I don't really understand what you mean by f(x + eps),but check this blog :http://codeforces.me/blog/entry/11497

It might be useful.

Going from f(x) to f(x + eps) or to f(x - eps) is neither binary nor ternary search. You don't get O(log) time in this case.

Binary*

Unimodal functions are functions with only one extreme point. Ternary search is, basically, a technique for finding the mode — e.g. the extreme point — of such a function. If the function is continuous and differentiable, which will probably be the case unless you work with discrete indices, the modified binary search basically searches for the x where f'(x) = 0, since (f(x + EPS) — f(x)) / EPS is a very good approximation to f'(x), for sufficiently small EPS. Because the derivative will always have a root at the extreme point, and there exists by definition of unimodality no other value that can have f'(x) = 0, the binary search you described is equivalent to ternary searching.

With infinitely precise calculation it should work.

the problem may be in that if you have x₁ ≈ x₂ = x₁ + ε then f(x₁) ≈ f(x₂) and you may then go in wrong direction. It may be also the case for 1/3-division, but it will only happen when your range is too small, so that it almost won't affect the result