Question

What are the points of inflection of `f(x)= 3sin2x - 2xcosx` on #x in [0, 2pi] #?

Answer 1

x `approx {0, 1.381, 2.931, 4.831}`

Explanation:

Begin by finding the first and second derivatives

`f(x) = 3sin(2x)-2xcos(x)`
`f'(x) = 2xsin(x)+6cos(2x)-2cos(x)`
`f''(x) = 2xcos(x) + 4sin(x) - 12sin(2x)`

Next go ahead and set `f''(x)` equal to zero

`2xcos(x) + 4sin(x) - 24sin(x)cos(x) = 0` (However I am unable to solve this equation for y = 0, expect for x = 0, if it is solvable at all)

Looking at the graphs of `f(x)`, `f'(x)`, and `f''(x)` we can make estimations of the zeros on the interval `[0, 2pi]`

`f(x)`
graph{3sin(2x)-2xcos(x) [0, 6.2831853071795864769, -5, 5]}
`f'(x)`
graph{ 2xsin(x)+6cos(2x)-2cos(x) [0, 6.2831853071795864769, -5, 5]}
`f''(x)`
graph{2xcos(x) + 4sin(x) - 12sin(2x) [0,6.2831853071795864769, -5, 5]}

For x `approx {0, 1.381, 2.931, 4.831}` we see that `f''(x)` is zero and that `f'(x)` has maximums and minimums as well as `f(x)` appears to be changing concavity.