Respuesta :

[tex]\bf \begin{cases} f(x)=6x\\ g(x)=4x+1 \end{cases}\qquad (f\circ g)(x)\iff f(\quad g(x)\quad ) \\\\\\ \textit{that is}\implies \begin{cases} f(\boxed{x})=6\boxed{x} \\\\ f(\boxed{g(x)})=6\boxed{g(x)}\\ --------------\\ g(x)=4x+1\qquad thus\\ --------------\\ f(\boxed{g(x)})=6(\boxed{4x+1})\iff (f\circ g)(x) \end{cases}[/tex]

distribute and simplify if needed, see what you get
I think it is A because if you do the math 4x+ 1 = 1 and A is the same hope that helps