For f(x) = e2sin(x) use your graphing calculator to find the number of zeros for f '(x) on the closed interval [0, 2π].

Answer 1

I will assume you meant
f(x) = e^2 sinx

e^2cosx = 0
then cosx = 0 , since e^2 is a constant
x = π/2, 3π/2

(the graph of y = a cosx crosses the x-axis at π/2 , (90°) and 3π/2 , (270°)
So there are two zeros for f ' (x)

Answer 2

no idea

Answer 3

Put the lime in the coconut and shake it all up