When $f$ is concave, $f''<0$, the max can be easily found using simply the first order condition.
What techniques should I apply if it is not?
I'm especially thinking of a function $f$ that has the following properties:
1) $f:[0,1]\rightarrow[0,1]$
2) $f(x)=1$ if $x<\bar x$ for some fixed and known $\bar x >0 $
3) $f'(x)<0$, $\forall x>\bar x$
So, the value of $f$ is 1 from 0 to some $\bar x$ and strictly decreases from $\bar x$ onward. We may assume $f$ is differentiable (except at $\bar x$).
Are there any techniques or some known results on maximizing this specific function?
For simplicity of notation, let's say that $g(x) = xf(x)$. The multiplication of the two continuous and differentiable (except at $\bar x$) functions $x$ and $f(x)$ is also continuous and differentiable, except at $\bar x$, with the derivative $g'(x) = f(x) + xf'(x)$ (multiplication rule).
$g(x) = x$ on $[0, \bar x]$, so we can throw out candidate maximum points on $[0, \bar x)$ and keep $\bar x$ as a candidate maximum point. And since $g(x)$ is continuous and differentiable on $(\bar x, 1)$, you can throw out points on that interval for which $g'(x) \neq 0$ as candidate maximum points. This leaves only $\bar x$, $1$, and points on $(\bar x, 1)$ for which $g'(x) = 0$; the maximum point(s) must be in that set.
It's hard to say without actually knowing $f$ or $\bar x$, but that's where I'd start.