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 $x\ge 0$, and the domain of $g(x)=x-5$ is all real numbers. Therefore, the domain of $f+g$ is $x\ge 0$.
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 $x\ge 0$.
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 $x\ge 0$ 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 $x\ge 0$ and $x\ne 5$ because the denominator cannot be zero.
5. Summarize the domains: $f+g$ and $f-g$ have domain $x\ge 0$, $fg$ has domain $x\ge 0$, and $\frac{f}{g}$ has domain $x\ge 0$ and $x\ne 5$.