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. 

Asked on by mllysmith

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embizze | High School Teacher | (Level 1) Educator Emeritus

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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 `(-.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:

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