}\], The values of the function at the endpoints are, \[{f\left( 4 \right) = \frac{{4 – 1}}{{4 – 3}} = 3,}\;\;\;\kern-0.3pt{f\left( 5 \right) = \frac{{5 – 1}}{{5 – 3}} = 2. This discussion on In [0,1] Lagranges Mean Value theorem is NOT applicable toa)b)c)f (x) = x|x|d)f (x) =|x|Correct answer is option 'A'. 1 answer. It states that if f (x) is a defined function which is continuous on the interval [a,b] and differentiable on (a,b), then there is at least one point c in the interval (a,b) (that is a