Question

Let f be the function defined by f(x) = x-2cosx on the closed interval [0,2pi]

Determine the value of x at which f has its
-Absolute Max
-Absolute Min

Answer 1

To find the absolute min and max, I know I have to take the derivative

I got: 1+2sinx = 0

sinx = -1/2
Now is this an arcsin problem?

x = arcsin(-1/2) ? ?

Where do I go from there?

Answer 2

I'm still here :)

Answer 3

the derivative is positive for every x which belongs to [0,2pi]. Means, the function is monotonically increasing. From that, you just find f(0) and f(2pi), and these are the min and the max..

Answer 4

sinx=-0.5 only for the angle -30 degrees (-pi/6), which is not between 0 to 2pi.

Answer 5

So I can make the derivatives equal to 0 and 2pi to find min and max?

Answer 6

No. You found out with the derivative that the function is monotonically increasing. So it's minimum is the left value=0, and the highest point of it's graph is the right value (2pi). To find these values, you put x=0 and x=2pi in the function, not in the derivative.

Answer 7

f(0)=0-2cos(0)=-2
f(2pi)=2pi-2cos(2pi)=2pi-2

Answer 8

Minimum: (0,-2) ; Maximum: (2pi,2-2pi)

Answer 9

Oh the only reason why I found the derivative to see if it was increasing or decreasing.
OH that's how I got those values.
OKay thank you :)

Answer 10

:D