\(\eqalign{& f\left( 0 \right) = {1 \over 2}.0 + 5 = 5 \cr & f\left( 2 \right) = {1 \over 2}.2 + 5 = 6 \cr & f\left( 3 \right) = {1 \over 2}.3 + 5 = {{13} \over 2} \cr & f\left( { - 2} \right) = {1 \over 2}.\left( { - 2} \right) + 5 = 4 \cr & f\left( { - 10} \right) = {1 \over 2}.\left( { - 10} \right) + 5 = 0 \cr} \)
\(\displaystyle f(1)={1 \over 2}.1 + 5=\dfrac{11}{2}\)
