How to find extreme values and where they occur? What are the extreme values of the function $$y = (x^2 - 1)^{1/2} = \sqrt{x^2 - 1}$$ and where do they occur?
I have gotten as far as finding that $$\frac{dy}{dx} = \frac{x}{\sqrt{x^2 - 1}}$$ What next? Set $dy/dx$ to $0$?
 A: Whenever you want to know something about a function's values, you first need to know what the domain of the function even is. In this case, the function is defined for $|x|\ge1$, that is, on $(-\infty,-1] \cup [1,\infty)$.
A wonderful property of continuous functions is that extreme values (local maxima and minima) can occur only at critical points and at endpoints of the domain. To find the critical points, you find everywhere in the domain that the derivative either vanishes or is undefined. Here, the derivative doesn't vanish anywhere in the domain; it is undefined at $x=\pm1$. The endpoints of the domain are at $x=\pm1$ anyway. So the only possible local extrema are at $x=-1$ and at $x=1$. And sure enough, those are local minima (the values there are $0$ and the function is positive elsewhere).
If by "extreme values" you mean global minima and maxima, then you need to check $\lim_{x\to\infty} y(x)$ and $\lim_{x\to-\infty} y(x)$ as well, since the domain extends without end in both directions. In this case, you'll find that the function has no global maximum, and $0$ is the global minimum.
