Find the Inverse square root of x-6

$\sqrt{x} - 6$

Step 1

Interchange the variables.

$x = \sqrt{y} - 6$

Step 2

Solve for $y$ .

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

$\sqrt{y} - 6 = x$

Add $6$ to both sides of the equation.

$\sqrt{y} = x + 6$

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

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

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

Simplify ${(x + 6)}^{2}$ .

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

$y = (x + 6) (x + 6)$

Expand $(x + 6) (x + 6)$ using the FOIL Method.

Apply the distributive property.

$y = x (x + 6) + 6 (x + 6)$

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

Move $6$ to the left of $x$ .

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

Multiply $6$ by $6$ .

$y = x^{2} + 6 x + 6 x + 36$

Add $6 x$ and $6 x$ .

$y = x^{2} + 12 x + 36$

Step 3

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

$f^{- 1} (x) = x^{2} + 12 x + 36$

Step 4

Verify if $f^{- 1} (x) = x^{2} + 12 x + 36$ is the inverse of $f (x) = \sqrt{x} - 6$ .

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

$f^{- 1} (\sqrt{x} - 6) = {(\sqrt{x} - 6)}^{2} + 12 (\sqrt{x} - 6) + 36$

Simplify each term.

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

$f^{- 1} (\sqrt{x} - 6) = (\sqrt{x} - 6) (\sqrt{x} - 6) + 12 (\sqrt{x} - 6) + 36$

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

Apply the distributive property.

$f^{- 1} (\sqrt{x} - 6) = \sqrt{x} (\sqrt{x} - 6) - 6 (\sqrt{x} - 6) + 12 (\sqrt{x} - 6) + 36$

Apply the distributive property.

$f^{- 1} (\sqrt{x} - 6) = \sqrt{x} \sqrt{x} + \sqrt{x} \cdot - 6 - 6 (\sqrt{x} - 6) + 12 (\sqrt{x} - 6) + 36$

Apply the distributive property.

$f^{- 1} (\sqrt{x} - 6) = \sqrt{x} \sqrt{x} + \sqrt{x} \cdot - 6 - 6 \sqrt{x} - 6 \cdot - 6 + 12 (\sqrt{x} - 6) + 36$

Simplify and combine like terms.

Simplify each term.

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

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

$f^{- 1} (\sqrt{x} - 6) = \sqrt{x} \sqrt{x} + \sqrt{x} \cdot - 6 - 6 \sqrt{x} - 6 \cdot - 6 + 12 (\sqrt{x} - 6) + 36$

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

$f^{- 1} (\sqrt{x} - 6) = \sqrt{x} \sqrt{x} + \sqrt{x} \cdot - 6 - 6 \sqrt{x} - 6 \cdot - 6 + 12 (\sqrt{x} - 6) + 36$

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

$f^{- 1} (\sqrt{x} - 6) = {\sqrt{x}}^{1 + 1} + \sqrt{x} \cdot - 6 - 6 \sqrt{x} - 6 \cdot - 6 + 12 (\sqrt{x} - 6) + 36$

Add $1$ and $1$ .

$f^{- 1} (\sqrt{x} - 6) = {\sqrt{x}}^{2} + \sqrt{x} \cdot - 6 - 6 \sqrt{x} - 6 \cdot - 6 + 12 (\sqrt{x} - 6) + 36$

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

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

$f^{- 1} (\sqrt{x} - 6) = {(x^{\frac{1}{2}})}^{2} + \sqrt{x} \cdot - 6 - 6 \sqrt{x} - 6 \cdot - 6 + 12 (\sqrt{x} - 6) + 36$

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

$f^{- 1} (\sqrt{x} - 6) = x^{\frac{1}{2} \cdot 2} + \sqrt{x} \cdot - 6 - 6 \sqrt{x} - 6 \cdot - 6 + 12 (\sqrt{x} - 6) + 36$

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

$f^{- 1} (\sqrt{x} - 6) = x^{\frac{2}{2}} + \sqrt{x} \cdot - 6 - 6 \sqrt{x} - 6 \cdot - 6 + 12 (\sqrt{x} - 6) + 36$

Cancel the common factor of $2$ .

