# Determine where the function is concave up and down and identify points of inflection. g(x)=sin(2x)+4sin(x) on the interval [0, 2pi]

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### 1 Answer

Given `g(x)=sin2x+4sinx` on `[0,2pi]` ; determine when the function is concave up and concave down, and identify points of inflection.

(1) To determine concavity, we examine the second derivative of the function. If the second derivative is positive, the function is concave up; if negative then concave down. If the second derivative at c is 0, and the second derivative changes sign at c then c is an inflection point.

(2) `g(x)=sin2x+4sinx`

`g'(x)=2cos2x+4cosx`

`g''(x)=-4sin2x-4sinx`

(3) Setting `g''(x)=0` we get:

`-4sin2x-4sinx=0`

`-4sin2x=4sinx`

` ``sin2x=-sinx`

`2sinxcosx=-sinx`

`2sinxcosx+sinx=0`

`sinx(2cosx+1)=0`

`=> sinx=0` or `cosx=-1/2`

(4) On `[0,2pi]` `sinx=0 => x=0,pi,2pi`

and `cosx=-1/2 => x=(2pi)/3,(4pi)/3`

So we check the sign of the second derivative on the indicated intervals by finding `g''(c)` where c is in the interval:

`(0,(2pi)/3): g''(pi/2)=-4<0 => g''(x)<0` on this interval thus conc. dwn

`((2pi)/3,pi):g''((5pi)/6)=2sqrt(3)-2>0` thus conc. up

`(pi,(4pi)/3):g''((7pi)/6)=-2sqrt(3)+2<0` thus concave down

`((4pi)/3,2pi):g''((3pi)/2)=4>0` thus concave up.

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**The points of inflection in the interval are `x=0,(2pi)/3,pi,(4pi)/3,2pi` .(I counted the endpoints as points of inflection because they will be for the unrestricted graph; if the domain is strictly `[0,2pi]` then neither 0 nor `2pi` is an inflection point.**)

**The function is concave up on `((2pi)/3,pi)` and `((4pi)/3,2pi)` .**

**The function is concave down on `(0,(2pi)/3)` and `(pi,(4pi)/3)` **

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