Find the Inverse 1+ square root of x-5

$1 + \sqrt{x - 5}$

Step 1

Interchange the variables.

$x = 1 + \sqrt{y - 5}$

Step 2

Solve for $y$ .

Rewrite the equation as $1 + \sqrt{y - 5} = x$ .

$1 + \sqrt{y - 5} = x$

Subtract $1$ from both sides of the equation.

$\sqrt{y - 5} = x - 1$

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

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

Simplify each side of the equation.

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

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

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, ${(a^{m})}^{n} = a^{m n}$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

Simplify ${(x - 1)}^{2}$ .

Rewrite ${(x - 1)}^{2}$ as $(x - 1) (x - 1)$ .

$y - 5 = (x - 1) (x - 1)$

Expand $(x - 1) (x - 1)$ using the FOIL Method.

Apply the distributive property.

$y - 5 = x (x - 1) - 1 (x - 1)$

Apply the distributive property.

$y - 5 = x \cdot x + x \cdot - 1 - 1 (x - 1)$

Apply the distributive property.

$y - 5 = x \cdot x + x \cdot - 1 - 1 x - 1 \cdot - 1$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$y - 5 = x^{2} + x \cdot - 1 - 1 x - 1 \cdot - 1$

Move $- 1$ to the left of $x$ .

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

Rewrite $- 1 x$ as $- x$ .

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

Rewrite $- 1 x$ as $- x$ .

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

Multiply $- 1$ by $- 1$ .

$y - 5 = x^{2} - x - x + 1$

Subtract $x$ from $- x$ .

$y - 5 = x^{2} - 2 x + 1$

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

Add $5$ to both sides of the equation.

$y = x^{2} - 2 x + 1 + 5$

Add $1$ and $5$ .

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

Step 3

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

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

Step 4

Verify if $f^{- 1} (x) = x^{2} - 2 x + 6$ is the inverse of $f (x) = 1 + \sqrt{x - 5}$ .

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

$f^{- 1} (1 + \sqrt{x - 5}) = {(1 + \sqrt{x - 5})}^{2} - 2 (1 + \sqrt{x - 5}) + 6$

Simplify each term.

Rewrite ${(1 + \sqrt{x - 5})}^{2}$ as $(1 + \sqrt{x - 5}) (1 + \sqrt{x - 5})$ .

$f^{- 1} (1 + \sqrt{x - 5}) = (1 + \sqrt{x - 5}) (1 + \sqrt{x - 5}) - 2 (1 + \sqrt{x - 5}) + 6$

Expand $(1 + \sqrt{x - 5}) (1 + \sqrt{x - 5})$ using the FOIL Method.

Apply the distributive property.

$f^{- 1} (1 + \sqrt{x - 5}) = 1 (1 + \sqrt{x - 5}) + \sqrt{x - 5} (1 + \sqrt{x - 5}) - 2 (1 + \sqrt{x - 5}) + 6$

Apply the distributive property.

$f^{- 1} (1 + \sqrt{x - 5}) = 1 \cdot 1 + 1 \sqrt{x - 5} + \sqrt{x - 5} (1 + \sqrt{x - 5}) - 2 (1 + \sqrt{x - 5}) + 6$

Apply the distributive property.

$f^{- 1} (1 + \sqrt{x - 5}) = 1 \cdot 1 + 1 \sqrt{x - 5} + \sqrt{x - 5} \cdot 1 + \sqrt{x - 5} \sqrt{x - 5} - 2 (1 + \sqrt{x - 5}) + 6$

Simplify and combine like terms.

Simplify each term.

Multiply $1$ by $1$ .

$f^{- 1} (1 + \sqrt{x - 5}) = 1 + 1 \sqrt{x - 5} + \sqrt{x - 5} \cdot 1 + \sqrt{x - 5} \sqrt{x - 5} - 2 (1 + \sqrt{x - 5}) + 6$

Multiply $\sqrt{x - 5}$ by $1$ .

$f^{- 1} (1 + \sqrt{x - 5}) = 1 + \sqrt{x - 5} + \sqrt{x - 5} \cdot 1 + \sqrt{x - 5} \sqrt{x - 5} - 2 (1 + \sqrt{x - 5}) + 6$

Multiply $\sqrt{x - 5}$ by $1$ .

$f^{- 1} (1 + \sqrt{x - 5}) = 1 + \sqrt{x - 5} + \sqrt{x - 5} + \sqrt{x - 5} \sqrt{x - 5} - 2 (1 + \sqrt{x - 5}) + 6$

Multiply $\sqrt{x - 5} \sqrt{x - 5}$ .

Raise $\sqrt{x - 5}$ to the power of $1$ .

