# What are the points of inflection of f(x)= 3sin2x - 2xcosx on x in [0, 2pi] ?

Dec 9, 2017

x $\approx \left\{0 , 1.381 , 2.931 , 4.831\right\}$

#### Explanation:

Begin by finding the first and second derivatives

$f \left(x\right) = 3 \sin \left(2 x\right) - 2 x \cos \left(x\right)$
$f ' \left(x\right) = 2 x \sin \left(x\right) + 6 \cos \left(2 x\right) - 2 \cos \left(x\right)$
$f ' ' \left(x\right) = 2 x \cos \left(x\right) + 4 \sin \left(x\right) - 12 \sin \left(2 x\right)$

Next go ahead and set $f ' ' \left(x\right)$ equal to zero

$2 x \cos \left(x\right) + 4 \sin \left(x\right) - 24 \sin \left(x\right) \cos \left(x\right) = 0$ (However I am unable to solve this equation for y = 0, expect for x = 0, if it is solvable at all)

Looking at the graphs of $f \left(x\right)$, $f ' \left(x\right)$, and $f ' ' \left(x\right)$ we can make estimations of the zeros on the interval $\left[0 , 2 \pi\right]$

$f \left(x\right)$
graph{3sin(2x)-2xcos(x) [0, 6.2831853071795864769, -5, 5]}
$f ' \left(x\right)$
graph{ 2xsin(x)+6cos(2x)-2cos(x) [0, 6.2831853071795864769, -5, 5]}
$f ' ' \left(x\right)$
graph{2xcos(x) + 4sin(x) - 12sin(2x) [0,6.2831853071795864769, -5, 5]}

For x $\approx \left\{0 , 1.381 , 2.931 , 4.831\right\}$ we see that $f ' ' \left(x\right)$ is zero and that $f ' \left(x\right)$ has maximums and minimums as well as $f \left(x\right)$ appears to be changing concavity.