Question

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​

Answer 1

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:

Answer 2

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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