<s>[INST] Is $x(1 - 2x) \le \frac{1}{8}$ and further, is $x(1 - ax) \le \frac{1}{4a}$ It is clear that $x(1-x) \le \frac{1}{4}$
Does it likewise follow that $x(1-2x) \le \frac{1}{8}$?
Here's my reasoning:
(1)  For $x < \frac{1}{4}$, $x(1-2x) < \frac{1}{8}$
(2)  For $\frac{1}{4} < x < \frac{1}{2}$, $x(1-2x) < \frac{1}{8}$
(3)  For $\frac{1}{2} < x$, $x(1-2x) < 0$
Further, can this be generalized to $x(1-ax) \le \frac{1}{4a}$
Since:
(1) For $x < \frac{1}{2a}$, $x(1-ax) < \left(\frac{1}{2a}\right)\left(\frac{1}{2}\right) = \frac{1}{4a}$
(2) For $\frac{1}{2a} < x < \frac{1}{a}$, $x(1-ax) < \left(\frac{1}{2a}\right)\left(\frac{1}{2}\right) = \frac{1}{4a}$
(3) For $\frac{1}{a} < x$, $x(1-ax) < 0$
Are both of these observations correct?  Is only one correct?  Is there an exception that I am missing?
 [/INST] Without any derivative, just high-school theory of quadratic equations:
A quadratic polynomial (with real coefficients) has a global extremum at the arithmetic mean of its roots. Further, this extremum is a maximum if its leading coefficient is negative, a minimum if the leading coefficient is positive.
So here, the extremum is attained  at $\dfrac1{2a}$ and the leading coefficient is $-a$. This extremum is a maximum if $a>0$, a minimum if $a<0$ and it is equal to
$$\frac1{2a}\biggl(1-a\,\frac1{2a}\biggr)=\frac1{4a}.$$
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