The problem in your image asks to find the inverse of the given function $f(x) = \sqrt{3 + x}, \, x \geq -3$.

Step-by-step solution:

1. Find the inverse function:
   Start by letting $y = f(x) = \sqrt{3 + x}$, which can be rewritten as: $y = \sqrt{3 + x}$ To find the inverse, solve for $x$ in terms of $y$: $y^2 = 3 + x$ $x = y^2 - 3$ Now, swap $x$ and $y$ to express the inverse function $f^{-1}(x)$: $f^{-1}(x) = x^2 - 3$

2. Domain and range:

   • For the original function $f(x) = \sqrt{3 + x}$, the domain is $x \geq -3$ since the expression inside the square root must be non-negative. The range is $f(x) \geq 0$, meaning $y \geq 0$.
   • For the inverse function $f^{-1}(x) = x^2 - 3$, the domain is $x \geq 0$ because the range of the original function becomes the domain of the inverse. The range of the inverse function is $y \geq -3$, as $f^{-1}(x) = x^2 - 3$.

Answer:

Looking at the multiple-choice options, the correct choice is:

• B. $f^{-1}(x) = x^2 - 3, \, x \geq 0$.

Would you like a detailed explanation of the inverse function process, or do you have any other questions?

Related Questions:

1. How do you check if a function is one-to-one?
2. Can every function have an inverse?
3. What is the difference between a one-to-one function and a many-to-one function?
4. How can you graphically determine if a function has an inverse?
5. Why does the domain of the original function become the range of the inverse function?

Tip:

Always remember that when finding the inverse, it's important to swap the domain and range of the original function for the inverse function.