Question

Identify the maximum and minimum values of the function y = 8 cos x in the interval [-2π, 2π]. Use your understanding of transformations, not your graphing calculator.

Answer 1

differentiate
y'=-8sin(x)
find critical values ( set the derivative equal to zero)
-8sin(x)=0

sin(x)=0
x=0,pi,

put in the function
y=8cos(0)=8
y=8cos(pi)=-8

so maximium at x=0 is 8
minimium at x =pi is -8

Answer 2

Where I am confused at is where did the y'=-8sin(x) come from?

Answer 3

8 is a constant so it comes along for the ride. Then look at your rules for:
\[\frac{ dy }{dx }\cos x\]

Know that one by heart? If not, here are some proofs to study: http://www.math.com/tables/derivatives/more/trig.htm