# What are the points of inflection of f(x)=x+cosx  on the interval x in [0,2pi]?

Oct 1, 2016

A point of inflection is a point in which the function switches from being concave to convex, or vice versa. Our tool to verify if a function is concave or convex is the second derivative, more precisely its sign: if $f ' ' \left(x\right) > 0$ then $f$ is concave in $x$, otherwise it's convex.

So, look for the convex/concave switch is the same as looking for the positive/negative switch for the second derivative.

So, first of all, let's compute it: since the derivative of a sum is the sum of the derivatives, we have

$f \left(x\right) = x + \cos \left(x\right)$

$f ' \left(x\right) = 1 - \sin \left(x\right)$

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

And since we need to refer to the $\left[0 , 2 \pi\right]$ interval, the cosine is positive in $\left[0 , \frac{\pi}{2}\right]$, negative in $\left[\frac{\pi}{2} , \frac{3 \pi}{2}\right]$, and again positive in $\left[\frac{3 \pi}{2} , 2 \pi\right]$.

The second derivative is $- \cos \left(x\right)$, so it will change negative areas with positive ones, but the switch points will still be $\frac{\pi}{2}$ and $\frac{3 \pi}{2}$