Q44. If f(x) =cosh (cosh x), xe
[-1,3], then the value of c for
which Lagrange's Mean Value
Theorem is satisfied is
Answers
Answered by
0
Answer:
Clearly, f(x)=
x
2
−4
is continuous on [2, 3] and differentiable on (2, 3).
So, by Lagrange's mean value theorem, there exists c ε(2, 3) such that
f
′
(c)=
3−2
f(3)−f(2)
⇒
c
2
−4
c
=
5
−0
[∵f(x)=
x
2
−4
⇒f
′
(x)=
x
2
−4
x
]
⇒c
2
=5(c
2
−4)
⇒4c
2
=20
⇒c=
5
Step-by-step explanation:
Answered by
0
Answer:
Step-by-step explanation:
ANSWER
Clearly, f(x)=
x
2
−4
is continuous on [2, 3] and differentiable on (2, 3).
So, by Lagrange's mean value theorem, there exists c ε(2, 3) such that
f
′
(c)=
3−2
f(3)−f(2)
⇒
c
2
−4
c
=
5
−0
[∵f(x)=
x
2
−4
⇒f
′
(x)=
x
2
−4
x
]
⇒c
2
=5(c
2
−4)
⇒4c
2
=20
⇒c=
5
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