Question

Find maxima of the following expression using calculus `(2x-sin2x)/x^2` ?

Answer 1

`x = -pi/2` minimum point
`x = pi/2` maximum point

Explanation:

Given `f(x) = (2 x - Sin(2 x))/x^2`

the stationary points are determined solving for `x` the condition

`d/(dx)f(x) = (4 cosx (sinx-xcos x))/x^3 = 0`

or

`{(cosx = 0),(tanx-x=0):}`

with solutions

`x = 0 uu pmpi/2+2kpi, k = 0,1,2,3,cdots`

Here `x=0` is the main solution of `tanx = x`

Now testing in the range `-2pi le x le 2pi` in

`d^2/(dx^2)f(x) = (2 (2 x + 4 x Cos(2 x) - 3 Sin(2 x) + 2 x^2 Sin(2 x)))/x^4`

we will obtain

`lim_(x->0)d^2/(dx^2)f(x)=0` inflexion point

and for `k = 0`

`d^2/(dx^2)f(-pi/2) = 32/pi^3` minimum point
`d^2/(dx^2)f(pi/2) = -32/pi^3` maximum point

there are infinite minima and maxima for `k=1,2,cdots` but their range is contained inside the former two for `k=0`.

Note:

We have obviate the limit determination for `x->0` in all the process steps.

For qualification justification see
https://socratic.org/questions/if-f-x-x-8-9-x-4-7-x-3-7-what-are-the-local-minima-and-maxima-of-f-x#300548

Attached the plot of `f(x)` for `-2pi le x le 2pi`