Answer
$$c = 2 \pm \sqrt 3 $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{{x^2} - 1}}{{x - 2}};{\text{ }}\left[ { - 1,1} \right],{\text{ }}a = - 1,{\text{ }}b = 1 \cr
& {\text{Calculating }}f'\left( x \right){\text{ using the quotient rule}} \cr
& f'\left( x \right) = \frac{{\left( {x - 2} \right)\left( {2x} \right) - \left( {{x^2} - 1} \right)\left( 1 \right)}}{{{{\left( {x - 2} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{2{x^2} - 4x - {x^2} + 1}}{{{{\left( {x - 2} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{{x^2} - 4x + 1}}{{{{\left( {x - 2} \right)}^2}}} \cr
& f'\left( c \right) = \frac{{{c^2} - 4c + 1}}{{{{\left( {c - 2} \right)}^2}}} \cr
& {\text{The function is continuous on the interval }}\left[ { - 1,1} \right] \cr
& f\left( a \right) = f\left( { - 1} \right) = \frac{{{{\left( { - 1} \right)}^2} - 1}}{{ - 1 - 2}} = 0 \cr
& f\left( b \right) = f\left( 1 \right) = \frac{{{{\left( 1 \right)}^2} - 1}}{{1 - 2}} = 0 \cr
& {\text{then there is at least one point }}c{\text{ in the interval}}\left( {a,b} \right){\text{ such that }} \cr
& f'\left( c \right) = 0 \cr
& \frac{{{c^2} - 4c + 1}}{{{{\left( {c - 2} \right)}^2}}} = 0 \cr
& {c^2} - 4c + 1 = 0 \cr
& {\text{By the quadratic formula}} \cr
& c = \frac{{4 \pm \sqrt {{{\left( { - 4} \right)}^2} - 4\left( 1 \right)\left( 1 \right)} }}{2} \cr
& c = 2 \pm \sqrt 3 \cr
& {\text{Therefore, }} \cr
& f'\left( {2 + \sqrt 3 } \right) = 0 \cr
& f'\left( {2 - \sqrt 3 } \right) = 0 \cr} $$
