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

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Answer 1 · 2016-08-29T21:32:44

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

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.

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

enter image source here