Let's define the functions:

• $f(x) = \sqrt{x + 1}$
• $g(x) = \sqrt{x - 3}$

Part 1: Find $f + g$, $f - g$, $fg$, and $\frac{f}{g}$

1. Sum: $f + g$ $f + g = \sqrt{x + 1} + \sqrt{x - 3}$

2. Difference: $f - g$ $f - g = \sqrt{x + 1} - \sqrt{x - 3}$

3. Product: $fg$ $fg = \sqrt{x + 1} \cdot \sqrt{x - 3} = \sqrt{(x + 1)(x - 3)} = \sqrt{x^2 - 2x - 3}$

4. Quotient: $\frac{f}{g}$ \frac{f}{g} = \frac{\sqrt{x + 1}}{\sqrt{x - 3}} = \sqrt{\frac{x + 1}{x - 3}} \quad \text{(assuming g(x) \neq 0)}

Domain Determination

To find the domain of each function, we must consider where the expressions under the square roots are non-negative.

1. Domain of $f(x) = \sqrt{x + 1}$: $x + 1 \geq 0 \implies x \geq -1$

2. Domain of $g(x) = \sqrt{x - 3}$: $x - 3 \geq 0 \implies x \geq 3$

3. Domain of $f + g$: The domain is the intersection of the domains of $f$ and $g$: $x \geq 3$

4. Domain of $f - g$: Same as for $f + g$: $x \geq 3$

5. Domain of $fg$: Same as for $f + g$: $x \geq 3$

6. Domain of $\frac{f}{g}$: Here, we need $g(x) \neq 0$: $x - 3 > 0 \implies x > 3$

Summary

• $f + g = \sqrt{x + 1} + \sqrt{x - 3}$, domain: $x \geq 3$
• $f - g = \sqrt{x + 1} - \sqrt{x - 3}$, domain: $x \geq 3$
• $fg = \sqrt{x^2 - 2x - 3}$, domain: $x \geq 3$
• $\frac{f}{g} = \sqrt{\frac{x + 1}{x - 3}}$, domain: $x > 3$

Would you like further details on any part of this, or do you have any questions? Here are some related questions:

1. How do we simplify expressions involving square roots?
2. What is the significance of the domain in functions?
3. Can you explain how to find the intersection of two domains?
4. How can we graph the functions $f(x)$ and $g(x)$?
5. What are some applications of square root functions in real life?

Tip: Always check the domain of your functions when dealing with square roots to avoid undefined expressions!