Find the Inverse g(x)=1/2 square root of x+1+4

$g (x) = \frac{1}{2} \sqrt[2]{x + 1} + 4$

Write $g (x) = \frac{1}{2} \sqrt[2]{x + 1} + 4$ as an equation.

$y = \frac{1}{2} \sqrt[2]{x + 1} + 4$

Interchange the variables.

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

Solve for $y$ .

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

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

Subtract $4$ from both sides of the equation.

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

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

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

Simplify each side of the equation.

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

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

Simplify the left side.

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

Combine fractions.

Combine $\frac{1}{2}$ and ${(y + 1)}^{\frac{1}{2}}$ .

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

Apply the product rule to $\frac{{(y + 1)}^{\frac{1}{2}}}{2}$ .

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

Simplify the numerator.

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

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

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

Simplify the right side.

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

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

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

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

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

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

Multiply $- 4$ by $- 4$ .

$\frac{y + 1}{4} = x^{2} - 4 x - 4 x + 16$

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

$\frac{y + 1}{4} = x^{2} - 8 x + 16$

Solve for $y$ .

Multiply both sides of the equation by $4$ .

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

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

$y + 1 = 4 (x^{2} - 8 x + 16)$

Simplify the right side.

Simplify $4 (x^{2} - 8 x + 16)$ .

Apply the distributive property.

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

Simplify.

Multiply $- 8$ by $4$ .

$y + 1 = 4 x^{2} - 32 x + 4 \cdot 16$

Multiply $4$ by $16$ .

$y + 1 = 4 x^{2} - 32 x + 64$

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

Subtract $1$ from both sides of the equation.

$y = 4 x^{2} - 32 x + 64 - 1$

Subtract $1$ from $64$ .

$y = 4 x^{2} - 32 x + 63$

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

$g^{- 1} (x) = 4 x^{2} - 32 x + 63$

Verify if $g^{- 1} (x) = 4 x^{2} - 32 x + 63$ is the inverse of $g (x) = \frac{1}{2} \cdot \sqrt{x + 1} + 4$ .

To verify the inverse, check if $g^{- 1} (g (x)) = x$ and $g (g^{- 1} (x)) = x$ .

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

Set up the composite result function.

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

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

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = 4 {(\frac{1}{2} \cdot \sqrt{x + 1} + 4)}^{2} - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

Simplify each term.

Combine $\frac{1}{2}$ and $\sqrt{x + 1}$ .

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = 4 {(\frac{\sqrt{x + 1}}{2} + 4)}^{2} - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

To write $4$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = 4 {(\frac{\sqrt{x + 1}}{2} + 4 \cdot \frac{2}{2})}^{2} - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

Combine $4$ and $\frac{2}{2}$ .

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = 4 {(\frac{\sqrt{x + 1}}{2} + \frac{4 \cdot 2}{2})}^{2} - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

Combine the numerators over the common denominator.

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = 4 {(\frac{\sqrt{x + 1} + 4 \cdot 2}{2})}^{2} - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

Multiply $4$ by $2$ .

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = 4 {(\frac{\sqrt{x + 1} + 8}{2})}^{2} - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

Apply the product rule to $\frac{\sqrt{x + 1} + 8}{2}$ .

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = 4 (\frac{{(\sqrt{x + 1} + 8)}^{2}}{2^{2}}) - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

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

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = 4 (\frac{{(\sqrt{x + 1} + 8)}^{2}}{4}) - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

Cancel the common factor of $4$ .

Cancel the common factor.

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = 4 (\frac{{(\sqrt{x + 1} + 8)}^{2}}{4}) - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

Rewrite the expression.

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = {(\sqrt{x + 1} + 8)}^{2} - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

Rewrite ${(\sqrt{x + 1} + 8)}^{2}$ as $(\sqrt{x + 1} + 8) (\sqrt{x + 1} + 8)$ .

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = (\sqrt{x + 1} + 8) (\sqrt{x + 1} + 8) - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

Expand $(\sqrt{x + 1} + 8) (\sqrt{x + 1} + 8)$ using the FOIL Method.

Apply the distributive property.

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = \sqrt{x + 1} (\sqrt{x + 1} + 8) + 8 (\sqrt{x + 1} + 8) - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

Apply the distributive property.

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = \sqrt{x + 1} \sqrt{x + 1} + \sqrt{x + 1} \cdot 8 + 8 (\sqrt{x + 1} + 8) - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

Apply the distributive property.

