Find the Inverse y=2- square root of x

$y = 2 - \sqrt{x}$

Interchange the variables.

$x = 2 - \sqrt{y}$

Solve for $y$ .

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

$2 - \sqrt{y} = x$

Subtract $2$ from both sides of the equation.

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

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

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

Simplify each side of the equation.

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

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

Simplify the left side.

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

Apply the product rule to $- y^{\frac{1}{2}}$ .

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

Raise $- 1$ to the power of $2$ .

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

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

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

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

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

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

Multiply $- 2$ by $- 2$ .

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

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

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

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

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

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

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

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

Simplify each term.

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

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

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $2$ by $2$ .

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

Multiply $- 1$ by $2$ .

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

Multiply $2$ by $- 1$ .

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

Multiply $- \sqrt{x} (- \sqrt{x})$ .

Multiply $- 1$ by $- 1$ .

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

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

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

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

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

Apply the distributive property.

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

Multiply $- 4$ by $2$ .

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

Multiply $- 1$ by $- 4$ .

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

Simplify by adding terms.

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

Add $- 4 \sqrt{x}$ and $4 \sqrt{x}$ .

$f^{- 1} (2 - \sqrt{x}) = 4 + x - 8 + 0 + 4$

Add $4 + x - 8$ and $0$ .

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

Subtract $8$ from $4$ .

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

Combine the opposite terms in $x - 4 + 4$ .

Add $- 4$ and $4$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Factor using the perfect square rule.

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

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

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.

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

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

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

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

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

Apply the distributive property.

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

Multiply $- 1$ by $- 2$ .

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

Add $2$ and $2$ .

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

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

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

Find the Inverse y=2- square root of x

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