Let f(x) = (x + 2)^2 - 2, x geq - 2. Then f^{ | vedclass

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(C) To find the inverse of the function $f(x) = (x + 2)^2 - 2$ for $x \geq -2$,we set $y = f(x)$.$y = (x + 2)^2 - 2$Add $2$ to both sides:$y + 2 = (x

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$\sqrt{2 + x} - 2$

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(C) To find the inverse of the function $f(x) = (x + 2)^2 - 2$ for $x \geq -2$,we set $y = f(x)$.$y = (x + 2)^2 - 2$Add $2$ to both sides:$y + 2 = (x + 2)^2$Since $x \geq -2$,we have $x + 2 \geq 0$. Taking the positive square root of both sides:$\sqrt{y + 2} = x + 2$Subtract $2$ from both sides to solve for $x$:$x = \sqrt{y + 2} - 2$By definition,$f^{-1}(y) = \sqrt{y + 2} - 2$. Replacing $y$ with $x$,we get:$f^{-1}(x) = \sqrt{x + 2} - 2$

Let $f(x) = (x + 2)^2 - 2, x \geq - 2$. Then $f^{-1}(x) =$

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