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

Apr 30, 2018

None. No points of inflection.

#### Explanation:

Points of inflection are slope extrema or $f ' \left(x\right)$ extrema. To find a maximum or minimum, take the second derivative and set equal to zero, to find where change in slope is zero.

$f \left(x\right) = 8 {x}^{2} + \sin \left(2 x - \pi\right)$
$f ' \left(x\right) = 16 x + 2 \cos \left(2 x - \pi\right)$
$f ' ' \left(x\right) = 16 - 4 \sin \left(2 x - \pi\right)$

Now set $f ' ' \left(x\right) = 0$:

$0 = 16 - 4 \sin \left(2 x - \pi\right)$
$- 16 = - 4 \sin \left(2 x - \pi\right)$
$4 = \sin \left(2 x - \pi\right)$
${\sin}^{-} 1 \left(4\right) = 2 x - \pi$
$\to$NO SOLUTION

There are no (real, I dunno about complex) solutions as sin(x) cannot be greater than 1.

Therefore there are no points of inflection.