Answer
$$c = 2\sqrt 3 $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = x - \frac{1}{x};{\text{ }}\left[ {3,4} \right],{\text{ }}a = 3,{\text{ }}b = 4 \cr
& {\text{Calculate }}f'\left( c \right) \cr
& f'\left( x \right) = 1 + \frac{1}{{{x^2}}} \cr
& f'\left( c \right) = 1 + \frac{1}{{{c^2}}} \cr
& f\left( x \right){\text{ is continuous and differentiable on the interval }}\left[ {3,4} \right], \cr
& {\text{Using the Mean - Value Theorem}} \cr
& f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}} \cr
& 1 + \frac{1}{{{c^2}}} = \frac{{f\left( 4 \right) - f\left( 3 \right)}}{{4 - 3}} \cr
& 1 + \frac{1}{{{c^2}}} = \frac{{4 - \frac{1}{4} - 3 + \frac{1}{3}}}{1} \cr
& 1 + \frac{1}{{{c^2}}} = \frac{{13}}{{12}} \cr
& \frac{1}{{{c^2}}} = \frac{1}{{12}} \cr
& {c^2} = 12 \cr
& c = \pm 2\sqrt 3 \cr
& {\text{but only }}c = 2\sqrt 3 {\text{ is in the interval }}\left( {3,4} \right),{\text{ then}} \cr
& c = 2\sqrt 3 \cr} $$
