g(f(x))>g(f(x-1))
f(g(x))>f(g(x-1))
f(g(x))>f(g(x+1))
g(f(x))<g(f(x+1))
Let f(x) = ax and g(x) = -bx where a, b > 0f(g(x)) → -abx (1)f(g(x+1)) → a(-b - bx) = -abx - ab (2) .So, from (1) and (2)f(g(x)) > f(g(x+1))