Cubic function range garauntees in 0-1 interval? Let's say i had a cubic function $y = Ax^3+Bx^2+Cx+D$ and I know that $A,B,C,D$ are all in the range $[0,1]$, but i don't know their value.
Is there any way to know any minimum or maximum value of $y$ of this function over the interval of $x$ being $0\dots1 $?
I know that $y$ will be between 0 and 4, but is there any way to get a tighter range?
 A: What you are essentially asking for is the $\sup$ and $\inf$ of 
$$\{Ax^3 + Bx^2 + Cx + D \mid A,B,C,D,x \in [0,1]\}.$$
Zero is the lower bound and infimum because the coefficients are positive and the polynomial is evaluated with positive inputs. Four is an upper bound because each term is maximized when the coefficient and input are 1 ($x^n \leq 1$ for $n \geq 1$, $0 \leq x \leq 1$).
A: In general, you would use the first derivative to find that
$y'=3Ax^2+2Bx+C$ and a maximum/minimum will happen when $y'=0$ so you have a quadratic equation.
Thus the extrema will be when $x=\frac{-2B±\sqrt{4B^2-12AC}}{6A}=\frac{-B±\sqrt{B^2-3AC}}{3A}$.
To find if an answer is a maximum or minimum, plug in these values for $x$ in to the second derivative, $y''=6Ax+2B$. Plugging in gives $y''=2(-B±\sqrt{B^2-3AC})+2B=±2\sqrt{B^2-3AC}$.
If $y''$ is positive then you have found a local min, and vice versa. Note that you should also check the endpoints to see if they have a value larger than these local extrema.
