find global maxima for f(x)=sinx.sin2x (+infinity,-infinity)?

Answer 1

SINCE THIS IS A PEROIDIC FUNC. with peroid = 2pie
THEREFORE global maxima lies in peroid (0,2pi)
f'(x)=2sinxcos2x + sin2xcosx
=2sinx(1-2cos^2x) + 2sinxcos^2(x)
=4sinx - 6sin^3x
for f(x) to attain maxima f'(x)=0
therefore f'(x)=4sinx- 6sin^3x=0
it comes out sinx=0,+-(2/3)^1/2
at x=0 it is minima whereas at other pts it is maxima
hence maxima is at x=sin^-1(+-k) k=(2/3)^(1/2)
sin2x=(1-cos^2(2x))
=(1-(1-2sin^2x)^2)= 2*k
therefore global maxima is =2*k*k=8/9

Answer 2

For f(x) = sin(x) the global maxima is +1 (and the minima is -1) since those are the greatest (and least) value that sin(x) assumes for -∞ < x < +∞. And it's the same for
f(x) = sin(2x). But, if it had been f(x) = 2*sin(x) then the maxima and minima would have been +2 and -2.

Hope that helps.

Doug

EDIT: Hy, I think they're looking at two different functions to see that the global maxima and minima are controlled by the function (in this case sin) rather than the value of the independent variable.

Or maybe I'm seeing it wrong and the function really is
f(x) = sin(x)*sin(2x) in which case you have the correct approach ☺

Doug

Answer 3

I don't think Doug is right.

f(x) is an odd function, and so its global minimum value is the negative of its global maximum value.

Differentiating gives
cosx sin2x + 2sinxcos2x, which is zero when
2sinx cos2x = - cosx sin2x
Dividing both sides by cosx cos2x gives
2tanx = -tan2x

Let t stand for tan x, and use the expansion for tan 2x, and this becomes
2t = -2t/(1-t^2)

t=0 is a solution of this, others are given by
-1/(1-t^2) = 1, which gives us
t = + or - sqrt(2)

The values of x which correspond to t = 0 (x = n*pi) all give
f(x) = 0

f"(x) = cosx cos2x ( 4 - 5tan x tan2x).
When tanx = sqrt(2), and tan 2x = -2sqrt(2). x is in either the first or the third quadrant. In each case cosx cos2x is negative, and
4 - 5tan x tan2x = 4 + 5*2*2, and so
f"(x) < 0 . Hence this gives a maximum value.
The value is 4/(3*sqrt(3)), and the minimum value is
-4/(3*sqrt(3))