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Find the inverse of the given function.
[tex]\[ f(x)=-\frac{1}{2} \sqrt{x+3}, \quad x \geq -3 \][/tex]

[tex]\[ f^{-1}(x) = \boxed{\square} x^2 = \boxed{\square}, \quad \text{for } x \leq \boxed{\square} \][/tex]


Sagot :

To find the inverse of the function [tex]\( f(x) = -\frac{1}{2} \sqrt{x+3} \)[/tex] with [tex]\( x \geq -3 \)[/tex], follow these steps:

1. Start by setting [tex]\( y = f(x) \)[/tex]:
[tex]\[ y = -\frac{1}{2} \sqrt{x+3} \][/tex]

2. Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]. Start by isolating the square root term:
[tex]\[ y = -\frac{1}{2} \sqrt{x+3} \][/tex]

Multiply both sides by -2 to get rid of the fraction and the negative sign:
[tex]\[ -2y = \sqrt{x+3} \][/tex]

3. Square both sides to eliminate the square root:
[tex]\[ (-2y)^2 = x + 3 \][/tex]
[tex]\[ 4y^2 = x + 3 \][/tex]

4. Solve for [tex]\( x \)[/tex]:
[tex]\[ x = 4y^2 - 3 \][/tex]

5. Now, express the inverse function [tex]\( f^{-1}(x) \)[/tex]:
[tex]\[ f^{-1}(x) = 4x^2 - 3 \][/tex]

6. To determine the domain of the inverse function, consider the range of the original function [tex]\( f(x) = -\frac{1}{2} \sqrt{x+3} \)[/tex], which is [tex]\( y \leq 0 \)[/tex] because the function is always non-positive. Therefore, the domain of the inverse function is [tex]\( x \leq 0 \)[/tex].

Summarizing:

- The inverse function is [tex]\( f^{-1}(x) = 4x^2 - 3 \)[/tex]
- The domain of the inverse function is [tex]\( x \leq 0 \)[/tex]

Thus, the correct answer is:

[tex]\[ f^{-1}(x) = 4x^2 - 3 \quad \text{, for } \quad x \leq 0 \][/tex]