x - 6$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x - 6}$ as ${(x - 6)}^{\frac{1}{2}}$ .

$f^{- 1} (\sqrt{x - 6} + 2) = {({(x - 6)}^{\frac{1}{2}})}^{2} + \sqrt{x - 6} \cdot 2 + 2 \sqrt{x - 6} + 2 \cdot 2 - 4 (\sqrt{x - 6} + 2) + 10$

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

$f^{- 1} (\sqrt{x - 6} + 2) = {(x - 6)}^{\frac{1}{2} \cdot 2} + \sqrt{x - 6} \cdot 2 + 2 \sqrt{x - 6} + 2 \cdot 2 - 4 (\sqrt{x - 6} + 2) + 10$

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

$f^{- 1} (\sqrt{x - 6} + 2) = {(x - 6)}^{\frac{2}{2}} + \sqrt{x - 6} \cdot 2 + 2 \sqrt{x - 6} + 2 \cdot 2 - 4 (\sqrt{x - 6} + 2) + 10$

Cancel the common factor of $2$ .

Cancel the common factor.

$f^{- 1} (\sqrt{x - 6} + 2) = {(x - 6)}^{\frac{2}{2}} + \sqrt{x - 6} \cdot 2 + 2 \sqrt{x - 6} + 2 \cdot 2 - 4 (\sqrt{x - 6} + 2) + 10$

Rewrite the expression.

$f^{- 1} (\sqrt{x - 6} + 2) = (x - 6) + \sqrt{x - 6} \cdot 2 + 2 \sqrt{x - 6} + 2 \cdot 2 - 4 (\sqrt{x - 6} + 2) + 10$

Simplify.

$f^{- 1} (\sqrt{x - 6} + 2) = x - 6 + \sqrt{x - 6} \cdot 2 + 2 \sqrt{x - 6} + 2 \cdot 2 - 4 (\sqrt{x - 6} + 2) + 10$

Move $2$ to the left of $\sqrt{x - 6}$ .

$f^{- 1} (\sqrt{x - 6} + 2) = x - 6 + 2 \cdot \sqrt{x - 6} + 2 \sqrt{x - 6} + 2 \cdot 2 - 4 (\sqrt{x - 6} + 2) + 10$

Multiply $2$ by $2$ .

$f^{- 1} (\sqrt{x - 6} + 2) = x - 6 + 2 \sqrt{x - 6} + 2 \sqrt{x - 6} + 4 - 4 (\sqrt{x - 6} + 2) + 10$

Add $- 6$ and $4$ .

$f^{- 1} (\sqrt{x - 6} + 2) = x - 2 + 2 \sqrt{x - 6} + 2 \sqrt{x - 6} - 4 (\sqrt{x - 6} + 2) + 10$

Add $2 \sqrt{x - 6}$ and $2 \sqrt{x - 6}$ .

$f^{- 1} (\sqrt{x - 6} + 2) = x - 2 + 4 \sqrt{x - 6} - 4 (\sqrt{x - 6} + 2) + 10$

Apply the distributive property.

$f^{- 1} (\sqrt{x - 6} + 2) = x - 2 + 4 \sqrt{x - 6} - 4 \sqrt{x - 6} - 4 \cdot 2 + 10$

Multiply $- 4$ by $2$ .

$f^{- 1} (\sqrt{x - 6} + 2) = x - 2 + 4 \sqrt{x - 6} - 4 \sqrt{x - 6} - 8 + 10$

Simplify by adding terms.

Combine the opposite terms in $x - 2 + 4 \sqrt{x - 6} - 4 \sqrt{x - 6} - 8 + 10$ .

Subtract $4 \sqrt{x - 6}$ from $4 \sqrt{x - 6}$ .

$f^{- 1} (\sqrt{x - 6} + 2) = x - 2 + 0 - 8 + 10$

Add $x - 2$ and $0$ .

$f^{- 1} (\sqrt{x - 6} + 2) = x - 2 - 8 + 10$

Subtract $8$ from $- 2$ .

$f^{- 1} (\sqrt{x - 6} + 2) = x - 10 + 10$

Combine the opposite terms in $x - 10 + 10$ .

Add $- 10$ and $10$ .

$f^{- 1} (\sqrt{x - 6} + 2) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (\sqrt{x - 6} + 2) = x$

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

Set up the composite result function.

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

Evaluate $f (x^{2} - 4 x + 10)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (x^{2} - 4 x + 10) = \sqrt{(x^{2} - 4 x + 10) - 6} + 2$

Simplify each term.

Subtract $6$ from $10$ .

$f (x^{2} - 4 x + 10) = \sqrt{x^{2} - 4 x + 4} + 2$

Factor using the perfect square rule.

Rewrite $4$ as $2^{2}$ .

$f (x^{2} - 4 x + 10) = \sqrt{x^{2} - 4 x + 2^{2}} + 2$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 x = 2 \cdot x \cdot 2$

Rewrite the polynomial.

$f (x^{2} - 4 x + 10) = \sqrt{x^{2} - 2 \cdot x \cdot 2 + 2^{2}} + 2$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 2$ .

$f (x^{2} - 4 x + 10) = \sqrt{{(x - 2)}^{2}} + 2$

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

$f (x^{2} - 4 x + 10) = x - 2 + 2$

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

Add $- 2$ and $2$ .

$f (x^{2} - 4 x + 10) = x + 0$

Add $x$ and $0$ .

$f (x^{2} - 4 x + 10) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = x^{2} - 4 x + 10$ is the inverse of $f (x) = \sqrt{x - 6} + 2$ .

$f^{- 1} (x) = x^{2} - 4 x + 10$

Find the Inverse y = square root of x-6+2

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