Let $f(x) = \sqrt{x}$ and $g(x) = x$ be two functions defined over the set of non-negative real numbers. Find $(f+g)(x)$,$(f-g)(x)$,$(fg)(x)$,and $(\frac{f}{g})(x)$.
Given $f(x) = \sqrt{x}$ and $g(x) = x$ for $x \ge 0$.
$(f+g)(x) = f(x) + g(x) = \sqrt{x} + x$.
$(f-g)(x) = f(x) - g(x) = \sqrt{x} - x$.
$(fg)(x) = f(x) \cdot g(x) = \sqrt{x} \cdot x = x^{\frac{1}{2}} \cdot x^1 = x^{\frac{3}{2}}$.
$(\frac{f}{g})(x) = \frac{f(x)}{g(x)} = \frac{\sqrt{x}}{x} = x^{\frac{1}{2} - 1} = x^{-\frac{1}{2}}$,where $x > 0$.
$(f+g)(x) = f(x) + g(x) = \sqrt{x} + x$.
$(f-g)(x) = f(x) - g(x) = \sqrt{x} - x$.
$(fg)(x) = f(x) \cdot g(x) = \sqrt{x} \cdot x = x^{\frac{1}{2}} \cdot x^1 = x^{\frac{3}{2}}$.
$(\frac{f}{g})(x) = \frac{f(x)}{g(x)} = \frac{\sqrt{x}}{x} = x^{\frac{1}{2} - 1} = x^{-\frac{1}{2}}$,where $x > 0$.
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