Chapter 3 - Applications of Differentiation - 3.1 Maximum and Minimum Values - 3.1 Execises - Page 211: 15

Absolute maximum = $4$ No local maximum. No absolute or local minimum.

The function is linear, its graph is a line. We need two points: $f(1)=\frac{1}{2}[3(1)-1]=1$ $f(3)$ = $\frac{1}{2}[3(3)-1]$ = $4$ We graph the line. Absolute maximum = $4$. No local maximum. No absolute or local minimum.