how do i find the range of f(x)=1/1-2cosx?

Answer 1

It´s periodic 2 pi
1/(1-2cos x)=k
so 1-2cosx=1/k and cosx = 1/2(1-1/k) = (k-1)/2k
so
-1<=(k-1)/2k<=1
(k-1)/2k +1>= 0 so (3k-1)/2k >=0 k<0 and k>=1/3

(k-1)/2k-1<=0 (-k-1)/2k<=0 so k>0 and k <= -1
so the range is f(x)<=-1 and f(x)>=1/3

Answer 2

Assuming you mean f(x)=1/(1-2cosx)

The range is every real number except 0.

You can make the function as large as you want by making (1-2cosx) as close to 0 as you need. So the closer cos(x) is to 1/2, the larger the function will be in absolute value.

Answer 3

This one is actually complicated to find. So i would recommend graphing this one. The graph should look like some U shape and upside down U shapes. So the range would be everything from -inf to the max of the upside down U and from the minimum value of the U to positive inf.

Answer 4

Ana M almost has it right.

1-2cos(x) has the range [-1,3]. That means the range of 1/(1-2cos(x)) can be, for negative values:

(-oo, -1]

and for positive values:

[1/3, oo)

Answer 5

Well, cos(x) takes values in the interval [-1,1].
Which means that 2cos(x) takes values in the interval [-2,2], while 1-2cos(x) takes values in the interval [-1,3]
Now, the problem is that in the interval [-1,3] you have 0. This means that your f(x) ends up taking values between minus infinity and plus infinity.