Answer:
[tex]\huge\boxed{k=\dfrac{2}{3}}[/tex]
Step-by-step explanation:
[tex]f(x)=4x+1\\\\g(x)=3x+k\\\\(f\circ g)(x)=f\bigg(g(x)\bigg)=4(3x+k)+1=(4)(3x)+(4)(k)+1=12x+4k+1\\\\(g\circ f)(x)=g\bigg(f(x)\bigg)=3(4x+1)+k=(3)(4x)+(3)(1)+k=12x+3+k\\\\(f\circ g)(x)=(g\circ f)(x)\iff12x+4k+1=12x+3+k\qquad|\text{cancel}\ 12x\\\\4k+1=3+k\qquad|\text{subtract 1 from both sides}\\\\4k=2+k\qquad|\text{subtract}\ k\ \text{from both sides}\\\\3k=2\qquad|\text{divide both sides by 3}\\\\k=\dfrac{2}{3}[/tex]
