Question

For the function f(x) = log_ex, x ∈ [1, 2], the value of c for the lagrange's mean value theorem is _______________.

Answer 1

The given function is f(x) = log_ex.

Now, f(x) = log_ex is differentiable and so continuous for all x > 0. So, f(x) is continuous on [1, 2] and differentiable on (1, 2). Thus, both the conditions of Lagrange's mean value theorem are satisfied.

So, there must exist at least one real number c ∈ (1, 2) such that

$f\text{'}\left(c\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}$

f(x) = log_ex

$\Rightarrow f\text{'}\left(x\right)=\frac{1}{x}$

$\therefore f\text{'}\left(c\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}$

$\Rightarrow \frac{1}{c}=\frac{{\log }_{e}2-{\log }_{e}1}{2-1}$

$\Rightarrow \frac{1}{c}={\log }_{e}2-0\left({\log }_{a}1=0,a>0\right)$

$\Rightarrow c=\frac{1}{{\log }_{e}2}={\log }_{2}e\left({\log }_{b}a=\frac{1}{{\log }_{a}b}\right)$

$\Rightarrow c={\log }_{2}e\in \left(1,2\right)\left(2<e<4\Rightarrow {\log }_{2}2<{\log }_{2}e<{\log }_{2}4\Rightarrow 1<{\log }_{2}e<2\right)$

Thus, $c={\log }_{2}e\in \left(1,2\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}$.

Hence, the value of c is ${\log }_{2}e$.


For the function f(x) = log_ex, x ∈ [1, 2], the value of c for the Lagrange's mean value theorem is $\underline{{\log }_{2}e}$.