Question

Find the maximum and minimum values of x + sin 2x on [0, 2π].

Answer 1

Let f (x) = x + sin2x, 0 ≤ x ≤ 2π

⇒ f' (x) = 1 + 2cos 2x

⇒ For critical points, let f' (x) = 0

⇒ 1 + cos 2x = 0

⇒ `cos 2x = -1/2`

⇒ `cos 2x = -cos  pi/3`

(If 0< x < 2π, then 0< 2x < 4π)

`⇒ cos 2x = cos (pi- pi/3), cos (pi + pi/2), cos (3pi - pi/3), cos (3pi + pi/3)`

⇒ `2x = (2pi)/3 , (4pi)/3, (8pi)/3, (10pi)/3`

⇒ `x = pi/3, (2pi)/3, (4pi)/3, (5pi)/3`

So, for finding maximum and minimum, we evaluate f (x) at `0, 2pi , pi/3, (2pi)/3, (4pi)/3, (5pi)/3`

Now f(0) = 0 + sin 0 = 

f (2π) = 2π + sin 4π = 2π + 0 = 2π

`f (pi/3) = pi/3 + sin  (2pi)/3 = pi/3 + sin (pi - pi/3)`

= `pi/3 + sin  pi/3 = pi/3 + sqrt3/2`

`f ((2pi)/3) = (2pi)/3 + sin  (4pi)/3 = (2pi)/3 + sin (pi + pi/3)`

= `(2pi)/3 -sin  pi/3 = (2pi)/3 - sqrt3/2`

`f((4pi)/3) = (4pi)/3 + sin  (8pi)/3 = (4pi)/3 + sin (2pi + (2pi)/3)`

= `(4pi)/3 + sin  (2pi)/3 = (4pi)/3 + sqrt3/2`

and `f ((5pi)/3) = (5pi)/3 + sin  (10pi)/3 = (5pi)/3 + sin (3pi + pi/3)`

= `(5pi)/3 -sin  pi/3 = (5pi)/3 - sqrt3/2`

Thus, maximum value of f (x) = 2π at x = 2π and minimum value of f (x) = 0 at x = 0.