Let's analyze and solve the questions visible in the uploaded image step by step. The task involves finding the functions $f + g$, $f - g$, $f \circ g$, and $f / g$, as well as determining their respective domains. The given functions are:

• $f(x) = x$
• $g(x) = \sqrt{x}$

Step 1: $(f + g)(x)$ and its domain

Formula:

$(f + g)(x) = f(x) + g(x) = x + \sqrt{x}$

Domain:

• $\sqrt{x}$ is defined when $x \geq 0$.
• Therefore, the domain of $(f + g)(x)$ is $[0, \infty)$.

Step 2: $(f - g)(x)$ and its domain

Formula:

$(f - g)(x) = f(x) - g(x) = x - \sqrt{x}$

Domain:

• The same condition applies as for $g(x) = \sqrt{x}$: $x \geq 0$.
• Thus, the domain of $(f - g)(x)$ is $[0, \infty)$.

Step 3: $(f \circ g)(x)$ and its domain

Formula:

$(f \circ g)(x) = f(g(x)) = f(\sqrt{x}) = \sqrt{x}$

Domain:

• $g(x) = \sqrt{x}$ requires $x \geq 0$, so the domain of $f \circ g$ is $[0, \infty)$.

Step 4: $(f / g)(x)$ and its domain

Formula:

$\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x}{\sqrt{x}}$

Simplify: \left(\frac{f}{g}\right)(x) = \sqrt{x} \quad \text{(for x > 0)}.

Domain:

• $g(x) = \sqrt{x}$ must be nonzero, which implies $x > 0$.
• Therefore, the domain of $\frac{f}{g}(x)$ is $(0, \infty)$.

Final Results:

1. $(f + g)(x) = x + \sqrt{x}, \quad \text{domain: } [0, \infty)$.
2. $(f - g)(x) = x - \sqrt{x}, \quad \text{domain: } [0, \infty)$.
3. $(f \circ g)(x) = \sqrt{x}, \quad \text{domain: } [0, \infty)$.
4. $\left(\frac{f}{g}\right)(x) = \sqrt{x}, \quad \text{domain: } (0, \infty)$.

Would you like additional steps or clarifications? Here are some related questions you might explore:

1. What is the graph of $f(x) + g(x)$ in its domain?
2. How do operations on $f(x)$ and $g(x)$ affect their domains?
3. How can you compute $(g \circ f)(x)$ instead of $(f \circ g)(x)$?
4. What happens to $\frac{f}{g}$ if $g(x)$ equals zero?
5. How can this problem generalize for other types of functions $f(x)$ and $g(x)$?

Tip: When finding the domain of a composite function, always consider both the domain of the inner and the outer function.