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

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

Interchange the variables.

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

Solve for $y$ .

Rewrite the equation as ${(\frac{y}{2})}^{2} = x$ .

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

Take the square root of both sides of the equation to eliminate the exponent on the left side.

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

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

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

Multiply both sides of the equation by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$y = 2 \sqrt{x}$

Next, use the negative value of the $\pm$ to find the second solution.

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

Multiply both sides of the equation by $2$ .

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

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$y = 2 (- \sqrt{x})$

Simplify the right side.

Multiply $- 1$ by $2$ .

$y = - 2 \sqrt{x}$

The complete solution is the result of both the positive and negative portions of the solution.

$y = 2 \sqrt{x}$

$y = - 2 \sqrt{x}$

$y = 2 \sqrt{x}$

$y = - 2 \sqrt{x}$

$y = 2 \sqrt{x}$

$y = - 2 \sqrt{x}$

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

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

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

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = {(\frac{x}{2})}^{2}$ and $f^{- 1} (x) = 2 \sqrt{x}, - 2 \sqrt{x}$ and compare them.

Find the range of $f (x) = {(\frac{x}{2})}^{2}$ .

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$[0, \infty)$

Find the domain of $2 \sqrt{x}$ .

Set the radicand in $\sqrt{x}$ greater than or equal to $0$ to find where the expression is defined.

$x \geq 0$

The domain is all values of $x$ that make the expression defined.

$[0, \infty)$

Find the domain of $f (x) = {(\frac{x}{2})}^{2}$ .

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

$(- \infty, \infty)$

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

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

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

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