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

Answers (1)

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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