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Question:

Let f and g be two differentiable functions on R such that f'(x) > 0 and g'(x) < 0, for all x ∈ R. Then for all x:

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))

Solution:

Let f(x) = ax and g(x) = -bx where a, b > 0
f(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))