 Hi and welcome to the session. I am Shashi and I am going to help you with the following question. The question says, find the maximum and minimum values of x plus sin 2x on closed interval 0 to pi. Now let us start the solution. Function f is given by fx equal to x plus sin 2x. Now differentiating both sides with respect to x we get f dash x equal to 1 plus 2 cos 2x. We know derivative of x is 1 and derivative of sin 2x is 2 cos 2x. So we get f dash x equal to 1 plus 2 cos 2x. Now first of all we will find all the critical points of the function in the given interval 0 to pi. So for finding critical points we will put f dash x equal to 0. Now we know f dash x is equal to 1 plus 2 cos 2x. So we can write 1 plus 2 cos 2x equal to 0. Now subtracting 1 from both sides we get 2 cos 2x equal to minus 1. Dividing both sides by 2 we get cos 2x equal to minus 1 upon 2. Now we know this is possible only when 2x is equal to 2 pi upon 3, 4 pi upon 3, 8 pi upon 3 and 10 pi upon 3. We know cos 2 pi upon 3 is equal to cos 4 pi upon 3 is equal to cos 8 pi upon 3 is equal to cos 10 pi upon 3 is equal to minus 1 upon 2. Now this implies x is equal to pi upon 3 or x is equal to 2 pi upon 3 or x is equal to 4 pi upon 3 or x is equal to 5 pi upon 3. Really we can see all these values of x lie in the closed interval 0 to pi. So we have neglected all the values of x at which the given function f is not defined. Now these are the critical points of the function f in the interval 0 to pi. Now we will find the value of function f at all these critical points and end points of the given interval that is 0 and 2 pi. So we will find value of function f at x equal to pi upon 3, x equal to 2 pi upon 3, x equal to 4 pi upon 3, x equal to 5 pi upon 3 and x equal to 0, x equal to 2 pi. First of all let us find out x f0 that is equal to 0 plus sin 2 multiplied by 0 this is equal to 0 plus sin 0. We know 2 multiplied by 0 is 0. Now we know sin 0 is equal to 0. So we get f0 has 0. Now let us find out x pi upon 3 this is equal to pi upon 3 plus sin 2 pi upon 3 this is equal to pi upon 3 plus root 3 upon 2. We know sin 2 pi upon 3 is equal to root 3 upon 2. Now let us find out value of function f at x equal to 2 pi upon 3. Now f 2 pi upon 3 is equal to 2 pi upon 3 plus sin 2 multiplied by 2 pi upon 3 this is equal to 2 pi upon 3 plus sin 4 pi upon 3. Now we know sin 4 pi upon 3 is equal to minus root 3 upon 2. So this is equal to 2 pi upon 3 minus root 3 upon 2. So we get f 2 pi upon 3 as 2 pi upon 3 minus root 3 upon 2. Now let us find out f 4 pi upon 3 this is equal to 4 pi upon 3 plus sin 2 multiplied by 4 pi upon 3 which is equal to 4 pi upon 3 plus sin 8 pi upon 3. Now this is equal to 4 pi upon 3 plus root 3 upon 2. We know sin 8 pi upon 3 is equal to root 3 upon 2. Now let us find out value of function f at x equal to 5 pi upon 3. So f 5 pi upon 3 is equal to 5 pi upon 3 plus sin 2 multiplied by 5 pi upon 3. This is equal to 5 pi upon 3 plus sin 10 pi upon 3. Now we know sin 10 pi upon 3 is equal to minus root 3 upon 2. So we get f 5 pi upon 3 as 5 pi upon 3 minus root 3 upon 2. Now we will find the value of function f to take as equal to 2 pi. So f 2 pi is equal to 2 pi plus sin 2 multiplied by 2 pi. This is equal to 2 pi plus sin 4 pi. Now we know sin 4 pi is equal to 0. So we can write 2 pi plus 0. So we get f 2 pi equal to 2 pi. We will identify the maximum and minimum value of function f. Clearly we can see minimum value of function f is equal to 0 which occurs at x equal to 0. And maximum value of function f is equal to 2 pi which occurs at x equal to 2 pi. So we can write minimum value of function x plus sin 2 x is 0 occurring at x equal to 0. And maximum value of x plus sin 2 x is 2 pi occurring at x equal to 2 pi. So this is our required answer. This completes the session. Hope you understood the session. Take care and have a nice day.