Since E is a countable subset of (a,b), let E = {e_{i}} with a < e_{1}< e_{2}< ... < b.

Then we have that f is continuos on [a, e_{1}], and differentiable on (a, e_{1}). The mean value theorem applies and we find that f(a) f(e_{1}). We can prove this by contradiction. Assuming f(a) > f(e_{1}), we would have by the MVT a c_{1}, s.t. f'(c_{1}) = (f(e_{1})-f(a))/(e_{1}-a) < 0, a contradiction to f'(x) 0.

Similiarly, we can find that f(e_{1}) f(e_{2}). Continuing in this way we can argue that for any e_{i}< e_{j}, f(e_{i}) f(e_{j}).

I think this may be a step toward proving the desired, f(a) f(b), but I haven't yet worked it completely. Will finish and post the rest soon, unless you or someone else can finish it.