Question

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

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 $f^{\prime }\left(c\right)={\displaystyle \frac{f\left(b\right)-f\left(a\right)}{b-a}}$, 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
$f^{\prime }\left(c\right)={\displaystyle \frac{f\left(b\right)-f\left(a\right)}{b-a}}$
Now given function is
$f\left(x\right)=\log x,x\in \left[1,e\right]$
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
$f^{\prime }\left(c\right)={\displaystyle \frac{f\left(b\right)-f\left(a\right)}{b-a}}$ ....................... (1) where, (a = 1, b = e)
Now differentiate f(x) we have,
$\Rightarrow {\displaystyle \frac{d}{dx}}f\left(x\right)={\displaystyle \frac{d}{dx}}\log x={\displaystyle \frac{1}{x}}$
$\Rightarrow f^{\prime }\left(x\right)={\displaystyle \frac{1}{x}}$
Now in place of x substitute (c) we have,
$\Rightarrow f^{\prime }\left(c\right)={\displaystyle \frac{1}{c}}$
Now from equation (1) we have,
$\Rightarrow {\displaystyle \frac{1}{c}}={\displaystyle \frac{\log e-\log 1}{e-1}}$
Now as we know the value of log e is 1 and the value of log 1 is zero so we have,
$\Rightarrow {\displaystyle \frac{1}{c}}={\displaystyle \frac{1-0}{e-1}}$
$\Rightarrow c=e-1$
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.