Find the critical points and maximum and minimum values

Question

Can I please get help I’m stuck. Thanks

Find all critical points of f(x)=e^-xcosx on [0,2pi]

find the absolute maximum and minimum values of f(x)=e^-xcosx on [0,2pi]. Where do these occur?

Answer 1

Critical points are found by setting the first derivative equal to zero and solving (plus looking at places on the interval where that derivative is not defined).

For f(x) = e^-xcos(x), since this is a product, we use the product rule: (u•v)' = u'v + uv' to find the derivative.

In this case, u = e^-x so u' = -e^-x and v = cos(x) so v' = -sin(x)

So f '(x) = (-e^-x)(cos(x)) + (e^-x)(-sin(x)) = (-e^-x)[cos(x) + sin(x)]

Setting f ' = 0 and solving:

(-e^-x)[cos(x) + sin(x)] = 0 means either (-e^-x) = 0 or [cos(x) + sin(x)] = 0

Since the exponential -e^-x can never equal zero, that portion has no solution to it. For cos(x) + sin(x) = 0 means cos(x) = -sin(x) which occurs at the 45° angles in Q2 and Q4 or x = 3π/4 and 7π/4. There are no places on [0, 2π] where f ' is undefined. So the critical points are x = 3π/4 and 7π/4.

To find the absolute max and min on the interval, find the function values at the endpoints and at the critical points and compare:

x f(x)

------------------

0 1

3π/4 -0.067

7π/4 0.003

2π 0.002

This means that the absolute max on the interval occurs at x = 0 and is 1. It also means the absolute min on the interval occurs at x = 3π/4 and is approx -0.067