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = \sqrt{x + 1} \sqrt{x + 1} + \sqrt{x + 1} \cdot 8 + 8 \sqrt{x + 1} + 8 \cdot 8 - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

Simplify and combine like terms.

Simplify each term.

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

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

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = \sqrt{x + 1} \sqrt{x + 1} + \sqrt{x + 1} \cdot 8 + 8 \sqrt{x + 1} + 8 \cdot 8 - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

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

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = \sqrt{x + 1} \sqrt{x + 1} + \sqrt{x + 1} \cdot 8 + 8 \sqrt{x + 1} + 8 \cdot 8 - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

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

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = {\sqrt{x + 1}}^{1 + 1} + \sqrt{x + 1} \cdot 8 + 8 \sqrt{x + 1} + 8 \cdot 8 - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

Add $1$ and $1$ .

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = {\sqrt{x + 1}}^{2} + \sqrt{x + 1} \cdot 8 + 8 \sqrt{x + 1} + 8 \cdot 8 - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

Rewrite ${\sqrt{x + 1}}^{2}$ as $x + 1$ .

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

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = {({(x + 1)}^{\frac{1}{2}})}^{2} + \sqrt{x + 1} \cdot 8 + 8 \sqrt{x + 1} + 8 \cdot 8 - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

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

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = {(x + 1)}^{\frac{1}{2} \cdot 2} + \sqrt{x + 1} \cdot 8 + 8 \sqrt{x + 1} + 8 \cdot 8 - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

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

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = {(x + 1)}^{\frac{2}{2}} + \sqrt{x + 1} \cdot 8 + 8 \sqrt{x + 1} + 8 \cdot 8 - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

Cancel the common factor of $2$ .

Cancel the common factor.

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = {(x + 1)}^{\frac{2}{2}} + \sqrt{x + 1} \cdot 8 + 8 \sqrt{x + 1} + 8 \cdot 8 - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

Rewrite the expression.

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = (x + 1) + \sqrt{x + 1} \cdot 8 + 8 \sqrt{x + 1} + 8 \cdot 8 - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

Simplify.

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = x + 1 + \sqrt{x + 1} \cdot 8 + 8 \sqrt{x + 1} + 8 \cdot 8 - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

Move $8$ to the left of $\sqrt{x + 1}$ .

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = x + 1 + 8 \cdot \sqrt{x + 1} + 8 \sqrt{x + 1} + 8 \cdot 8 - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

Multiply $8$ by $8$ .

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = x + 1 + 8 \sqrt{x + 1} + 8 \sqrt{x + 1} + 64 - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

Add $1$ and $64$ .

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = x + 65 + 8 \sqrt{x + 1} + 8 \sqrt{x + 1} - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

Add $8 \sqrt{x + 1}$ and $8 \sqrt{x + 1}$ .

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = x + 65 + 16 \sqrt{x + 1} - 32 (\frac{1}{2} \cdot \sqrt{x + 1} + 4) + 63$

Combine $\frac{1}{2}$ and $\sqrt{x + 1}$ .

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = x + 65 + 16 \sqrt{x + 1} - 32 (\frac{\sqrt{x + 1}}{2} + 4) + 63$

To write $4$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = x + 65 + 16 \sqrt{x + 1} - 32 (\frac{\sqrt{x + 1}}{2} + 4 \cdot \frac{2}{2}) + 63$

Combine $4$ and $\frac{2}{2}$ .

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = x + 65 + 16 \sqrt{x + 1} - 32 (\frac{\sqrt{x + 1}}{2} + \frac{4 \cdot 2}{2}) + 63$

Combine the numerators over the common denominator.

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = x + 65 + 16 \sqrt{x + 1} - 32 \frac{\sqrt{x + 1} + 4 \cdot 2}{2} + 63$

Multiply $4$ by $2$ .

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = x + 65 + 16 \sqrt{x + 1} - 32 \frac{\sqrt{x + 1} + 8}{2} + 63$

Cancel the common factor of $2$ .

Factor $2$ out of $- 32$ .

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = x + 65 + 16 \sqrt{x + 1} + 2 (- 16) (\frac{\sqrt{x + 1} + 8}{2}) + 63$

Cancel the common factor.

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = x + 65 + 16 \sqrt{x + 1} + 2 \cdot (- 16 \frac{\sqrt{x + 1} + 8}{2}) + 63$

Rewrite the expression.

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = x + 65 + 16 \sqrt{x + 1} - 16 (\sqrt{x + 1} + 8) + 63$

Apply the distributive property.

