Originally Posted by
red_dog $\displaystyle 3x^2-x^3=k\Leftrightarrow x^3-3x^2+k=0$
Let $\displaystyle f(x)=x^3-3x^2+k, \ f:\mathbb{R}\to\mathbb{R}$
$\displaystyle f'(x)=3x^2-6x$
$\displaystyle f'(x)=0\Rightarrow x_1=0, \ x_2=2$
$\displaystyle f(0)=k, \ f(2)=k-4, \ \lim_{x\to-\infty}f(x)=-\infty, \ \lim_{x\to\infty}f(x)=\infty$
The Rolle's sequence is: $\displaystyle -\infty \ \ k \ \ k-4 \ \ \infty$
To have 3 real solutions the following inequalities must hold:
$\displaystyle \left\{\begin{array}{ll}k\geq 0\\k-4\leq 0\end{array}\right.\Rightarrow k\in[0,4]$