# f(x) = 8 sin(x) + cot(x) −pi ≤ x ≤ piA. 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...

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.

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### 1 Answer

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 `(-.3752857,0)`

`f'(x)<0` on `(0,.3752857)`

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

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

**Thus the function decreases on `(-pi,-1.443401)` achieving a local minimum at `x~~-1.443401` , increases on `(-1.443401,-.3752857)` achieving a local maximum at `x~~.3752857` , and decreases on `(-.3752857,0)` with the function failing to exist at x=0. The graph decreases on `(0,.3752857)` achieving a local minimum at `x~~.3752857` , increases on `(.3752857,1.443401)` achieving a local maximum at `x~~1.443401)` , and decreases on `(1.443401,pi)` .**

(2) The inflection points occur when the second derivative equals zero at a point c, and changes sign at point c.

**Again using technology we find the points where `f''(x)=0` to be `x~~-.718938` and `x~~.718938` . Checking we find the second derivative changes sign at these points, thus they are inflection points.**

(3) The function is concave up if the second derivative is positive; concave down if the second derivative is negative.

`f''(x)>0` on `(-pi,-.718938)`

`f''(x)<0` on `(-.718938,0)`

`f''(x)>0` on `(0,.718938)`

`f''(x)<0` on `(.718938,pi)`

**Thus the function is concave up , then concave down, concave up, then concave down on the intervals indicated.**

(4) The graph: