jmc

Clearly, $f(x)$ must be quadratic. Let $f(x)=a{x}^2+bx+c.$ Then $$\begin{array}{rl}g(f(x))& =g(a{x}^2+bx+c)\\ & =(a{x}^2+bx+c{)}^2-11(a{x}^2+bx+c)+30\\ & =a^2{x}^4+2ab{x}^3+(2ac+b^2-11a)x^2+(2bc-11b)x+c^2-11c+30\\ & =x^4-14x^3+62x^2-91x+42.\end{array}$$Matching coefficients, we get $$\begin{array}{rl}{a}^2& =1,\\ 2ab& =-14,\\ 2ac+b^2-11a& =62,\\ 2bc-11b& =-91,\\ c^2-11c+30& =42.\end{array}$$Since the leading coefficient $a$ is positive, $a=1.$ Then $2b=-14,$ so $b=-7.$ From the fourth equation, $$-14c+77=-91,$$so $c=12.$ We can verify that all the equations are satisfied, so $f(x)=\boxed{x^2-7x+12}.$