So for this case, [tex](f\circ g)(x)=f(g(x))\ \textsf{and}\ (g\circ f)(x)=g(f(x))[/tex]
So with the first question, plug the value of g(x) into the x variable of f(x) and solve as such:
[tex](f\circ g)(x)=7(x^2-4x+6)-5\\(f\circ g)(x)=7x^2-28x+42-5\\(f\circ g)(x)=7x^2-28x+37[/tex]
Your first answer is [tex](f\circ g)(x)=7x^2-28x+37[/tex]
Now with the second question, plug the value of f(x) into the x variable of g(x) and solve as such:
[tex](g\circ f)(x)=(7x-5)^2-4(7x-5)+6\\(g\circ f)(x)=(7x-5)(7x-5)-28x+20+6\\(g\circ f)(x)=49x^2-35x-35x+25-28x+20+6\\(g\circ f)(x)=49x^2-98x+51[/tex]
Your second answer is [tex](g\circ f)(x)=49x^2-98x+51[/tex]
