Question

find the global maximum and minimum for the function on the closed interval f(x)= 6 e^-x sinx
0 is less than or equal to x and x is less than or equal to 2 pi

Answer 1

Global min and max occur when the derivative of f(x) = 0
The derivative of f(x) = 6 e^(-x) sin(x) is a simple application of the chain rule
d(f(x))/dx = -6 e^(-x) sin(x) + 6 e^(-x) cos(x) = 6 e^(-x) (cos(x) - sin(x))
let this expression equal 0 and solve for x
6 e^(-x) (cos(x) - sin(x)) = 0; divide both sides by 6 e^(-x)
cos(x) - sin(x) = 0
and cos(x) = sin(x)
from Trigonometry we know that this happens for x = pi / 4 and x = 5 pi / 4
plug both values in
f(pi/4) = 1.9343 and f(5pi/4) = -.0835
the first is the max value, and the second is the minimum value

Answer 2

To verify this, plot the function on a graphing calculator. The plot looks like this: