Find the Inverse y=x^(1/2)+4

$y = x^{\frac{1}{2}} + 4$

Step 1

Interchange the variables.

$x = y^{\frac{1}{2}} + 4$

Step 2

Solve for $y$ .

Rewrite the equation as $y^{\frac{1}{2}} + 4 = x$ .

$y^{\frac{1}{2}} + 4 = x$

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

Subtract $4$ from both sides of the equation.

$y^{\frac{1}{2}} - x = - 4$

Add $x$ to both sides of the equation.

$y^{\frac{1}{2}} = - 4 + x$

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

Simplify ${(- 4 + x)}^{2}$ .

Rewrite ${(- 4 + x)}^{2}$ as $(- 4 + x) (- 4 + x)$ .

$y = (- 4 + x) (- 4 + x)$

Expand $(- 4 + x) (- 4 + x)$ using the FOIL Method.

Apply the distributive property.

$y = - 4 (- 4 + x) + x (- 4 + x)$

Apply the distributive property.

$y = - 4 \cdot - 4 - 4 x + x (- 4 + x)$

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $- 4$ by $- 4$ .

$y = 16 - 4 x + x \cdot - 4 + x \cdot x$

Move $- 4$ to the left of $x$ .

$y = 16 - 4 x - 4 \cdot x + x \cdot x$

Multiply $x$ by $x$ .

$y = 16 - 4 x - 4 x + x^{2}$

Subtract $4 x$ from $- 4 x$ .

$y = 16 - 8 x + x^{2}$

Simplify $16 - 8 x + x^{2}$ .

Move $16$ .

$y = - 8 x + x^{2} + 16$

Reorder $- 8 x$ and $x^{2}$ .

$y = x^{2} - 8 x + 16$

Step 3

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

$f^{- 1} (x) = x^{2} - 8 x + 16$

Step 4

Verify if $f^{- 1} (x) = x^{2} - 8 x + 16$ is the inverse of $f (x) = x^{\frac{1}{2}} + 4$ .

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

$f^{- 1} (x^{\frac{1}{2}} + 4) = {(x^{\frac{1}{2}} + 4)}^{2} - 8 (x^{\frac{1}{2}} + 4) + 16$

Simplify each term.

Rewrite ${(x^{\frac{1}{2}} + 4)}^{2}$ as $(x^{\frac{1}{2}} + 4) (x^{\frac{1}{2}} + 4)$ .

$f^{- 1} (x^{\frac{1}{2}} + 4) = (x^{\frac{1}{2}} + 4) (x^{\frac{1}{2}} + 4) - 8 (x^{\frac{1}{2}} + 4) + 16$

Expand $(x^{\frac{1}{2}} + 4) (x^{\frac{1}{2}} + 4)$ using the FOIL Method.

Apply the distributive property.

$f^{- 1} (x^{\frac{1}{2}} + 4) = x^{\frac{1}{2}} (x^{\frac{1}{2}} + 4) + 4 (x^{\frac{1}{2}} + 4) - 8 (x^{\frac{1}{2}} + 4) + 16$

Apply the distributive property.

$f^{- 1} (x^{\frac{1}{2}} + 4) = x^{\frac{1}{2}} x^{\frac{1}{2}} + x^{\frac{1}{2}} \cdot 4 + 4 (x^{\frac{1}{2}} + 4) - 8 (x^{\frac{1}{2}} + 4) + 16$

Apply the distributive property.

$f^{- 1} (x^{\frac{1}{2}} + 4) = x^{\frac{1}{2}} x^{\frac{1}{2}} + x^{\frac{1}{2}} \cdot 4 + 4 x^{\frac{1}{2}} + 4 \cdot 4 - 8 (x^{\frac{1}{2}} + 4) + 16$

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.

$f^{- 1} (x^{\frac{1}{2}} + 4) = x^{\frac{1}{2} + \frac{1}{2}} + x^{\frac{1}{2}} \cdot 4 + 4 x^{\frac{1}{2}} + 4 \cdot 4 - 8 (x^{\frac{1}{2}} + 4) + 16$

Combine the numerators over the common denominator.

$f^{- 1} (x^{\frac{1}{2}} + 4) = x^{\frac{1 + 1}{2}} + x^{\frac{1}{2}} \cdot 4 + 4 x^{\frac{1}{2}} + 4 \cdot 4 - 8 (x^{\frac{1}{2}} + 4) + 16$

Add $1$ and $1$ .

