Find the Inverse f(x)=(x-12)^2 , x<=12

$f (x) = {(x - 12)}^{2}$ , $x \leq 12$

Step 1

Find the range of the given function.

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

$[0, \infty)$

Convert $[0, \infty)$ to an inequality.

$y \geq 0$

Step 2

Find the inverse.

Interchange the variables.

$x = {(y - 12)}^{2}$

Solve for $y$ .

Rewrite the equation as ${(y - 12)}^{2} = x$ .

${(y - 12)}^{2} = x$

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

$y - 12 = \pm \sqrt{x}$

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.

$y - 12 = \sqrt{x}$

Add $12$ to both sides of the equation.

$y = \sqrt{x} + 12$

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

$y - 12 = - \sqrt{x}$

Add $12$ to both sides of the equation.

$y = - \sqrt{x} + 12$

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

$y = \sqrt{x} + 12$

$y = - \sqrt{x} + 12$

$y = \sqrt{x} + 12$

$y = - \sqrt{x} + 12$

$y = \sqrt{x} + 12$

$y = - \sqrt{x} + 12$

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

$f^{- 1} (x) = \sqrt{x} + 12, - \sqrt{x} + 12$

Step 3

Find the inverse using the domain and the range of the original function.

$f^{- 1} (x) = - \sqrt{x} + 12, x \geq 0$

Step 4