Chapter 3 - Applications of Differentiation - 3.1 Exercises - Page 167: 35

absolute minimum $= -3$ at $x= \pi$ absolute maximum $= 3 $ at $x= 0$ and $x=2\pi$

$y= 3cos(x)$ the derivative of the function y: $y'= -3sin(x)$ To find the critical points put $y'=0$ $-3sin(x)=0$ $sin(x) = 0$ $ x= 0,\pi,2\pi$ Now, plugging the values of x in the function, $y(0)= 3cos(0)= 3$ $y(\pi)= 3cos(\pi)= -3$ $y(2\pi)= 3cos(2\pi)= 3$