Question

What are the points of inflection of `f(x)=x^3sin^2x` on the interval `x in [0,2pi]`?

Answer 1

The successive graphs reveal POI at x = 0, +-1.24, +-2.55, +-3.68, +-4.82, ....

Explanation:

`f=1/2x^3(1-cos 2x)`, giving x-intercepts `0, 0, +-pi, +-2pi, +-3pi, ...`

`f''=1/2((x^3)''-(x^3cos2x)'')`

`=x(3-(3cos2x-6xsin2x-2x^2cos2x))`

Excepting x = 0, the other zeros of f'', in infinitude, cannot be

obtained in mathematical exactitude.

Yet, we can locate them in pairs in between

`(kpi. (k+1)pi) and (-(k+1)pi, -kpi), k = 1, 2, 3, ---`, respectively.

There are 4 POI, for `x in (0. 2pi)`.

They are nearly 1.24, 2.55, 3.58 and 4.82, nearly..

Some locations near O have been obtained by graphical method,

using appropriate sub domains and ranges.

graph{x^3sinxsinx [-20, 20, -10, 10]}

Graph 1 : Graph is symmetrical about O.

graph{x^3sinxsinx [0, 6.28, -50, 300]}
Graph 2 : A revelation of four POI and the power-growth of local maxima of y

graph{x^3sinxsinx [1.24 1.26, -10, 10]}
Graph 3 : locates a POI, near 1.24.

graph{x^3sinxsinx [2.54, 2.56,-10, 10]}
Graph 4 : Locates a POII, near 2.55.
graph{x^3sinxsinx [3.56 3.59,-10, 10]}

graph{x^3sinxsinx [4.81 4.83, 0, 200]}