Find the maximum and minimum values of x + sin 2x on [0,2π].
Answers
Answered by
0
Answer:
Let
f(x)=x+sin2x
f
′
(x)=1+2cos2x
For critical points,
1+2cos2x=0
Or
2cos2x=−1
cos2x=
2
−1
2x=
3
2π
,
3
4π
x=
3
π
,
3
2π
Now
f(
3
π
)=
3
π
+sin(
3
2π
)
=
3
π
+
2
3
...(i)
f(
3
2π
)=
3
2π
+sin(
3
4π
)
=
3
2π
−
2
3
...(ii)
Hence
Maximum value is
3
π
+
2
3
and Minimum Value is
3
2π
−
2
3
Similar questions