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 minima
But the answer key for the exercise containing the above question shows x = 2(pi) to be the point of maxima and 2(pi) to be the maximum value and x = 0 to be the point of minima with 0 as the minimum value
This 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.
You will not arrive at the answer key solutions using calculus
because the maximum and minimum values of the function for the given domain,
do not occur at the turning points.
The curve is rising at the maximum and minimum within the domain.