Quickly find an upper bound for a simple function Given a simple function, what would be a good method to quickly find an upper bound ?
For instance, let $$f\left(
\begin{array}{}     
    \mathbb{R} & \longrightarrow &  \mathbb{R}\\
    x  & \longmapsto & x(1-x)
\end{array}\right)$$
To show it has an upper bound, you can :

Derivate $f$ :
$$f'(x) = 1 - 2x$$
Then either conclude with the variations (which is looooonnggggg) or find the zeroes :
Here, $\frac{1}{2}$ is the only zero, so you can conclude that $\sup\limits_{\mathbb{R}}f$ is either the zero of the derivative, or one of the bound of the study domain (so here, it is $f(\frac{1}{2}) = \frac{1}{4}$).
The advantage of this method is that it gives you a sup and not only an upper bound.

Say $\lim\limits_{-\infty}f=-\infty$ and $\lim\limits_{+\infty}f=-\infty$, so :
$$\exists x_a < 0, \forall x < x_a, f(x) < 0$$
$$\exists x_b > 0, \forall x > x_b, f(x) < 0$$
And $f$ is continuous on $[x_a, x_b]$ so $f[x_a, x_b]=[m,M]$.
Therefore :
$$\forall x \in \mathbb{R}, f(x) \leq \max(0, M) \leq |M|$$
$$i.e. \ f \text{ has an upper bound on } \mathbb{R}$$
This method is slightly faster (even tho is it much lesser precise on the upper bound, but I don't really care about this).

When the function is some low degree polynomials, just as $f$, you can use $\Delta$, supposing that for some c, $f(x) - c$ doesn't have a root in $\mathbb{R}$ (i.e. $\forall x \in \mathbb{R}, f(x) - c < 0$):
$$f(x) - c = x - x^2 - c$$
$$\Delta = 1^2 + 4c$$
Hence :
$$f(x) - c \text{ has no real root } \iff \Delta < 0$$
$$\iff 1-4c < 0 \iff \frac{1}{4}<c$$
This one is somehow the fastest I think but it is pretty limited.

This one is the best one I could think of :
Using the geometric/arithmetic mean inequality :
$$\sqrt{x(1-x)} \leq \frac{1}{2}(x+1-x)$$
$$\iff x(1-x) \leq \frac{1}{4}$$
However I'm not sure it is applicable as $x$ and $1-x$ aren't necessarly positive numbers (but I'm sure you can manage to do it with a some case study).

Are there some better methods ? Or other, more general methods ?

Rewrite $f$ as 
$$x-x^2 = \frac{1}{4} - (\frac{1}{2}-x)^2 .$$
