$a<0$
$a\le 0$
$a\ge 0$
$a$
can take any real value

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The function will be crescent on a [a,b] interval if the derivative of is positive in every x that is on [a,b]

$f(x)=x^3-3ax^2+b$

$f'(x)=3x^2-6ax+b$

The problem asks what can be said if the

derivativeof the function is strictly increasing $\forall x>0$ , so the $f''(x)$ is always > 0 when this happens$f''(x)=6x-6a>0$

$x-a>0$

$x>a$

Lemma: if $x>n$ for every $x>k$ , where k is constant, then $n\leq k$

Proof: suppose $n>k$

Let $x=\frac{n+k}{2}$ , then $x>k$

However, $x>n$ does not hold, so it is impossible for $n>k$

Therefore $n\leq k$

If $x>a$ and $x>0$ ,then, by my lemma, $\boxed{a \leq 0}$