Question

F(x) = 9 sin(x) + cot(x), −π ≤ x ≤ π find the interval of increase.

Answer 1

We are given the function
F(x) = 9 sinx + cot x

We need to take the first derivative of the given function so,
F' (x) = 9 cos x - csc² x

Next, we equate the first derivative of the function to 0 and solve for the values of x
0 = 9 cos x - csc² x
Solving for x
x = 2.04

Picking out an arbitrary value between 2.04 and π, say 3 and substituting in F(x)
F(3) = 9 sin 3 + cot 3 = 19.55

Therefore, the interval where the function is increasing is from 2.04 to π
Consequently, the interval where the function is decreasing is from -π to 2.04<span />