Hi All,

I have a very unique maxima-minima problem in differential calculus:-

Find the point of absolute maxima and minima for the function given as :-

y = x + sin(2x), 0 <= x <= 2(pi)

Also find the respective maximum and minimum values.

It seems to be a very simple problem which can be easily solved by the Second Derivative test. the steps that I have already carried out are:-

y = x + sin(2x) = x + 2sin(x)cos(x)

dy/dx = 1 + 2 (cos^2(x) - sin^2(x)) = 1 + 2(2cos^2(x) - 1) = 4cos^2(x) - 1

For y to be maximum or minimum, dy/dx = 0

4cos^2(x) - 1 = 0

4cos^2(x) = 1

cos^2(x) = (1/4)

cos(x) = + (1/2) or - (1/2)

x = (pi)/3, (5(pi))/3 or x = (2(pi))/3, (4(pi))/3

d2y/dx2 = -8cos(x)sin(x)

So, d2y/dx2 at x=(pi)/3 is = -8X(+ve)X(+ve) < 0

So, d2y/dx2 at x=5(pi)/3 is = -8X(+ve)X(-ve) > 0

So, d2y/dx2 at x=2(pi)/3 is = -8X(-ve)X(+ve) > 0

So, d2y/dx2 at x=4(pi)/3 is = -8X(-ve)X(-ve) < 0

So, according to this, x = (pi)/3 and 4(pi)/3 needs to be the points of local maxima and one of them to be the absolute maxima

Similarly,x = 2(pi)/3 and 5(pi)/3 needs to be the points of local minima and one of them to be the absolute minimaBut the answer key for the exercise containing the above question shows

x = 2(pi)to be the point of maximaand2(pi) to be the maximum valueandx = 0to be thepoint of minimawith0 as the minimum valueThis answer (

x = 2(pi) and x = 0) if we put the values in the function y = x + sin(2x) seems to be the logical solution but we are not able to arrive at it using derivatives.

Can anyone help me find the solution to this problem? or is this a limitation of finding a solution using derivatives?

Please let me know.

amahindroo