$f^{- 1} (1 + \sqrt{x - 5}) = 1 + \sqrt{x - 5} + \sqrt{x - 5} + \sqrt{x - 5} \sqrt{x - 5} - 2 (1 + \sqrt{x - 5}) + 6$

Raise $\sqrt{x - 5}$ to the power of $1$ .

$f^{- 1} (1 + \sqrt{x - 5}) = 1 + \sqrt{x - 5} + \sqrt{x - 5} + \sqrt{x - 5} \sqrt{x - 5} - 2 (1 + \sqrt{x - 5}) + 6$

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

$f^{- 1} (1 + \sqrt{x - 5}) = 1 + \sqrt{x - 5} + \sqrt{x - 5} + {\sqrt{x - 5}}^{1 + 1} - 2 (1 + \sqrt{x - 5}) + 6$

Add $1$ and $1$ .

$f^{- 1} (1 + \sqrt{x - 5}) = 1 + \sqrt{x - 5} + \sqrt{x - 5} + {\sqrt{x - 5}}^{2} - 2 (1 + \sqrt{x - 5}) + 6$

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

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

$f^{- 1} (1 + \sqrt{x - 5}) = 1 + \sqrt{x - 5} + \sqrt{x - 5} + {({(x - 5)}^{\frac{1}{2}})}^{2} - 2 (1 + \sqrt{x - 5}) + 6$

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

$f^{- 1} (1 + \sqrt{x - 5}) = 1 + \sqrt{x - 5} + \sqrt{x - 5} + {(x - 5)}^{\frac{1}{2} \cdot 2} - 2 (1 + \sqrt{x - 5}) + 6$

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

$f^{- 1} (1 + \sqrt{x - 5}) = 1 + \sqrt{x - 5} + \sqrt{x - 5} + {(x - 5)}^{\frac{2}{2}} - 2 (1 + \sqrt{x - 5}) + 6$

Cancel the common factor of $2$ .

Cancel the common factor.

$f^{- 1} (1 + \sqrt{x - 5}) = 1 + \sqrt{x - 5} + \sqrt{x - 5} + {(x - 5)}^{\frac{2}{2}} - 2 (1 + \sqrt{x - 5}) + 6$

Rewrite the expression.

$f^{- 1} (1 + \sqrt{x - 5}) = 1 + \sqrt{x - 5} + \sqrt{x - 5} + x - 5 - 2 (1 + \sqrt{x - 5}) + 6$

Simplify.

$f^{- 1} (1 + \sqrt{x - 5}) = 1 + \sqrt{x - 5} + \sqrt{x - 5} + x - 5 - 2 (1 + \sqrt{x - 5}) + 6$

Subtract $5$ from $1$ .

$f^{- 1} (1 + \sqrt{x - 5}) = - 4 + \sqrt{x - 5} + \sqrt{x - 5} + x - 2 (1 + \sqrt{x - 5}) + 6$

Add $\sqrt{x - 5}$ and $\sqrt{x - 5}$ .

$f^{- 1} (1 + \sqrt{x - 5}) = - 4 + 2 \sqrt{x - 5} + x - 2 (1 + \sqrt{x - 5}) + 6$

Apply the distributive property.

$f^{- 1} (1 + \sqrt{x - 5}) = - 4 + 2 \sqrt{x - 5} + x - 2 \cdot 1 - 2 \sqrt{x - 5} + 6$

Multiply $- 2$ by $1$ .

$f^{- 1} (1 + \sqrt{x - 5}) = - 4 + 2 \sqrt{x - 5} + x - 2 - 2 \sqrt{x - 5} + 6$

Simplify by adding terms.

Combine the opposite terms in $- 4 + 2 \sqrt{x - 5} + x - 2 - 2 \sqrt{x - 5} + 6$ .

Subtract $2 \sqrt{x - 5}$ from $2 \sqrt{x - 5}$ .

$f^{- 1} (1 + \sqrt{x - 5}) = - 4 + x - 2 + 0 + 6$

Add $- 4 + x - 2$ and $0$ .

$f^{- 1} (1 + \sqrt{x - 5}) = - 4 + x - 2 + 6$

Subtract $2$ from $- 4$ .

$f^{- 1} (1 + \sqrt{x - 5}) = x - 6 + 6$

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

Add $- 6$ and $6$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

$f (x^{2} - 2 x + 6) = 1 + \sqrt{(x^{2} - 2 x + 6) - 5}$

Simplify each term.

Subtract $5$ from $6$ .

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

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

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

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

$2 x = 2 \cdot x \cdot 1$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 1$ .

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

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

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

Combine the opposite terms in $1 + x - 1$ .

Subtract $1$ from $1$ .

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

Add $x$ and $0$ .

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

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = x^{2} - 2 x + 6$ is the inverse of $f (x) = 1 + \sqrt{x - 5}$ .

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

Find the Inverse 1+ square root of x-5

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