Find the Inverse y=4 square root of x+3

$y = 4 \sqrt{x + 3}$

Interchange the variables.

$x = 4 \sqrt{y + 3}$

Solve for $y$ .

Rewrite the equation as $4 \sqrt{y + 3} = x$ .

$4 \sqrt{y + 3} = x$

Divide each term in $4 \sqrt{y + 3} = x$ by $4$ and simplify.

Divide each term in $4 \sqrt{y + 3} = x$ by $4$ .

$\frac{4 \sqrt{y + 3}}{4} = \frac{x}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 \sqrt{y + 3}}{4} = \frac{x}{4}$

Divide $\sqrt{y + 3}$ by $1$ .

$\sqrt{y + 3} = \frac{x}{4}$

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

${\sqrt{y + 3}}^{2} = {(\frac{x}{4})}^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{y + 3}$ as ${(y + 3)}^{\frac{1}{2}}$ .

${({(y + 3)}^{\frac{1}{2}})}^{2} = {(\frac{x}{4})}^{2}$

Simplify the left side.

Simplify ${({(y + 3)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(y + 3)}^{\frac{1}{2}})}^{2}$ .

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

${(y + 3)}^{\frac{1}{2} \cdot 2} = {(\frac{x}{4})}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(y + 3)}^{\frac{1}{2} \cdot 2} = {(\frac{x}{4})}^{2}$

Rewrite the expression.

${(y + 3)}^{1} = {(\frac{x}{4})}^{2}$

Simplify.

$y + 3 = {(\frac{x}{4})}^{2}$

Simplify the right side.

Simplify ${(\frac{x}{4})}^{2}$ .

Apply the product rule to $\frac{x}{4}$ .

$y + 3 = \frac{x^{2}}{4^{2}}$

Raise $4$ to the power of $2$ .

$y + 3 = \frac{x^{2}}{16}$

Subtract $3$ from both sides of the equation.

$y = \frac{x^{2}}{16} - 3$

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

$f^{- 1} (x) = \frac{x^{2}}{16} - 3$

Verify if $f^{- 1} (x) = \frac{x^{2}}{16} - 3$ is the inverse of $f (x) = 4 \sqrt{x + 3}$ .

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 $f^{- 1} (4 \sqrt{x + 3})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (4 \sqrt{x + 3}) = \frac{{(4 \sqrt{x + 3})}^{2}}{16} - 3$

Simplify each term.

Simplify the numerator.

Apply the product rule to $4 \sqrt{x + 3}$ .

$f^{- 1} (4 \sqrt{x + 3}) = \frac{4^{2} {\sqrt{x + 3}}^{2}}{16} - 3$

Raise $4$ to the power of $2$ .

$f^{- 1} (4 \sqrt{x + 3}) = \frac{16 {\sqrt{x + 3}}^{2}}{16} - 3$

Rewrite ${\sqrt{x + 3}}^{2}$ as $x + 3$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 3}$ as ${(x + 3)}^{\frac{1}{2}}$ .

$f^{- 1} (4 \sqrt{x + 3}) = \frac{16 {({(x + 3)}^{\frac{1}{2}})}^{2}}{16} - 3$

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

$f^{- 1} (4 \sqrt{x + 3}) = \frac{16 {(x + 3)}^{\frac{1}{2} \cdot 2}}{16} - 3$

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

$f^{- 1} (4 \sqrt{x + 3}) = \frac{16 {(x + 3)}^{\frac{2}{2}}}{16} - 3$

Cancel the common factor of $2$ .

Cancel the common factor.

$f^{- 1} (4 \sqrt{x + 3}) = \frac{16 {(x + 3)}^{\frac{2}{2}}}{16} - 3$

Rewrite the expression.

$f^{- 1} (4 \sqrt{x + 3}) = \frac{16 (x + 3)}{16} - 3$

Simplify.

$f^{- 1} (4 \sqrt{x + 3}) = \frac{16 (x + 3)}{16} - 3$

Cancel the common factor of $16$ .

Cancel the common factor.

$f^{- 1} (4 \sqrt{x + 3}) = \frac{16 (x + 3)}{16} - 3$

Divide $x + 3$ by $1$ .

$f^{- 1} (4 \sqrt{x + 3}) = x + 3 - 3$

Combine the opposite terms in $x + 3 - 3$ .

Subtract $3$ from $3$ .

$f^{- 1} (4 \sqrt{x + 3}) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (4 \sqrt{x + 3}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{x^{2}}{16} - 3)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{x^{2}}{16} - 3) = 4 \sqrt{(\frac{x^{2}}{16} - 3) + 3}$

Add $- 3$ and $3$ .

$f (\frac{x^{2}}{16} - 3) = 4 \sqrt{\frac{x^{2}}{16} + 0}$

Add $\frac{x^{2}}{16}$ and $0$ .

$f (\frac{x^{2}}{16} - 3) = 4 \sqrt{\frac{x^{2}}{16}}$

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

$f (\frac{x^{2}}{16} - 3) = 4 \sqrt{\frac{x^{2}}{4^{2}}}$

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

$f (\frac{x^{2}}{16} - 3) = 4 \sqrt{{(\frac{x}{4})}^{2}}$

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

$f (\frac{x^{2}}{16} - 3) = 4 (\frac{x}{4})$

Cancel the common factor of $4$ .

Cancel the common factor.

$f (\frac{x^{2}}{16} - 3) = 4 (\frac{x}{4})$

Rewrite the expression.

$f (\frac{x^{2}}{16} - 3) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x^{2}}{16} - 3$ is the inverse of $f (x) = 4 \sqrt{x + 3}$ .

$f^{- 1} (x) = \frac{x^{2}}{16} - 3$

Find the Inverse y=4 square root of x+3

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