To find the inverse \( f^{-1}(x) \) of the function \( f(x) = -\frac{1}{2} \sqrt{x + 3} \) where \( x \geq -3 \), we need to follow these steps:
1. Express the function in terms of \( y \):
[tex]\[
y = -\frac{1}{2} \sqrt{x + 3}
\][/tex]
2. Solve for \( x \) in terms of \( y \):
[tex]\[
y = -\frac{1}{2} \sqrt{x + 3}
\][/tex]
Multiply both sides by \(-2\):
[tex]\[
-2y = \sqrt{x + 3}
\][/tex]
Square both sides to eliminate the square root:
[tex]\[
(-2y)^2 = x + 3
\][/tex]
Simplify:
[tex]\[
4y^2 = x + 3
\][/tex]
Solve for \( x \):
[tex]\[
x = 4y^2 - 3
\][/tex]
3. Write the inverse function:
When solving for the inverse function, replace \( y \) with \( x \):
[tex]\[
f^{-1}(x) = 4x^2 - 3
\][/tex]
4. Determine the domain of the inverse function:
The domain of \( f^{-1}(x) \) is the range of the original function \( f(x) \).
For \( f(x) = -\frac{1}{2} \sqrt{x + 3} \), where \( x \geq -3 \), we find the range by checking the end behavior of \( f(x) \):
- When \( x = -3 \), \( f(-3) = -\frac{1}{2} \sqrt{0} = 0 \).
- As \( x \to \infty \), \( f(x) \to -\infty \) since the square root function increases without bound but is scaled and reflected.
Therefore, the range of \( f(x) \) is \( (-\infty, 0] \). This range becomes the domain of the inverse function \( f^{-1}(x) \).
So, the inverse function is given by:
[tex]\[
f^{-1}(x) = 4x^2 - 3 \quad \text{for} \quad x \leq 0
\][/tex]
Thus, the correct answers to fill in the boxes are:
[tex]\[
f^{-1}(x) = 4x^2 - 3, \text{ for } x \leq 0
\][/tex]
