I am trying to build a cubic polynomial $y = ax^3 + bx^2 + cx + d$ with the following conditions:
$y(0) = 3$
$y'(0) = 2$
$y'(1) = 0$
$y(1) = 4$
Unfortunately, solving the required system yields $a = 0$. I am trying to understanding geometrically why this must be the case. I want a local max at $(1,4)$ and the slope at the y-intercept to equal $2$.
This problem builds on a simpler problem of finding a quadratic that satisfies the first three conditions, which gives $y = -x^2 + 2x + 3$, thereby yielding the fourth. Thus, I am looking for a different value of $b$ in the cubic, and I would like to compensate for it with $a$.
For the quadratic, we have $y'' = -2$.
For the cubic, we have $y'' = 6ax + 2b$. I want $y'$ to also have a net change of $-2$ over the interval $[0,1]$, while $y$ has a net change of $+1$.
Can anyone explain why, due to growth rates, that this is not possible with a cubic, or if there's something I'm missing?
Update:
The general case of this same question:
$y'(\alpha) = 2$
$y'(\alpha+1) = 0$
$\displaystyle \int_{\alpha}^{\alpha + 1} y' dy = 1$
for a cubic $y$ yields $a = 0$. As the average value of $y'$ is $1$, the only way to satisfy the third condition is if $y'$ is indeed linear. Have I answered my own question, and should I be ashamed to be a Calculus teacher?
With $n$ constraints of this type, one of which must be on $f$ and not $f'$ or any other derivative, you are guaranteed to create a single polynomial with order at most $n-1$. Since solving these constraints is solving a system of linear equations, there is no practical restraint on getting 0s for the highest-order coefficients.
For a more extreme instance, obviously the constraints $f(0)=2$, $f'(0)=f'(1)=f'(50)=3$, you're only ever going to get $f(x)=3x+2$, a linear function, despite the four inputs.