함수의 증가와 감소

Since we can't take the square root of a negative number, $f(x)$ is only defined when $x \ge 1$. Furthermore, since we can't have zero as a denominator, $x=-2$ is not in the domain — but as long as $x \ge 1$, we also have $x \ne -2$. So, the domain of the function is $[1,\infty)$.

In order to find where $f(x)$ is increasing, we find where $f'(x)$ is positive. $$f'(x)=\frac{\frac{2x+4}{2\sqrt{x-1}}-2\sqrt{x-1}}{(2x+4)^2} =\frac{(x+2)-2(x-1)}{\sqrt{x-1}(2x+4)^2} =\frac{-x+4}{\sqrt{x-1}(2x+4)^2}$$ The denominator is never negative, so $f(x)$ is increasing when the numerator of $f'(x)$ is positive, i.e. when $4-x>0$, or $x<4$. Recalling that the domain of definition for $f(x)$ is $[1,+\infty)$, we conclude that $f(x)$ is increasing on the open interval $(1,4)$.