A Very Unique Maxima - Minima problem in Differential calculus

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 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.

amahindroo

Re: A Very Unique Maxima - Minima problem in Differential calculus

Quote:

Originally Posted by

**amahindroo** **Find the point of absolute maxima and minima for the function given as**

$\displaystyle y = x + \sin(2x),~ 0 \le x\le 2\pi$

Also find the respective maximum and minimum values.

Why not just use it as given?

Then $\displaystyle y' = 1 + 2\cos(2x)$ which has two zeros in $\displaystyle 0\le x\le 2\pi$.

Re: A Very Unique Maxima - Minima problem in Differential calculus

Quote:

Originally Posted by

**Plato** Why not just use it as given?

Then $\displaystyle y' = 1 + 2\cos(2x)$ which has two zeros in $\displaystyle 0\le x\le 2\pi$.

Well, I had tried that also, but could get the 2 points as **x = (pi)/3** and **x = 2(pi)/3**, which still is not as per the answer key given for that question.

Re: A Very Unique Maxima - Minima problem in Differential calculus

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.

Re: A Very Unique Maxima - Minima problem in Differential calculus

On a closed interval, function extrema also occur at the interval's endpoints.