# What are the points of inflection of #f(x)=8x + 3 - 2sinx # on # x in [0, 2pi]#?

##### 1 Answer

#### Explanation:

Since

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The first step in finding the points of inflection is taking the second derivative of the function. So let's do that:

#f(x) = 8x + 3 - 2sinx#

#f'(x) = 8 - 2cosx#

#f''(x) = 2sinx#

The points of inflection are points where the second derivative equals 0.

#f''(x) = 0#

#2sinx = 0#

#sinx = 0#

On the closed interval

#x in {0, pi, 2pi}#

Just to make sure that the concavity actually DOES change at these points, let's use some test points to look at the concavity:

#f''(-pi/2) = 2sin(-pi/2) = color(red)(-2#

#f''(pi/2) = 2sin(pi/2) = color(limegreen)2#

#f''((3pi)/2) = 2sin((3pi)/2) = color(red)(-2#

#f''((5pi)/2) = 2sin((5pi)/2) = color(limegreen)(2#

These points tell us that:

Below 0, the concavity is negative, and between 0 and

#pi# , the concavity is positive.Therefore, 0 is an inflection point.Between 0 and

#pi# the concavity is positive, and between#pi# and#2pi# the concavity is negative.Therefore,#pi# is an inflection point.Between

#pi# and#2pi# the concavity is negative, and above#2pi# the concavity is positive.Therefore,#2pi# is an inflection point.

So our three points of inflection are

*Final Answer*