Sure! To find the inverse of the function [tex]\( f(x) = -\frac{1}{2} \sqrt{x + 3} \)[/tex] for [tex]\( x \geq -3 \)[/tex]:
1. Start with the function [tex]\( y = -\frac{1}{2} \sqrt{x + 3} \)[/tex].
2. Isolate the square root term:
[tex]\[
y = -\frac{1}{2} \sqrt{x + 3}
\][/tex]
Multiply both sides by -2:
[tex]\[
-2y = \sqrt{x + 3}
\][/tex]
3. Square both sides to remove the square root:
[tex]\[
4y^2 = x + 3
\][/tex]
4. Solve for [tex]\( x \)[/tex]:
[tex]\[
x = 4y^2 - 3
\][/tex]
5. Switch [tex]\( x \)[/tex] and [tex]\( y \)[/tex] to find the inverse function [tex]\( f^{-1}(x) \)[/tex]:
[tex]\[
f^{-1}(x) = 4x^2 - 3
\][/tex]
6. The domain of the original function [tex]\( f(x) \)[/tex] was [tex]\( x \geq -3 \)[/tex], which implies the range of [tex]\( f(x) \)[/tex] is [tex]\( y \leq 0 \)[/tex]. Thus, for the inverse function, the domain must be [tex]\( x \leq 0 \)[/tex].
So the inverse function is:
[tex]\[
f^{-1}(x) = 4x^2 - 3, \text{ for } x \leq 0
\][/tex]
Thus, the correct answer to fill in the box is [tex]\( 0 \)[/tex].
[tex]\[
f^{-1}(x) = 4x^2 - 3, \text{ for } x \leq 0
\][/tex]
