Find the Inverse y=2 square root of x

$y = 2 \sqrt{x}$

Step 1

Interchange the variables.

$x = 2 \sqrt{y}$

Step 2

Solve for $y$ .

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

$2 \sqrt{y} = x$

Divide each term in $2 \sqrt{y} = x$ by $2$ and simplify.

Divide each term in $2 \sqrt{y} = x$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

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

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

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

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

Simplify each side of the equation.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

Apply the product rule to $\frac{x}{2}$ .

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

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

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

Step 3

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

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

Step 4

Verify if $f^{- 1} (x) = \frac{x^{2}}{4}$ 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 $f^{- 1}$ .

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

Simplify the numerator.

Apply the product rule to $2 \sqrt{x}$ .

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

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

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

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Remove parentheses.

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Find the Inverse y=2 square root of x

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