Question 8.1.1: Find possible maximum and minimum points for: (a) f (x) = 3 ......

(a) Because (x − 2)^{2} ≥ 0 for all x, it follows that f (x) ≤ 3 for all x. But f (x) = 3 when (x − 2)^{2}= 0 at x = 2. Therefore, x = 2 is a maximum point for f . Because f (x)→−∞ as x→∞, f has no minimum.
(b) Since \sqrt{x − 5} is ≥ 0 for all x ≥ 5, it follows that f (x) ≥ −100 for all x ≥ 5. Since f (5) = −100, we conclude that x = 5 is a minimum point. Since f (x)→∞ as  x→∞, f has no maximum.