Answer
At $x=-1$, $y=0$ absolute and local minimum.
At $x=1$, $y=0$ absolute and local minimum.
Work Step by Step
Step 1. Given the function $y=\sqrt {x^2-1}=(x^2-1)^{1/2}$, take its derivative to get $y′=\frac{1}{2}(x^2-1)^{-1/2}(2x)=\frac{x}{\sqrt {x^2-1}}$
Step 2. The critical points can be found when $y'=0$ or undefined, we have $y′=\frac{x}{\sqrt {x^2-1}}=0$ which gives $x=0, \pm1$. Discard $x=0$ as it is not in the domain.
Step 3. At $x=-1$, $y=0$ and we can identify this as an absolute and local minimum (use local test points and $x\to\pm\infty, y\to\infty$) as shown in the figure.
Step 4. At $x=1$, $y=0$ and we can identify this as an absolute and local minimum (use local test points and $x\to\pm\infty, y\to\infty$) as shown in the figure.
