what is the domain and range of f(x)= -4_/1 + cos 2x?

Question

negative 4 square root of 1 + cos2x..plz help with pre cal

Answer 1

domain is all values except x cannot equal 90 degrees. (Assuming the interval is 0 to 360.) That would give you a zero in the denominator.

The denominator will always be between 0 and 2. (including 2, not including 0.)

So, the range will be less than or equal to -2.

Answer 2

y = -4(1+cos 2x)^.5
-1 =< cos2x =< 1
Thus 0 =< (1 +cos 2x)^.5 =< sqrt(2)
Thus 0 >= -4(1+cos 2x)^.5 >= -4 sqrt(2)
So range is 0 >= y >= -4sqrt(2)
Domain is all real numbers

Answer 3

Any point in which (1+cos(2x)) is different from 0:

1+cos(2x) =/= 0
cos(2x) =/= -1
2x =/= pi+2n pi
x =/= pi/2 + n pi
x =/= (1/2 + n) pi

where n is integer

Answer 4

f(x) = -4sqrt(1+cos2x)

1+cos2x >= 0 cos2x <= 1
R: all real number
D: [-4sqrt(2), 0]