f(x)=x+2f−1(x)=?y=x+2|x↔yx=y+2y=x−2f−1(x)=x−2g(x)=x3g−1(x)=?y=x3|x↔yx=y3y=3xg−1(x)=3x
f(19)=19+2=21g(21)=213=7f−1(7)=7−2=5f−1(5)=5−2=3g−1(3)=3∗3=9f(9)=9+2=11
f(g−1(f−1(f−1(g(f(19))))))=11