Question

find the inverse of the following function. also list the domain restriction for the inverse
$f(x) = -x^2 - 3$, for $x \geq 0$
$f^{-1}(x) = $
for

Explanation:

Step1: Replace \( f(x) \) with \( y \)

We start with the function \( f(x)=-x^{2}-2 \) for \( x\geq0 \). Replace \( f(x) \) with \( y \), so we have \( y = -x^{2}-2 \).

Step2: Solve for \( x \) in terms of \( y \)

First, we add 2 to both sides of the equation: \( y + 2=-x^{2} \). Then multiply both sides by - 1: \( -y - 2=x^{2} \). Since \( x\geq0 \), we take the square root of both sides, and we get \( x=\sqrt{-y - 2} \) (we take the non - negative square root because \( x\geq0 \)).

Step3: Swap \( x \) and \( y \) to find the inverse function

Swap \( x \) and \( y \) in the equation \( x=\sqrt{-y - 2} \). We get \( y=\sqrt{-x - 2} \). So the inverse function \( f^{-1}(x)=\sqrt{-x - 2} \).

Step4: Determine the domain restriction for the inverse function

The domain of the inverse function is the range of the original function. For the original function \( y=-x^{2}-2 \) with \( x\geq0 \), when \( x = 0 \), \( y=-0^{2}-2=-2 \). As \( x\) increases (since \( x\geq0 \) and the coefficient of \( x^{2}\) is negative), \( y\) decreases. So the range of the original function is \( y\leq - 2 \). Therefore, the domain of the inverse function \( f^{-1}(x) \) is \( x\leq - 2 \).

Answer:

\( f^{-1}(x)=\sqrt{-x - 2} \) for \( x\leq - 2 \)