Answer
$$c = \root 3 \of 3 $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{1}{x},{\text{ }}g\left( x \right) = {x^2} - 4{\text{ }}\underbrace {\left[ {1,2} \right]}_{\left[ {a,b} \right]} \cr
& {\text{Differentiate both functions}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{x}} \right] = - \frac{1}{{{x^2}}} \cr
& g'\left( x \right) = \frac{d}{{dx}}\left[ {{x^2} - 4} \right] = 2x \cr
& {\text{Apply the Extended Mean Value Theorem}} \cr
& \frac{{f'\left( c \right)}}{{g'\left( c \right)}} = \frac{{f\left( b \right) - f\left( a \right)}}{{g\left( b \right) - g\left( a \right)}} \cr
& {\text{,then}} \cr
& \frac{{ - \frac{1}{{{{\left( c \right)}^2}}}}}{{2\left( c \right)}} = \frac{{\frac{1}{2} - \frac{1}{1}}}{{\left[ {{{\left( 2 \right)}^2} - 4} \right] - \left[ {{{\left( 1 \right)}^2} - 4} \right]}} \cr
& {\text{Simplifying}} \cr
& - \frac{1}{{2{c^3}}} = \frac{{ - 1/2}}{3} \cr
& \frac{1}{{2{c^3}}} = \frac{1}{6} \cr
& {c^3} = 3 \cr
& c = \root 3 \of 3 \cr} $$
