Answer
$f+g=x+\sqrt{x}$
$D=\{x|x\geq0\}$ or $D=[0,\infty)$
$f-g=x-\sqrt{x}$
$D=\{x|x\geq0\}$ or $D=[0,\infty)$
$fg=x\sqrt{x}$
$D=\{x|x\geq0\}$ or $D=[0,\infty)$
$f/g=\dfrac{x}{\sqrt{x}}$
$D=\{x|x\gt0\}$ or $D=(0,\infty)$
Work Step by Step
$f(x)=x,$ $g(x)=\sqrt{x}$
$f+g$
$f(x)+g(x)=x+\sqrt{x}$
The domain of this function doesn't include the values of $x$ that make the expression inside the square root a negative number. Therefore, the domain of this function only has the numbers greater than or equal to $0$. The domain can be expressed as follows:
$D=\{x|x\geq0\}$ or $D=[0,\infty)$
$f-g$
$f(x)-g(x)=x-\sqrt{x}$
The domain of this function doesn't include the values of $x$ that make the expression inside the square root a negative number. Therefore, the domain of this function only has the numbers greater than or equal to $0$. The domain can be expressed as follows:
$D=\{x|x\geq0\}$ or $D=[0,\infty)$
$fg$
$f(x)\cdot g(x)=x\sqrt{x}$
The domain of this function doesn't include the values of $x$ that make the expression inside the square root a negative number. Therefore, the domain of this function only has the numbers greater than or equal to $0$. The domain can be expressed as follows:
$D=\{x|x\geq0\}$ or $D=[0,\infty)$
$f/g$
$\dfrac{f(x)}{g(x)}=\dfrac{x}{\sqrt{x}}$
This function is undefined for the values of $x$ that make the denominator equal to $0$ and for the values of $x$ that make the expression inside the square root a negative number. Therefore, the domain only includes the numbers greater than $0$. The domain can be expressed as follows:
$D=\{x|x\gt0\}$ or $D=(0,\infty)$