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Examine the applicability of Mean Value Theorem for all three functions given in the above exercise 2.

 Q6 Examine the applicability of Mean Value Theorem for all three functions given in
       the above exercise 2.

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According to Mean value theorem function 
f:[a,b]\rightarrow R must be
a )  continuous in given closed interval say [a,b] 
b ) differentiable in given open interval say (a,b)
Then their exist a c \ \epsilon \ (x,y)  such that 
 f^{'}(c)= \frac{f(b)-f(a)}{b-a}
If all these conditions are satisfies then we can verify  mean value theorem 
Given function is 
 f (x) = [x] 
It is clear that Given function f (x) = [x] is not  continuous for each and every point in [5,9]
Now, lets check differentiability of f(x)
L.H.L. at x = n , n \ \epsilon \ [5,9]
\lim_{h\rightarrow 0^-}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^-}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^-}\frac{n-1-n}{h} = \lim_{h\rightarrow 0^-}\frac{-1}{h}= -\infty
                                                                                                                                ( [n+h]=n-1 \because h < 0 \ therefore \ (n+h)<n)
Now,
R.H.L. at x = n , n \ \epsilon \ [5,9]
\lim_{h\rightarrow 0^+}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^+}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^+}\frac{n-n}{h} = \lim_{h\rightarrow 0^-}\frac{0}{h}=0
                                                                                                                                   ( [n+h]=n \because h > 0 \ therefore \ (n+h)>n)
We can clearly see that R.H.L. is not equal to L.H.L.
Therefore, function is not differential in (5,9)
Hence, Mean value theorem is not applicable for given function   f (x) = [x]    ,  x \ \epsilon \ [5,9]

Similaly,
Given function is 
 f (x) = [x] 
It is clear that Given function f (x) = [x] is not  continuous for each and every point in [-2,2]
Now, lets check differentiability of f(x)
L.H.L. at x = n , n \ \epsilon \ [-2,2]
\lim_{h\rightarrow 0^-}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^-}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^-}\frac{n-1-n}{h} = \lim_{h\rightarrow 0^-}\frac{-1}{h}= -\infty
                                                                                                                                ( [n+h]=n-1 \because h < 0 \ therefore \ (n+h)<n)
Now,
R.H.L. at x = n , n \ \epsilon \ [-2,2]
\lim_{h\rightarrow 0^+}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^+}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^+}\frac{n-n}{h} = \lim_{h\rightarrow 0^-}\frac{0}{h}=0
                                                                                                                                   ( [n+h]=n \because h > 0 \ therefore \ (n+h)>n)
We can clearly see that R.H.L. is not equal to L.H.L.
Therefore, function is not differential in (-2,2)
Hence, Mean value  theorem is not applicable for given function   f (x) = [x]    ,  x \ \epsilon \ [-2,2]

Similarly,
Given function is 
 f (x) = x^2-1 
Now, being a polynomial , function f (x) = x^2-1 is continuous in [1,2] and differentiable in(1,2)
Now,
f(1)=1^2-1 = 1-1 = 0
And
f(2)=2^2-1 = 4-1 = 3
Now,
f^{'}(c)= \frac{f(b)-f(a)}{b-a}
f^{'}(c)= \frac{f(2)-f(1)}{2-1}\\ f^{'}(c)=\frac{3-0}{1}\\ f^{'}(c)= 3
Now,
f^{'}(x)= 2x\\ f^{'}(c)=2c\\ 3=2c\\ c=\frac{3}{2}
And c=\frac{3}{2} \ \epsilon \ (1,2)
Therefore, mean value theorem is applicable for the function f (x) = x^2-1
 

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