Question

What are the points of inflection, if any, of `f(x)=xsin^2x` on [0,2pi]?

Answer 1

points corresponding to
`f''(x)=0,`
between
`x=0.7 and 0.8`
`x=1. 1and 1.2`
`x=2.3 and 2.4`
`x= 2.5 and 2.6`
`x=3.0 and 3.4`
`x=4.0 and 4.1`
`x=5.4 and 5.5`
`x=5.5 and 5.6`
correct to one decimal place

`x=pi/2`
and
`x=3pi/2`

Explanation:

Given: `f(x)=xsin^2x`
Evaluating the curvature,
`f'(x)=x(2sinx)cosx+sin^2x`
=`2xsinxcosx+sin^2x`
=`xsin2x+sin^2x`
Since, `2sinxcosx=sin2x`
Thus, `f'(x)=xsin2x+sin^2x`
`f''(x)=xcos2x(2)+sin2x(1)+2sinxcosx`
=`2xcos2x+sin2x+sin2x`
Now, `f''(x)=2xcos2x+2sin2x`
`f''(x)=0`
implies
`2xcos2x+2sin2x=0`
Taking `2cos2x` common
`cos2x(x+tan2x)=0`
Either `cos2z=0`
where, x is clearly `pi/2 and 3pi/2`
or
`x+tan2x=0`
By trial and error,

Inspecting for values between 0 and 2pi in the intervals of 0.1pi
At `x=0, f''(x)=0`
`f''(x)=0,`
between
`x=0.7 and 0.8`
`x=1. 1and 1.2`
`x=2.3 and 2.4`
`x= 2.5 and 2.6`
`x=3.0 and 3.4`
`x=4.0 and 4.1`
`x=5.4 and 5.5`
`x=5.5 and 5.6`
correct to one decimal place
MY thanks for Fleur for reminding