$$ 1 \leftarrow $$ Determine whether the given function is one-to-one. If it is one-to-one, find a formula for the $$ f(x)=\sqrt{x-3} $$ A. $$ f^{-1}(x)=x^{2}+3, x \geq 0 $$ B. $$ f^{-1}(x)=(x-3)^{2} $$ C. $$ N^{-1}(x)=\sqrt{x+3} $$

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To determine whether the function $$ f(x) = \sqrt{x - 3} $$ is one-to-one, we need to check if it passes the horizontal line test or if it is strictly increasing or decreasing.

1. **Domain of $$ f(x) $$**: The function $$ f(x) = \sqrt{x - 3} $$ is defined for $$ x \geq 3 $$ because the expression under the square root must be non-negative.

2. **Behavior of $$ f(x) $$**: The function $$ f(x) $$ is a square root function, which is always non-negative and increases as $$ x $$ increases. Specifically, as $$ x $$ increases from 3 to infinity, $$ f(x) $$ also increases from 0 to infinity. This indicates that the function is strictly increasing.

Since $$ f(x) $$ is strictly increasing, it is one-to-one.

3. **Finding the inverse**: To find the inverse of $$ f(x) $$, we set $$ y = f(x) $$:
   $$
   y = \sqrt{x - 3}
   $$
   Now, we solve for $$ x $$:
   $$
   y^2 = x - 3
   $$
   $$
   x = y^2 + 3
   $$
   Thus, the inverse function is:
   $$
   f^{-1}(x) = x^2 + 3
   $$
   However, since the original function $$ f(x) $$ is defined for $$ x \geq 3 $$, the inverse function will be defined for $$ y \geq 0 $$ (since $$ f(x) $$ outputs non-negative values).

4. **Conclusion**: The correct inverse function is:
   $$
   f^{-1}(x) = x^2 + 3, \quad x \geq 0
   $$

Thus, the answer is:
**A. $$ f^{-1}(x) = x^2 + 3, x \geq 0 $$**.
The inverse function of $$ f(x) = \sqrt{x - 3} $$ is $$ f^{-1}(x) = x^2 + 3, x \geq 0 $$.

Determine whether the given function is one-to-one. If it is one-to-one, find a formula for the

A.
B.
C.

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