Q4     Verify Mean Value Theorem, if f (x) = x^2 - 4x - 3in the interval [a, b], where
          a = 1 and b = 4.

Answers (1)

Condition for M.V.T.
If f ; [ a ,b] \rightarrow R
a ) f is continuous in [a,b]
b ) f is differentiable in (a,b)
Then, there exist a c  in (a,b) such that
f^{'}(c) = \frac{f(b)-f(a)}{b-a}
It is given that
 f (x) = x^2 - 4x - 3  and interval is [1,4]
Now, f is a polynomial function , f (x) = x^2 - 4x - 3 is continuous in[1,4] and  differentiable in (1,4) 
And
f(1)= 1^2-4(1)-3= 1-7= -6
and
f(4)= 4^2-4(4)-3= 16-16-3= 16-19=-3
Then, by Mean value theorem we can say that their exist a c  in (1,4) such that
f^{'}(c) = \frac{f(b)-f(a)}{b-a}
f^{'}(c) = \frac{f(4)-f(1)}{4-1}\\ f^{'}(c)= \frac{-3-(-6)}{3}\\ f^{'}(c)= \frac{3}{3}\\ f^{'}(c)= 1
Now,
f^{'}(x) =2x-4\\ f^{'}(c)-2c-4\\ 1=2c-4\\ 2c=5\\ c=\frac{5}{2}
And c=\frac{5}{2} \ \epsilon \ (1,4)
Hence, mean value theorem is verified for the function f (x) = x^2 - 4x - 3

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