Answers:
[tex](f \circ g)(x) = \sqrt{\frac{5x-3}{x-1}}\\\\\\(g \circ f)(x) = \frac{1}{2\sqrt{4x+5}-2}\\\\[/tex]
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Work Shown for part 1
[tex](f \circ g)(x) = f(g(x))\\\\f(x) = \sqrt{4x+5}\\\\f(g(x)) = \sqrt{4*g(x)+5}\\\\f(g(x)) = \sqrt{4*\frac{1}{2x-2}+5}\\\\f(g(x)) = \sqrt{\frac{4}{2x-2}+5}\\\\f(g(x)) = \sqrt{\frac{4}{2x-2}+5*\frac{2x-2}{2x-2}}\\\\f(g(x)) = \sqrt{\frac{4}{2x-2}+\frac{5(2x-2)}{2x-2}}\\\\f(g(x)) = \sqrt{\frac{4+5(2x-2)}{2x-2}}\\\\f(g(x)) = \sqrt{\frac{4+10x-10}{2x-2}}\\\\f(g(x)) = \sqrt{\frac{10x-6}{2x-2}}\\\\f(g(x)) = \sqrt{\frac{2(5x-3)}{2(x-1)}}\\\\f(g(x)) = \sqrt{\frac{5x-3}{x-1}}\\\\[/tex]
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Work Shown for part 2
[tex](g \circ f)(x) = g(f(x))\\\\g(x) = \frac{1}{2x-2}\\\\g(f(x)) = \frac{1}{2*f(x)-2}\\\\g(f(x)) = \frac{1}{2\sqrt{4x+5}-2}\\\\[/tex]
We could optionally rationalize the denominator, but I think that is just extra busy work in my opinion.
