Question

Verify LMVT (Lagrange’s mean value theorem) for the function $f\left( x \right) = \log x,{\text{ }}x \in \left[ {1,e} \right]$.

Answer 1

Hint: Use the concept that log x is both differentiable as well as continuous in the interval [1, e], so according to Lagrange’s mean value theorem there exists a point c such that $f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$, where b=e and a=1.

Complete Step-by-Step solution:
Lagrange’s mean value theorem (LMVT) states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x = c on this interval, such that
$f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$
Now given function is
$f\left( x \right) = \log x,{\text{ }}x \in \left[ {1,e} \right]$
The graph of log x is shown above which is true for $x \in \left[ {1,e} \right]$
Now as we know that log x is differentiable as well as continuous in the interval [1, e] so there exists a point x = c such that
$f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$ ....................... (1) where, (a = 1, b = e)
Now differentiate f(x) we have,
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = \dfrac{d}{{dx}}\log x = \dfrac{1}{x}$
$ \Rightarrow f'\left( x \right) = \dfrac{1}{x}$
Now in place of x substitute (c) we have,
$ \Rightarrow f'\left( c \right) = \dfrac{1}{c}$
Now from equation (1) we have,
$ \Rightarrow \dfrac{1}{c} = \dfrac{{\log e - \log 1}}{{e - 1}}$
Now as we know the value of log e is 1 and the value of log 1 is zero so we have,
$ \Rightarrow \dfrac{1}{c} = \dfrac{{1 - 0}}{{e - 1}}$
$ \Rightarrow c = e - 1$
Hence c is belongs between (1, e)
Hence LMVT is verified.

Note: If a function is continuous at some points then it may or may not be differentiable at those points, but if a function is differentiable at some points that we can say with certainty that it has to be continuous. That is differentiability is a sure condition for continuity however converse is not true. These tricks help commenting upon continuity and differentiability while solving problems of such kind.