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

Answer 1

TimGNO is correct, although a pedant would point out that he hasn't shown this point is a maximum. But it is common sense. See my response to the second time you asked the question, I got the same answer (foolishly mucking around with tangents), but make a couple of other observations you may wish to look at.

Answer 2

The global maxima (and minima) for a function is defined to be the greatest (least) value taken on by the function over the range -∞ < x < +∞. Since the value of sin(x) is limited to -1 ≤ sin(x) ≤ +1 the global maxima for sin(x) is +1 (and the global minima is -1). And it's the same for
sin(2x). But, had the function been 2*sin(x) the maxima and minima would have been +2, -2.


Doug

Answer 3

The maxima of this function:

f(x) = sin(x)sin(2x)

can be found by solving the traditional first derivative set to 0. Of course, sine being periodic results in such maxima being non-unique. We merely seek to characterize them in some way.

We wish to solve:

f'(x) = cos(x)sin(2x) + 2sin(x)cos(2x) = 0

Invoking the double angle identities, this can be simplified:

0 = f'(x) = cos(x)[2cos(x)sin(x)] + 2sin(x)[2cos^2(x) - 1]
= 2cos^2(x)sin(x) + 4cos^2(x)sin(x) - 2sin(x)

Thus,
sin(x) = 3cos^2(x)sin(x)

It is understood that sin(x) does not equal 0 at our points of interest, so we divide both sides:

1/3 = cos^2(x)
sqrt (1/3) = cos(x)

This yields the solution(s) x = 0.95 (+2npi) radians. The function f evaluated at these points yields a maximum value of approximately 0.816, if my calculations are correct.

Hope this helps!