# How do you find all points of inflection given y=-sin(2x)?

Oct 21, 2017

Points of inflection would occur every $\frac{\pi}{2}$.

#### Explanation:

To find points of inflection, we need to find all the points on the graph at which the second derivatives will have a value of 0:

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

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

Using chain rule:

$u = 2 x$

$\frac{d}{\mathrm{du}} \left(- \sin \left(u\right)\right) = - \cos \left(u\right)$

$\frac{\mathrm{du}}{\mathrm{dx}} = 2$

$\frac{d}{\mathrm{dx}} = \frac{d}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}} = - 2 \cos \left(u\right) = - 2 \cos \left(2 x\right) = f ' \left(x\right)$

Using same principles to differentiate again:

$u = 2 x$

$\frac{d}{\mathrm{du}} \left(- 2 \cos \left(u\right)\right) = 2 \sin \left(u\right)$

$\frac{\mathrm{du}}{\mathrm{dx}} = 2$

$\frac{d}{\mathrm{dx}} = \frac{d}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}} = 4 \sin \left(u\right) = f ' ' \left(x\right)$

$f ' ' \left(x\right) = 4 \sin \left(2 x\right)$

And now to make $f ' ' \left(x\right) = 0$:

$4 \sin \left(2 x\right) = 0$

$\sin \left(2 x\right) = 0$

$2 x = \arcsin \left(0\right) = 0 + n \pi$

$x = \frac{0 + n \pi}{2} = n \frac{\pi}{2}$