Question

Let f = {(1, −1), (4, −2), (9, −3), (16, 4)} and g = {(−1, −2), (−2, −4), (−3, −6), (4, 8)}. Show that gof is defined while fog is not defined. Also, find gof.

Answer 1

f = {(1, −1), (4, −2), (9, −3), (16, 4)} and g = {(−1, −2), (−2, −4), (−3, −6), (4, 8)}
f : {1, 4, 9, 16} → {-1, -2, -3, 4} and g : {-1, -2, -3, 4} → {-2, -4, -6, 8}
Co-domain of f = domain of g
So, gof exists and gof : {1, 4, 9, 16} → {-2, -4, -6, 8}
$$\begin{gathered} \left(gof\right)\left(1\right)=g\left(f\left(1\right)\right)=g\left(-1\right)=-2 \\ \left(gof\right)\left(4\right)=g\left(f\left(4\right)\right)=g\left(-2\right)=-4 \\ \left(gof\right)\left(9\right)=g\left(f\left(9\right)\right)=g\left(-3\right)=-6 \\ \left(gof\right)\left(16\right)=g\left(f\left(16\right)\right)=g\left(4\right)=8 \\ \mathrm{So},gof=\left\{\left(1,-2\right),\left(4,-4\right),\left(9,-6\right),\left(16,8\right)\right\} \end{gathered}$$

But the co-domain of g is not same as the domain of f.
So, fog does not exist.