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

1 Answer
Aug 29, 2016

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

enter image source here