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = x + 65 + 16 \sqrt{x + 1} - 16 \sqrt{x + 1} - 16 \cdot 8 + 63$

Multiply $- 16$ by $8$ .

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = x + 65 + 16 \sqrt{x + 1} - 16 \sqrt{x + 1} - 128 + 63$

Simplify by adding terms.

Combine the opposite terms in $x + 65 + 16 \sqrt{x + 1} - 16 \sqrt{x + 1} - 128 + 63$ .

Subtract $16 \sqrt{x + 1}$ from $16 \sqrt{x + 1}$ .

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = x + 65 + 0 - 128 + 63$

Add $x + 65$ and $0$ .

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = x + 65 - 128 + 63$

Subtract $128$ from $65$ .

$g^{- 1} (\frac{1}{2} \cdot \sqrt{x + 1} + 4) = x - 63 + 63$

Combine the opposite terms in $x - 63 + 63$ .

Add $- 63$ and $63$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

Evaluate $g (4 x^{2} - 32 x + 63)$ by substituting in the value of $g^{- 1}$ into $g$ .

$g (4 x^{2} - 32 x + 63) = \frac{1}{2} \cdot \sqrt{(4 x^{2} - 32 x + 63) + 1} + 4$

Simplify each term.

Add $63$ and $1$ .

$g (4 x^{2} - 32 x + 63) = \frac{1}{2} \cdot \sqrt{4 x^{2} - 32 x + 64} + 4$

Factor $4$ out of $4 x^{2} - 32 x + 64$ .

Factor $4$ out of $4 x^{2}$ .

$g (4 x^{2} - 32 x + 63) = \frac{1}{2} \cdot \sqrt{4 (x^{2}) - 32 x + 64} + 4$

Factor $4$ out of $- 32 x$ .

$g (4 x^{2} - 32 x + 63) = \frac{1}{2} \cdot \sqrt{4 (x^{2}) + 4 (- 8 x) + 64} + 4$

Factor $4$ out of $64$ .

$g (4 x^{2} - 32 x + 63) = \frac{1}{2} \cdot \sqrt{4 x^{2} + 4 (- 8 x) + 4 \cdot 16} + 4$

Factor $4$ out of $4 x^{2} + 4 (- 8 x)$ .

$g (4 x^{2} - 32 x + 63) = \frac{1}{2} \cdot \sqrt{4 (x^{2} - 8 x) + 4 \cdot 16} + 4$

Factor $4$ out of $4 (x^{2} - 8 x) + 4 \cdot 16$ .

$g (4 x^{2} - 32 x + 63) = \frac{1}{2} \cdot \sqrt{4 (x^{2} - 8 x + 16)} + 4$

Factor using the perfect square rule.

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

$g (4 x^{2} - 32 x + 63) = \frac{1}{2} \cdot \sqrt{4 (x^{2} - 8 x + 4^{2})} + 4$

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.

$g (4 x^{2} - 32 x + 63) = \frac{1}{2} \cdot \sqrt{4 (x^{2} - 2 \cdot x \cdot 4 + 4^{2})} + 4$

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

$g (4 x^{2} - 32 x + 63) = \frac{1}{2} \cdot \sqrt{4 {(x - 4)}^{2}} + 4$

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

$g (4 x^{2} - 32 x + 63) = \frac{1}{2} \cdot \sqrt{{(2 (x - 4))}^{2}} + 4$

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

$g (4 x^{2} - 32 x + 63) = \frac{1}{2} \cdot (2 (x - 4)) + 4$

Cancel the common factor of $2$ .

Cancel the common factor.

$g (4 x^{2} - 32 x + 63) = \frac{1}{2} \cdot (2 (x - 4)) + 4$

Rewrite the expression.

$g (4 x^{2} - 32 x + 63) = x - 4 + 4$

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

Add $- 4$ and $4$ .

$g (4 x^{2} - 32 x + 63) = x + 0$

Add $x$ and $0$ .

$g (4 x^{2} - 32 x + 63) = x$

Since $g^{- 1} (g (x)) = x$ and $g (g^{- 1} (x)) = x$ , then $g^{- 1} (x) = 4 x^{2} - 32 x + 63$ is the inverse of $g (x) = \frac{1}{2} \cdot \sqrt{x + 1} + 4$ .

$g^{- 1} (x) = 4 x^{2} - 32 x + 63$

Find the Inverse g(x)=1/2 square root of x+1+4

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