Solution We have
f (x) = sin 3x
or
f ′(x) = 3cos 3x
Therefore, f′(x) = 0 gives cos 3x = 0 which in turn gives 
3
3
2,
2
x
π
π
=
 (as x ∈

0 2
, π
implies 
3
3
0, 2
x
π


∈ 



) So 
6
x
=π
 and 2
π The point 
6
x
=π
 divides the interval 0 2
, π


into two disjoint intervals 0, 6
π



  and π π
6 2
,


 Now, 
( )
0
f
′x >
 for all 
0, 6
x
π


∈


  as 0
0
3
6
2
x
x
π
π
≤
<
⇒
≤
<
 and 
( )
0
f
′x <
 for
all 
,
6 2
x
π π

∈


  as 
3
3
6
2
2
2
x
x
π
π
π
π
<
<
⇒
<
<