$f^{- 1} (x^{\frac{1}{2}} + 4) = x^{\frac{2}{2}} + x^{\frac{1}{2}} \cdot 4 + 4 x^{\frac{1}{2}} + 4 \cdot 4 - 8 (x^{\frac{1}{2}} + 4) + 16$

Divide $2$ by $2$ .

$f^{- 1} (x^{\frac{1}{2}} + 4) = x + x^{\frac{1}{2}} \cdot 4 + 4 x^{\frac{1}{2}} + 4 \cdot 4 - 8 (x^{\frac{1}{2}} + 4) + 16$

Simplify $x^{1}$ .

$f^{- 1} (x^{\frac{1}{2}} + 4) = x + x^{\frac{1}{2}} \cdot 4 + 4 x^{\frac{1}{2}} + 4 \cdot 4 - 8 (x^{\frac{1}{2}} + 4) + 16$

Move $4$ to the left of $x^{\frac{1}{2}}$ .

$f^{- 1} (x^{\frac{1}{2}} + 4) = x + 4 \cdot x^{\frac{1}{2}} + 4 x^{\frac{1}{2}} + 4 \cdot 4 - 8 (x^{\frac{1}{2}} + 4) + 16$

Multiply $4$ by $4$ .

$f^{- 1} (x^{\frac{1}{2}} + 4) = x + 4 x^{\frac{1}{2}} + 4 x^{\frac{1}{2}} + 16 - 8 (x^{\frac{1}{2}} + 4) + 16$

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

$f^{- 1} (x^{\frac{1}{2}} + 4) = x + 8 x^{\frac{1}{2}} + 16 - 8 (x^{\frac{1}{2}} + 4) + 16$

Apply the distributive property.

$f^{- 1} (x^{\frac{1}{2}} + 4) = x + 8 x^{\frac{1}{2}} + 16 - 8 x^{\frac{1}{2}} - 8 \cdot 4 + 16$

Multiply $- 8$ by $4$ .

$f^{- 1} (x^{\frac{1}{2}} + 4) = x + 8 x^{\frac{1}{2}} + 16 - 8 x^{\frac{1}{2}} - 32 + 16$

Simplify by adding terms.

Combine the opposite terms in $x + 8 x^{\frac{1}{2}} + 16 - 8 x^{\frac{1}{2}} - 32 + 16$ .

Subtract $8 x^{\frac{1}{2}}$ from $8 x^{\frac{1}{2}}$ .

$f^{- 1} (x^{\frac{1}{2}} + 4) = x + 0 + 16 - 32 + 16$

Add $x$ and $0$ .

$f^{- 1} (x^{\frac{1}{2}} + 4) = x + 16 - 32 + 16$

Subtract $32$ from $16$ .

$f^{- 1} (x^{\frac{1}{2}} + 4) = x - 16 + 16$

Combine the opposite terms in $x - 16 + 16$ .

Add $- 16$ and $16$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

Evaluate $f (x^{2} - 8 x + 16)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (x^{2} - 8 x + 16) = {(x^{2} - 8 x + 16)}^{\frac{1}{2}} + 4$

Simplify each term.

Factor using the perfect square rule.

Rewrite $16$ as $4^{2}$ .

$f (x^{2} - 8 x + 16) = {(x^{2} - 8 x + 4^{2})}^{\frac{1}{2}} + 4$

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

$8 x = 2 \cdot x \cdot 4$

Rewrite the polynomial.

$f (x^{2} - 8 x + 16) = {(x^{2} - 2 \cdot x \cdot 4 + 4^{2})}^{\frac{1}{2}} + 4$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 4$ .

$f (x^{2} - 8 x + 16) = {({(x - 4)}^{2})}^{\frac{1}{2}} + 4$

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

$f (x^{2} - 8 x + 16) = {(x - 4)}^{2 (\frac{1}{2})} + 4$

Cancel the common factor of $2$ .

Cancel the common factor.

$f (x^{2} - 8 x + 16) = {(x - 4)}^{2 (\frac{1}{2})} + 4$

Rewrite the expression.

$f (x^{2} - 8 x + 16) = (x - 4) + 4$

Simplify.

$f (x^{2} - 8 x + 16) = x - 4 + 4$

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

Add $- 4$ and $4$ .

$f (x^{2} - 8 x + 16) = x + 0$

Add $x$ and $0$ .

$f (x^{2} - 8 x + 16) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = x^{2} - 8 x + 16$ is the inverse of $f (x) = x^{\frac{1}{2}} + 4$ .

$f^{- 1} (x) = x^{2} - 8 x + 16$

Find the Inverse y=x^(1/2)+4

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