Finding $p'(0)$ for the polynomial of least degree which has a local maximum at $x=1$ and a local minimum at $x=3$ The question goes as follows:

Let $p(x)$ be a real polynomial of least degree which has a local maximum at $x=1$ and a local minimum at $x=3$. If $p(1)=6$ and $p(3)=2$, then $p'(0)$ is...

What I did first, naturally, was consider that $p(x)$ is a cubic. But, all conditions cannot be satisfied simultaneously if $p(x)$ is a cubic.
Next, I considered $p(x)$ to be a $4$ degree polynomial, solved and got $p'(0)=15$. But the answer provided was $p'(0)=9$.
 A: Basically you need to setup system of equations and solve the coefficients of $p(x)$
Since $p(x)$ has two turning points, the minimum possible degree is $3$
Say $p(x) = ax^3+bx^2+cx+d$
$\implies p'(x) = 3ax^2+2bx+c$
from the problem we have :
$p(1) = 6$
$p(3) = 2$
$p'(1) = 0$
$p'(3)=0$
four equations and four unknowns - can be easily solved
A: $p'(1)=0, p'(3)=0, p(1)=6, p(3)=2$.
We must have $\partial p' \ge 2$, so try $p'(x) = c(x-1)(x-3)$. This gives
$p(x) = p(1)+c\int_1^x p'(t)dt = 6+{c \over 3} (x-4)(x-1)^2$. Setting $p(3) = 2$ gives $c=3$ and so we have
$p(x) = 
6+(x-4)(x-1)^2$.
Computing $p'(0)$ gives 9.
A: As an alternative to finding the coefficients of $p$ explicitly, start with the "standard" zig-zagging cubic
$$ f(x) = 2x^3 - 3x^2 $$
which has a maximum at $(0,0)$ and a minimum at $(1,-1)$.
Now just scale and translate to get $p(x) = 6 + 4f(\frac{x-1}2)$. Then you don't even need to simplify to find $p'(0)$, just the chain rule.
A: Hint:
$p(x)$ is in fact a cubic.
The system of equations you have to solve is:
$$\begin{eqnarray}p'(1) = 0&\Rightarrow&3a& + &2b& + &c& & &=& 0\\p(1) = 6&\Rightarrow&a&+&b&+&c&+&d &=& 6\\p'(3) = 0&\Rightarrow&27a&+&6b&+&c& & &=&0\\ p'(3) = 2&\Rightarrow&27a&+&9b&+&3c&+&d&=&2 \end{eqnarray}$$
Note that $p'(0) = c$
(You should end up with $p(x) = x^3-6x^2+9x+2$)
