Find the Inverse f(x)=(x+3)^2-2 , x<=-3

$f (x) = {(x + 3)}^{2} - 2$ , $x \leq - 3$

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.

$[- 2, \infty)$

Convert $[- 2, \infty)$ to an inequality.

$y \geq - 2$

Step 2

Find the inverse.

Interchange the variables.

$x = {(y + 3)}^{2} - 2$

Solve for $y$ .

Rewrite the equation as ${(y + 3)}^{2} - 2 = x$ .

${(y + 3)}^{2} - 2 = x$

Add $2$ to both sides of the equation.

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

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

$y + 3 = \pm \sqrt{x + 2}$

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.

$y + 3 = \sqrt{x + 2}$

Subtract $3$ from both sides of the equation.

$y = \sqrt{x + 2} - 3$

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

$y + 3 = - \sqrt{x + 2}$

Subtract $3$ from both sides of the equation.

$y = - \sqrt{x + 2} - 3$

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

$y = \sqrt{x + 2} - 3$

$y = - \sqrt{x + 2} - 3$

$y = \sqrt{x + 2} - 3$

$y = - \sqrt{x + 2} - 3$

$y = \sqrt{x + 2} - 3$

$y = - \sqrt{x + 2} - 3$

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

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

Step 3

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

$f^{- 1} (x) = - \sqrt{x + 2} - 3, x \geq - 2$

Step 4