Question

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Explanation:

Step 1: Replace \( f(x) \) with \( y \)

We start with the function \( f(x) = -\sqrt{x + 2} \) for \( x \geq -2 \). Replace \( f(x) \) with \( y \), so we have \( y = -\sqrt{x + 2} \), where \( x \geq -2 \).

Step 2: Swap \( x \) and \( y \)

To find the inverse, we swap the roles of \( x \) and \( y \). This gives us \( x = -\sqrt{y + 2} \).

Step 3: Solve for \( y \)

First, multiply both sides by -1 to get \( -x = \sqrt{y + 2} \). Then, square both sides to eliminate the square root: \( (-x)^2 = (\sqrt{y + 2})^2 \), which simplifies to \( x^2 = y + 2 \). Now, subtract 2 from both sides to solve for \( y \): \( y = x^2 - 2 \). So the inverse function \( f^{-1}(x) = x^2 - 2 \).

Step 4: Determine the domain and range restrictions for the inverse

For the original function \( f(x) = -\sqrt{x + 2} \) with \( x \geq -2 \):

• The domain of \( f(x) \) is \( x \geq -2 \), so the range of \( f(x) \) (which becomes the domain of \( f^{-1}(x) \)): since the square root \( \sqrt{x + 2} \geq 0 \), multiplying by -1 gives \( -\sqrt{x + 2} \leq 0 \), and as \( x \geq -2 \), the range of \( f(x) \) is \( y \leq 0 \). So the domain of \( f^{-1}(x) \) is \( x \leq 0 \).
• The range of \( f^{-1}(x) \) is the domain of \( f(x) \), which is \( y \geq -2 \).

Answer:

\( f^{-1}(x) = \boldsymbol{x^2 - 2} \) for \( \boldsymbol{x \leq 0} \) and \( \boldsymbol{y \geq -2} \)