Find interval of increase and decrease of f(x) = 8 sin(x) + cot(x), −π ≤ x ≤ π

Given f(x)=8sin(x)+cot(x) for -pi<x<pi :

Note that:

f'(x)=8cos(x)-csc^2(x)

f''(x)=-8sin(x)+2csc^2(x)cot(x)

(1) To find the intervals where f(x) is increasing or decreasing we use the first derivative test; if the first derivative is positive on an interval the functio is increasing, negative implies the functio is decreasing.

Using technology we find the approximate zeros of f'(x) on -pi<x<pi :

x~~-1.443401

x~~-.3752857

x~~.3752857

x~~1.443401

Plugging in test values on the intervals yields:

f'(x)<0 on (-pi,-1.443401)

f'(x)>0 on (-1.443401,-.3752857)

f'(x)<0 on

Plz correct me if wrong

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