Question

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

Answer 1

`x={0,0.76,1.997,3.416,4.91}`

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(x)=8x^2sin(2x-pi)=-8x^2sin2x`

and `(df)/(dx)=-16xsin2x-16x^2cos2x`

and `(d^2f)/(dx^2)=-16sin2x-32xcos2x-32xcos2x+32x^2sin2x`

= `-16sin2x-64xcos2x+32x^2sin2x`

Now `-16sin2x-64xcos2x+32x^2sin2x=0`

`=>sin2x+4xcos2x-2x^2sin2x=0`

It is apparent that at `x=0`, we have a point of inflection and other points are `x={0.76,1.997,3.416,4.91}` as can be seen from the graph of `sin2x+4xcos2x-2x^2sin2x`.
graph{sin(2x)+4xcos(2x)-2x^2sin(2x) [-2.167, 7.833, -2.48, 2.52]}