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

May 29, 2017

$x = \left\{0 , 0.76 , 1.997 , 3.416 , 4.91\right\}$

#### Explanation:

Point of inflection appears where a curve changes from concave to convex or vice versa and at this point second derivative of curve is $0$.

As $f \left(x\right) = 8 {x}^{2} \sin \left(2 x - \pi\right) = - 8 {x}^{2} \sin 2 x$

and $\frac{\mathrm{df}}{\mathrm{dx}} = - 16 x \sin 2 x - 16 {x}^{2} \cos 2 x$

and $\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = - 16 \sin 2 x - 32 x \cos 2 x - 32 x \cos 2 x + 32 {x}^{2} \sin 2 x$

= $- 16 \sin 2 x - 64 x \cos 2 x + 32 {x}^{2} \sin 2 x$

Now $- 16 \sin 2 x - 64 x \cos 2 x + 32 {x}^{2} \sin 2 x = 0$

$\implies \sin 2 x + 4 x \cos 2 x - 2 {x}^{2} \sin 2 x = 0$

It is apparent that at $x = 0$, we have a point of inflection and other points are $x = \left\{0.76 , 1.997 , 3.416 , 4.91\right\}$ as can be seen from the graph of $\sin 2 x + 4 x \cos 2 x - 2 {x}^{2} \sin 2 x$.
graph{sin(2x)+4xcos(2x)-2x^2sin(2x) [-2.167, 7.833, -2.48, 2.52]}