Answer:
Step-by-step explanation:
[tex]g(x)=12x-12,\ f(x)=\dfrac{x+12}{12}\\\\g\bigg(f(x)\bigg)-\text{Exchange}\ x\ \text{in}\ g (x)\ \text{to the expression}\ \dfrac{x+12}{12}:\\\\ g\bigg(f(x)\bigg)=g\left(\dfrac{x+12}{12}\right)=12\!\!\!\!\!\diagup^1\cdot\dfrac{x+12}{12\!\!\!\!\!\diagup_1}-12=x+12-12=x[/tex]