f(x) = 8 sin(x) + cot(x) −pi ≤ x ≤ pi A. Find the interval of increase and decrease.  b. Find the inflection points of the function (2) (x,y) C. Find the interval where the function is concave up and concave down. 

Expert Answers

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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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