We are given the following functions and asked to find:

• $f + g$
• $f - g$
• $fg$
• $\frac{f}{g}$

The functions are:

• $f(x) = \sqrt{x}$
• $g(x) = x - 17$

We'll also determine the domain of each resulting function.

1. $f + g$

$(f + g)(x) = f(x) + g(x) = \sqrt{x} + (x - 17)$ Domain:

• $f(x) = \sqrt{x}$ requires $x \geq 0$ because we cannot take the square root of a negative number.
• $g(x) = x - 17$ is defined for all real numbers.

Thus, the domain of $f + g$ is $[0, \infty)$.

2. $f - g$

$(f - g)(x) = f(x) - g(x) = \sqrt{x} - (x - 17) = \sqrt{x} - x + 17$ Domain:

• Same as $f(x)$, $x \geq 0$.

Thus, the domain of $f - g$ is $[0, \infty)$.

3. $fg$

$(fg)(x) = f(x) \cdot g(x) = \sqrt{x} \cdot (x - 17) = \sqrt{x}(x - 17)$ Domain:

• Again, the square root function $\sqrt{x}$ requires $x \geq 0$.

Thus, the domain of $fg$ is $[0, \infty)$.

4. $\frac{f}{g}$

$\left(\frac{f}{g}\right)(x) = \frac{\sqrt{x}}{x - 17}$ Domain:

• $f(x) = \sqrt{x}$ requires $x \geq 0$.
• The denominator $g(x) = x - 17 \neq 0$, so $x \neq 17$.

Thus, the domain of $\frac{f}{g}$ is $[0, \infty) \setminus \{17\}$, meaning $x \geq 0$ but $x \neq 17$.

Summary:

• $f + g = \sqrt{x} + x - 17$, domain: $[0, \infty)$
• $f - g = \sqrt{x} - x + 17$, domain: $[0, \infty)$
• $fg = \sqrt{x}(x - 17)$, domain: $[0, \infty)$
• $\frac{f}{g} = \frac{\sqrt{x}}{x - 17}$, domain: $[0, \infty) \setminus \{17\}$

Would you like more details or have any further questions?
Here are some related questions:

1. What is the difference between the domain and range of a function?
2. How can you determine if a function is continuous or discontinuous?
3. How do we handle square root functions when combined with other operations?
4. Can you explain how division affects the domain of a function?
5. How would you graph these combined functions to visualize their domains?

Tip: When dealing with square roots, always ensure the expression under the root is non-negative to determine the function's domain.