F(x)=4sinx-2x-xcosx/2+cosx
Answers
Answer:
f given by f(x)=4sinx−2x−xcosx2+cosxis increasing decreasing, x∈(0,2π)
Step-by-step explanation:f(x)=
2+cosx
4sinx−2x−xcosx
=
2+cosx
4sinx
−
2+cosx
2x+xcosx
=
2+cosx
4sinx
−
(2+cosx)
x(2+cosx)
=
2+cosx
4sinx
−x
Now let's find out f
′
(x)
f
′
(x)=
(2+cosx)
2
4cosx(2+cosx)+sinx(4sinx)
−1
=
(2+cosx)
2
8cosx+4(cos
2
x+sin
2
x)
−1
=
(2+cosx)
2
8cosx+4−(2+cosx)
2
=
(2+cosx)
2
8cosx+4−4−4cosx−cos
2
x
=
(2+cosx)
2
cosx(4−cosx)
Now,(2+cosx)
2
>0∀xϵR
Similarly,4−cosx>0∀xϵR
as cosxϵ[−1,1]
⇒4−cosxϵ[3,5]>0
∴ the function f(x) will be increasing if f
′
(x)>0⇒cosx>0 and decreasing if f
′
(x)<0⇒cosx<0
(i) Increasing
f
′
(x)>0
⇒cosx>0
∴xϵ(0,
4
π
)∪(
4
3π
,2π)
(ii) Decreasing
f
′
(x)<0
⇒cosx<0
∴xϵ(
4
π
,π)∪(π,
2
3π
)