Cancel the common factor.

$f^{- 1} (\sqrt{x} - 6) = x^{\frac{2}{2}} + \sqrt{x} \cdot - 6 - 6 \sqrt{x} - 6 \cdot - 6 + 12 (\sqrt{x} - 6) + 36$

Rewrite the expression.

$f^{- 1} (\sqrt{x} - 6) = x + \sqrt{x} \cdot - 6 - 6 \sqrt{x} - 6 \cdot - 6 + 12 (\sqrt{x} - 6) + 36$

Simplify.

$f^{- 1} (\sqrt{x} - 6) = x + \sqrt{x} \cdot - 6 - 6 \sqrt{x} - 6 \cdot - 6 + 12 (\sqrt{x} - 6) + 36$

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

$f^{- 1} (\sqrt{x} - 6) = x - 6 \cdot \sqrt{x} - 6 \sqrt{x} - 6 \cdot - 6 + 12 (\sqrt{x} - 6) + 36$

Multiply $- 6$ by $- 6$ .

$f^{- 1} (\sqrt{x} - 6) = x - 6 \sqrt{x} - 6 \sqrt{x} + 36 + 12 (\sqrt{x} - 6) + 36$

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

$f^{- 1} (\sqrt{x} - 6) = x - 12 \sqrt{x} + 36 + 12 (\sqrt{x} - 6) + 36$

Apply the distributive property.

$f^{- 1} (\sqrt{x} - 6) = x - 12 \sqrt{x} + 36 + 12 \sqrt{x} + 12 \cdot - 6 + 36$

Multiply $12$ by $- 6$ .

$f^{- 1} (\sqrt{x} - 6) = x - 12 \sqrt{x} + 36 + 12 \sqrt{x} - 72 + 36$

Simplify by adding terms.

Combine the opposite terms in $x - 12 \sqrt{x} + 36 + 12 \sqrt{x} - 72 + 36$ .

Add $- 12 \sqrt{x}$ and $12 \sqrt{x}$ .

$f^{- 1} (\sqrt{x} - 6) = x + 0 + 36 - 72 + 36$

Add $x$ and $0$ .

$f^{- 1} (\sqrt{x} - 6) = x + 36 - 72 + 36$

Subtract $72$ from $36$ .

$f^{- 1} (\sqrt{x} - 6) = x - 36 + 36$

Combine the opposite terms in $x - 36 + 36$ .

Add $- 36$ and $36$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

Evaluate $f (x^{2} + 12 x + 36)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (x^{2} + 12 x + 36) = \sqrt{x^{2} + 12 x + 36} - 6$

Remove parentheses.

$f (x^{2} + 12 x + 36) = \sqrt{x^{2} + 12 x + 36} - 6$

Simplify each term.

Factor using the perfect square rule.

Rewrite $36$ as $6^{2}$ .

$f (x^{2} + 12 x + 36) = \sqrt{x^{2} + 12 x + 6^{2}} - 6$

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

$12 x = 2 \cdot x \cdot 6$

Rewrite the polynomial.

$f (x^{2} + 12 x + 36) = \sqrt{x^{2} + 2 \cdot x \cdot 6 + 6^{2}} - 6$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 6$ .

$f (x^{2} + 12 x + 36) = \sqrt{{(x + 6)}^{2}} - 6$

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

$f (x^{2} + 12 x + 36) = x + 6 - 6$

Combine the opposite terms in $x + 6 - 6$ .

Subtract $6$ from $6$ .

$f (x^{2} + 12 x + 36) = x + 0$

Add $x$ and $0$ .

$f (x^{2} + 12 x + 36) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = x^{2} + 12 x + 36$ is the inverse of $f (x) = \sqrt{x} - 6$ .

$f^{- 1} (x) = x^{2} + 12 x + 36$

Find the Inverse square root of x-6

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