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

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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