Find the Inverse A(r)=pir^2

$A (r) = π r^{2}$

Step 1

Write $A (r) = π r^{2}$ as an equation.

$y = π r^{2}$

Step 2

Interchange the variables.

$r = π y^{2}$

Step 3

Solve for $y$ .

Rewrite the equation as $π y^{2} = r$ .

$π y^{2} = r$

Divide each term in $π y^{2} = r$ by $π$ and simplify.

Divide each term in $π y^{2} = r$ by $π$ .

$\frac{π y^{2}}{π} = \frac{r}{π}$

Simplify the left side.

Cancel the common factor of $π$ .

Cancel the common factor.

$\frac{π y^{2}}{π} = \frac{r}{π}$

Divide $y^{2}$ by $1$ .

$y^{2} = \frac{r}{π}$

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

$y = \pm \sqrt{\frac{r}{π}}$

Simplify $\pm \sqrt{\frac{r}{π}}$ .

Rewrite $\sqrt{\frac{r}{π}}$ as $\frac{\sqrt{r}}{\sqrt{π}}$ .

$y = \pm \frac{\sqrt{r}}{\sqrt{π}}$

Multiply $\frac{\sqrt{r}}{\sqrt{π}}$ by $\frac{\sqrt{π}}{\sqrt{π}}$ .

$y = \pm \frac{\sqrt{r}}{\sqrt{π}} \cdot \frac{\sqrt{π}}{\sqrt{π}}$

Combine and simplify the denominator.

Multiply $\frac{\sqrt{r}}{\sqrt{π}}$ by $\frac{\sqrt{π}}{\sqrt{π}}$ .

$y = \pm \frac{\sqrt{r} \sqrt{π}}{\sqrt{π} \sqrt{π}}$

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

$y = \pm \frac{\sqrt{r} \sqrt{π}}{{\sqrt{π}}^{1} \sqrt{π}}$

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

$y = \pm \frac{\sqrt{r} \sqrt{π}}{{\sqrt{π}}^{1} {\sqrt{π}}^{1}}$

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

$y = \pm \frac{\sqrt{r} \sqrt{π}}{{\sqrt{π}}^{1 + 1}}$

Add $1$ and $1$ .

$y = \pm \frac{\sqrt{r} \sqrt{π}}{{\sqrt{π}}^{2}}$

Rewrite ${\sqrt{π}}^{2}$ as $π$ .

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

$y = \pm \frac{\sqrt{r} \sqrt{π}}{{(π^{\frac{1}{2}})}^{2}}$

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

$y = \pm \frac{\sqrt{r} \sqrt{π}}{π^{\frac{1}{2} \cdot 2}}$

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

$y = \pm \frac{\sqrt{r} \sqrt{π}}{π^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$y = \pm \frac{\sqrt{r} \sqrt{π}}{π^{\frac{2}{2}}}$

Rewrite the expression.

$y = \pm \frac{\sqrt{r} \sqrt{π}}{π^{1}}$

Simplify.

$y = \pm \frac{\sqrt{r} \sqrt{π}}{π}$

Combine using the product rule for radicals.

$y = \pm \frac{\sqrt{r π}}{π}$

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.

$y = \frac{\sqrt{r π}}{π}$

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

$y = - \frac{\sqrt{r π}}{π}$

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

$y = \frac{\sqrt{r π}}{π}$

$y = - \frac{\sqrt{r π}}{π}$

$y = \frac{\sqrt{r π}}{π}$

$y = - \frac{\sqrt{r π}}{π}$

$y = \frac{\sqrt{r π}}{π}$

$y = - \frac{\sqrt{r π}}{π}$

Step 4

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

$A^{- 1} (r) = \frac{\sqrt{r π}}{π}, - \frac{\sqrt{r π}}{π}$

Step 5

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

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

Find the range of $A (r) = π r^{2}$ .

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 $\frac{\sqrt{r π}}{π}$ .

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

$r π \geq 0$

Divide each term in $r π \geq 0$ by $π$ and simplify.

Divide each term in $r π \geq 0$ by $π$ .

$\frac{r π}{π} \geq \frac{0}{π}$

Simplify the left side.

Cancel the common factor of $π$ .

Cancel the common factor.

$\frac{r π}{π} \geq \frac{0}{π}$

Divide $r$ by $1$ .

$r \geq \frac{0}{π}$

Simplify the right side.

Divide $0$ by $π$ .

$r \geq 0$

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

$[0, \infty)$

Find the domain of $A (r) = π r^{2}$ .

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 $A^{- 1} (r) = \frac{\sqrt{r π}}{π}, - \frac{\sqrt{r π}}{π}$ is the range of $A (r) = π r^{2}$ and the range of $A^{- 1} (r) = \frac{\sqrt{r π}}{π}, - \frac{\sqrt{r π}}{π}$ is the domain of $A (r) = π r^{2}$ , then $A^{- 1} (r) = \frac{\sqrt{r π}}{π}, - \frac{\sqrt{r π}}{π}$ is the inverse of $A (r) = π r^{2}$ .

$A^{- 1} (r) = \frac{\sqrt{r π}}{π}, - \frac{\sqrt{r π}}{π}$

Step 6

Find the Inverse A(r)=pir^2

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