Answer
a. $f(3)=0$ minimum.
b. $f(3)=0$ absolute minimum, no absolute maximum.
c. See graph.
Work Step by Step
a. Step 1. Given the function $f(x)=\sqrt {x^2-2x-3}, 3\leq x\lt\infty$, we have $f'(x)=\frac{2x-2}{2\sqrt {x^2-2x-3}}=\frac{2x-2}{2\sqrt {(x-3)(x+1)}}$.
Step 2. The extrema happen when $f'(x)=0$, undefined, or at endpoints; thus the critical points are $x=\pm1, 3$. However, as $x\geq3$, we choose $x=3$ only.
Step 3. We have $f(3)=0$. Test signs of $f'(x)$ across critical points $(3)..(+)$; thus we have $f(3)$ as a local minimum.
b. We can identify that at $f(3)=0$ is an absolute minimum and there is no absolute maximum because $x\to\infty, f(x)\to \infty$
c. See graph.
