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

(The entire section contains 295 words.)

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