Given: f(x)=tanx−4x where x∈(0,2π)
A function is strictly decreasing when its derivative is negative.
Taking the derivative:
f′(x)=sec2x−4
For the function to be strictly decreasing:
f′(x)<0
sec2x−4<0
sec2x<4
Since sec2x=cos2x1:
cos2x1<4
1<4cos2x
cos2x>41
Taking the square root (note that cosx is positive in (0,2π)):
cosx>21
The value cosx=21 at x=3π
Since cosine decreases from 1 to 0 as x goes from 0 to 2π, the inequality cosx>21 holds when x<3π.
Combined with the given domain:
0<x<3π
Therefore, the function is strictly decreasing in the interval (0,3π).