Question

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

Answer 1

Given function is,

f( x )=x+sin2x

Differentiate the function with respect to x,

f ′ ( x )=1+2cos2x

Put f ′ ( x )=0,

1+2cos2x=0 2cos2x=−1 cos2x=− 1 2 cos2x=cos 2π 3

Further simplify,

2x=2nπ± 2π 3 x=nπ± π 3 x= π 3 , 2π 3 , 4π 3 , 5π 3

The value of the function is calculated by substituting the critical point and the end points of the interval.

At x=0,

f( 0 )=0+sin0 =0

At x= π 3 ,

f( π 3 )= π 3 +sin 2π 3 = π 3 + 3 2

At x= 2π 3 ,

f( 2π 3 )= 2π 3 +sin 4π 3 = 2π 3 − 3 2

At x= 4π 3 ,

f( 4π 3 )= 4π 3 +sin 8π 3 = 4π 3 + 3 2

At x= 5π 3 ,

f( 5π 3 )= 5π 3 +sin 10π 3 = 5π 3 − 3 2

At x=2π,

f( 2π )=2π+sin4π =2π

Therefore, the given function f( x )=x+sin2x is maximum at x=2π with the maximum value of the function 2π and the function is minimum at x=0 with the minimum value of the function 0.