Find the Inverse y=x^(1/2)-2

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

Rewrite the equation as $y^{\frac{1}{2}} - 2 = x$ .

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

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

Add $2$ to both sides of the equation.

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

Add $x$ to both sides of the equation.

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

Raise each side of the equation to the power of $2$ to eliminate the fractional exponent on the left side.

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

Simplify the exponent.

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

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

$y = (2 + x) (2 + x)$

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

Apply the distributive property.

$y = 2 (2 + x) + x (2 + x)$

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $2$ by $2$ .

$y = 4 + 2 x + x \cdot 2 + x \cdot x$

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

$y = 4 + 2 x + 2 \cdot x + x \cdot x$

Multiply $x$ by $x$ .

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

Add $2 x$ and $2 x$ .

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

Simplify $4 + 4 x + x^{2}$ .

Move $4$ .

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

Reorder $4 x$ and $x^{2}$ .

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

Step 3

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

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

Step 4

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

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

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

Simplify each term.

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

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

Expand $(x^{\frac{1}{2}} - 2) (x^{\frac{1}{2}} - 2)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $x^{\frac{1}{2}}$ by $x^{\frac{1}{2}}$ by adding the exponents.

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

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

Combine the numerators over the common denominator.

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

Add $1$ and $1$ .

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

Divide $2$ by $2$ .

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

Simplify $x^{1}$ .

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

Move $- 2$ to the left of $x^{\frac{1}{2}}$ .

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

Multiply $- 2$ by $- 2$ .

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

Subtract $2 x^{\frac{1}{2}}$ from $- 2 x^{\frac{1}{2}}$ .

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

Apply the distributive property.

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

Multiply $4$ by $- 2$ .

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

Simplify by adding terms.

Combine the opposite terms in $x - 4 x^{\frac{1}{2}} + 4 + 4 x^{\frac{1}{2}} - 8 + 4$ .

Add $- 4 x^{\frac{1}{2}}$ and $4 x^{\frac{1}{2}}$ .

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

Add $x$ and $0$ .

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

Subtract $8$ from $4$ .

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

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

Add $- 4$ and $4$ .

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

Add $x$ and $0$ .

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

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 $f$ .

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

Simplify each term.

Factor using the perfect square rule.

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

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

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.

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

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

Subtract $2$ from $2$ .

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

Add $x$ and $0$ .

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

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

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

Find the Inverse y=x^(1/2)-2

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