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

Jan 24, 2018

$0$, $\pi$, and $2 \pi$

#### Explanation:

Since $8 x$, $3$, and $2 \sin x$ are all continuous functions for all real numbers, we don't have to worry about checking for undefined points or asymptotes. So, we can just take the second derivative and set it equal to zero.

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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:

$f \left(x\right) = 8 x + 3 - 2 \sin x$

$f ' \left(x\right) = 8 - 2 \cos x$

$f ' ' \left(x\right) = 2 \sin x$

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

$f ' ' \left(x\right) = 0$

$2 \sin x = 0$

$\sin x = 0$

On the closed interval $\left[0 , 2 \pi\right]$, there are 3 points where $\sin x = 0$:

$x \in \left\{0 , \pi , 2 \pi\right\}$

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

$f ' ' \left(\frac{\pi}{2}\right) = 2 \sin \left(\frac{\pi}{2}\right) = \textcolor{\lim e g r e e n}{2}$

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 $2 \pi$ the concavity is negative. Therefore, $\pi$ is an inflection point.

Between $\pi$ and $2 \pi$ the concavity is negative, and above $2 \pi$ the concavity is positive. Therefore, $2 \pi$ is an inflection point.

So our three points of inflection are $0$, $\pi$, and $2 \pi$.