What are the points of inflection of f(x)=8x2sin(2x−π) on x∈[0,2π]?

1 Answer
Shwetank Mauria
May 29, 2017

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)=8x2sin(2x−π)=−8x2sin2x

and dfdx=−16xsin2x−16x2cos2x

and d2fdx2=−16sin2x−32xcos2x−32xcos2x+32x2sin2x

= −16sin2x−64xcos2x+32x2sin2x

Now −16sin2x−64xcos2x+32x2sin2x=0

⇒sin2x+4xcos2x−2x2sin2x=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−2x2sin2x.
graph{sin(2x)+4xcos(2x)-2x^2sin(2x) [-2.167, 7.833, -2.48, 2.52]}

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