verify rolle's theorem for |cosx| on [0,pi]
Answers
Step-by-step explanation:
We know that
(i) f (x) = cos x is a trigonometric function which is continuous
Hence,
f (x) = cos x is continuous on [-π/2, π/2]
(ii) f’(x) = – sin x exist in [-π/2, π/2] Hence,
f (x) = cos x is differentiable on (-π/2, π/2)
(iii) We know that f (-π/2) = cos (-π/2) = 0
Similarly f (π/2) = cos (π/2) = 0
Here f (-π/2) = f (π/2) The conditions of Rolle’s Theorem are satisfied.
There exist at least one c ϵ (-π/2, π/2) where f’(c) = 0 – sin c = 0
Which gives c = 0 We know that value of c = 0 ϵ (-π/2, π/2)
Therefore, Rolle’s Theorem is satisfied.
Answer: Verifyrolle
′
stheoremforwemayress
itf(x)={
cosx0x
2
π
−cosx
2
π
<xπ
clearly,f(0)=cos0
∘
=1
⇒f(π)=−cosπ=1
∴f(0)=f(π)=1
⇒Rf
′
(
2
π
+h)=lim
h→0
f
h
(
2
π
+h)−f(
2
π
)
=lim
h→0
h
∣−sin1muh∣−0
=1
⇒Lf
′
(
2
π
)=lim
h→0
=
−h
f(
2
π
−h)−f(
2
π
)
=lim
h→0
=
−h
∣sin1mu1muh∣−0
=lim
h→0
=
h
−sin1mu1muh
=−1
∴Rf
′
(o)
=Lf
′
(o)
Step-by-step explanation: Hope it helps you please mark me as brainliest.