The graphs of the functions f and g are represented in In the figure g

\(? = g\left( {g\left( { - 1} \right) + 1} \right) + f\left( {g\left( 3 \right) + 1} \right) = g\left( {A + 1} \right) + f\left( {B + 1} \right)\)


\(A\,\, = \,\,g\left( { - 1} \right) = 4\,\,\,\left[ {{\rm{figure}}} \right]\)

\({?_{{\rm{temp1}}}} = g\left( {A + 1} \right) = g\left( 5 \right)\)

\({\rm{line}}\,\,L\,\,:\,\,\left\{ \matrix{\\
\,{\rm{slop}}{{\rm{e}}_L} = {{3 - 0} \over {1 - 4}} = - 1 \hfill \cr \\
\,\left( {0,4} \right) \in \,\,L\,\,\,\, \Rightarrow \,\,\,\,{{y - 4} \over {x - 0}} = - 1 \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,L\,\,:\,\,y = 4 - x\)

\(\left( {5,g\left( 5 \right)} \right)\,\, \in \,\,L\,\,\,\,\, \Rightarrow \,\,\,\,\,{?_{{\rm{temp1}}}} = g\left( 5 \right) = 4 - 5 = - 1\)


\(B = g\left( 3 \right)\,\,\,:\,\,\,\,\left( {3,g\left( 3 \right)} \right) \in \,\,L\,\,\,\,\, \Rightarrow \,\,\,\,\,g\left( 3 \right) = 4 - 3 = 1\)

\({?_{{\rm{temp2}}}} = f\left( {B + 1} \right) = f\left( 2 \right) = {c^2}\)

\(\left( {1,3} \right) \in {\rm{graph}}\left( f \right)\,\,\,\, \Rightarrow \,\,\,\,3 = f\left( 1 \right) = {c^1}\,\,\,\,\, \Rightarrow \,\,\,\,\,c = 3\)

\({?_{{\rm{temp2}}}} = f\left( 2 \right) = 9\)


\(?\,\,\, = \,\,\,{?_{{\rm{temp1}}}}\,\, + \,\,{?_{{\rm{temp2}}}}\,\,\, = \,\,\, - 1 + 9\,\,\, = \,\,\,8\)


The correct answer is (E).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.