If the functions are defined as f(x) = sqrt{x | vedclass

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(C) For $f(x) = \sqrt{x}$,the domain is $[0, \infty)$.For $g(x) = \sqrt{1-x}$,the domain is $(-\infty, 1]$.The domain of $f+g, f-g,$ and $g-f$ is the

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$0 < x < 1$

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(C) For $f(x) = \sqrt{x}$,the domain is $[0, \infty)$.For $g(x) = \sqrt{1-x}$,the domain is $(-\infty, 1]$.The domain of $f+g, f-g,$ and $g-f$ is the intersection of the domains of $f$ and $g$,which is $[0, 1]$.For $f/g$,we require $g(x) 
eq 0$,so $1-x 
eq 0 \implies x 
eq 1$. The domain is $[0, 1)$.For $g/f$,we require $f(x) 
eq 0$,so $x 
eq 0$. The domain is $(0, 1]$.The common domain for all these functions is the intersection of $[0, 1], [0, 1),$ and $(0, 1]$,which is $(0, 1)$.

If the functions are defined as $f(x) = \sqrt{x}$ and $g(x) = \sqrt{1-x}$,then what is the common domain of the following functions: $f+g, f-g, f/g, g/f, g-f$ where $(f \pm g)(x) = f(x) \pm g(x)$ and $(f/g)(x) = \frac{f(x)}{g(x)}$?

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