A value of c for which conclusion of Mean Value Theorem holds for the function \[f\left( x \right) = {\log _e}x\] on the interval \[\left[ {1,3} \right]\] is
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Official solution
Using Lagrange's Mean Value Theorem
Let f(x) be a function defined on \[\left[ {a,b} \right]\]
then, \[f'\left( c \right) = {{f\left( b \right) - f\left( a \right)} \over {b - a}}\,\,\,\,\,\,\,\,\,\,\,\,....\left( i \right)\]
\[c\,\, \in \left[ {a,b} \right]\]
\[\therefore\] Given \[f\left( x \right) = {\log _e}x\]
\[\therefore\] \[f'\left( x \right) = {1 \over x}\]
\[\therefore\] equation (i) become \[{1 \over c} = {{f\left( 3 \right) - f\left( 1 \right)} \over {3 - 1}}\]
\[ \Rightarrow {1 \over c} = {{{{\log }_e}3 - {{\log }_e}1} \over 2} = {{{{\log }_e}3} \over 2}\]
\[ \Rightarrow c = {2 \over {{{\log }_e}3}} \Rightarrow c = 2\,{\log _3}e\]