Find $f+g$, $f-g$, $fg$, and $\frac{f}{g}$. Determine the domain for each function.
$f\left(x\right)=\sqrt{x}$, $g\left(x\right)=x-5$

Verified step by step guidance

1. To find \( (f+g)(x) \), add the functions: \( f(x) + g(x) = \sqrt{x} + (x - 5) \). Simplify the expression to get \( \sqrt{x} + x - 5 \). The domain of \( f(x) = \sqrt{x} \) is \( x \geq 0 \), and the domain of \( g(x) = x - 5 \) is all real numbers. Therefore, the domain of \( f+g \) is \( x \geq 0 \).
2. To find \( (f-g)(x) \), subtract the functions: \( f(x) - g(x) = \sqrt{x} - (x - 5) \). Simplify the expression to get \( \sqrt{x} - x + 5 \). The domain is the same as \( f+g \), which is \( x \geq 0 \).
3. To find \( (fg)(x) \), multiply the functions: \( f(x) \cdot g(x) = \sqrt{x} \cdot (x - 5) \). This simplifies to \( x\sqrt{x} - 5\sqrt{x} \). The domain is \( x \geq 0 \) because \( \sqrt{x} \) is only defined for non-negative \( x \).
4. To find \( \left(\frac{f}{g}\right)(x) \), divide the functions: \( \frac{f(x)}{g(x)} = \frac{\sqrt{x}}{x - 5} \). The domain is \( x \geq 0 \) and \( x \neq 5 \) because the denominator cannot be zero.
5. Summarize the domains: \( f+g \) and \( f-g \) have domain \( x \geq 0 \), \( fg \) has domain \( x \geq 0 \), and \( \frac{f}{g} \) has domain \( x \geq 0 \) and \( x \neq 5 \).