Consider the function $$ f(x)=x^{2}+3 $$ for the domain $$ [0, \infty) $$. Find $$ f^{-1}(x) $$, where $$ f^{-1} $$ is the inverse of $$ f $$. Also state the domain of $$ f^{-1} $$ in interval notation. $$ f^{-1}(x)=\prod $$ for the domain

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1. Start with the function:
   $$
   y = x^2 + 3
   $$
   with the domain $$ x \in [0, \infty) $$.

2. To find the inverse, solve for $$ x $$ in terms of $$ y $$:
   $$
   x^2 = y - 3
   $$

3. Since $$ x \geq 0 $$, take the positive square root:
   $$
   x = \sqrt{y - 3}
   $$

4. Replace $$ y $$ with $$ x $$ in the inverse function notation:
   $$
   f^{-1}(x) = \sqrt{x - 3}
   $$

5. The domain of $$ f^{-1}(x) $$ corresponds to the range of $$ f(x) $$. Since $$ f(x) = x^2 + 3 $$ and $$ x \geq 0 $$, the minimum value of $$ f(x) $$ occurs when $$ x = 0 $$:
   $$
   f(0) = 0^2 + 3 = 3
   $$
   Hence, the range of $$ f $$ is $$[3, \infty)$$, which becomes the domain of $$ f^{-1} $$.

6. Summary of the results:
   $$
   f^{-1}(x) = \sqrt{x - 3} \quad \text{with domain } [3, \infty)
   $$
The inverse function is $$ f^{-1}(x) = \sqrt{x - 3} $$ with the domain $$[3, \infty)$$.

Consider the function \( f(x)=x^{2}+3 \) for the domain \( [0, \infty) \). Find \( f^{-1}(x) \), where \( f^{-1} \) is the inverse of \( f \). Also state the domain of \( f^{-1} \) in interval notation. \( f^{-1}(x)=\prod \) for the domain

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