By the property [tex] \sqrt{\dfrac{a}{b}}=\dfrac{\sqrt a}{\sqrt b } [/tex], it may seem that both functions are the same, but they're not, because they have different domains.
[tex] f(x)=\sqrt{\dfrac{x+1}{x-1}}\\\dfrac{x+1}{x-1}\geq 0\wedge x\not=1\\(x+1)(x-1)\geq0\\x\in(-\infty-1\rangle\cup\langle1,\infty)\\\boxed{D:x\in(-\infty-1\rangle\cup(1,\infty)}\\\\\\g(x)=\dfrac{\sqrt{x+1}}{\sqrt{x-1}}\\x+1 \geq 0 \wedge x-1\geq 0 \wedge \sqrt{x-1}\not =0\\x\geq -1 \wedge x\geq 1 \wedge x\not =1\\\boxed{D:x>1} [/tex]
[tex] D_{f(x)}\not=D_{g(x)} [/tex] so the two functions are not the same.
If f(x) and g(x) were the same function, they would produce identical outputs from identical inputs. If we can find a single example where they don't, we've proven that they're distinct.
Take x = 3, for instance. For f(3), we find
[tex] f(3)=\sqrt{\frac{3+1}{3-1}} =\sqrt{\frac{4}{2}} =\sqrt{2} [/tex]
while for g(3):
[tex] g(3)=\frac{\sqrt{3+1}}{\sqrt{3-1}} =\frac{\sqrt{4}}{\sqrt{2}} =\frac{2}{\sqrt{2}} [/tex]
From this example, we can see that we're dealing with two nonequivalent functions.
EDIT: I've realized that I made a pretty big gaffe here, since
[tex] \frac{2}{\sqrt{2}} \times\frac{\sqrt{2}}{\sqrt{2}}=\frac{2\sqrt{2}}{2}=\sqrt{2} [/tex]
My argument would actually be incorrect in this case, then, and Konrad's would be more appropriate.
